TEST BANK for Genetics: A Conceptual Approach 7th Edition Benjamin Pierce. ISBN-13 978-1319216801.

Name:

Class:

Date:

Chap 01_7e Indicate the answer choice that best completes the statement or answers the question.

1. What commonsense observation makes the theory of acquired characteristics UNLIKELY? a. An individual may inherit traits found in both of his or her parents. b. Anatomical changes such as loss of a limb are not seen in the offspring of an individual. c. Evolution requires genetic change in populations. d. Alleles that result in abnormal phenotypes may be less common in some populations than in others. e. Offspring often look more like their parents than unrelated individuals.

2. Which one of the following topics of research belongs to the discipline of transmission genetics? a. inheritance pattern of gene alleles b. mechanism of DNA replication c. gene expression patterns d. evolution e. chemical modification of nucleic acids 3. The first complete DNA sequence of a nonviral, free-living organism was obtained for a. a bacterium in 1900. b. a bacterium in 1945. c. a bacterium in 1995. d. humans in 1990. e. humans in 2000.

4. Identify a FALSE statement from the following descriptions of genetics. a. Humans first applied genetics to the domestication of plants and animals between approximately 10,000 and 12,000 years ago. b. Some viruses use RNA to carry their genetic information. c. Albinism results from a mutation in the genes that control the synthesis and storage of melanin. d. All human traits that display blending inheritance are affected by a single gene. e. The process by which genetic information is copied and decoded is similar for all forms of life.

5. The contribution of Gilbert and Sanger to modern genetics was to a. develop the PCR technique. b. discover DNA in the nucleus of cells. c. describe the structure of DNA. d. show that genes were made of DNA. e. develop a method for sequencing DNA.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 01_7e 6. In the late 1990s, what important discovery in genetics was made? a. The three-dimensional structure of DNA was described, which showed how DNA might be replicated. b. The first recombinant DNA experiments were performed that started the biotechnology field. c. DNA sequencing methods were first discovered. d. Genes were found to be located on chromosomes. e. Tiny RNAs were discovered that play important roles in the regulation of gene expression.

7. The golden mutation in the zebrafish was useful because of which of the following results? a. It led to the discovery of a similar gene in humans that is involved in skin pigmentation. b. It led to the development of new varieties of wheat. c. It led to the ability to identify many of the genes that result in an increase in heart attacks. d. It allowed the zebrafish to be grown in captivity and become commercially profitable. e. It became the first gene in a model organism to be sequenced.

8. Which of the following scientists contributed significantly to the foundations of molecular genetics? a. James Watson b. Thomas Hunt Morgan c. John B. S. Haldane d. Charles Darwin e. Sewall Wright 9. Identify a TRUE statement. a. Genetic influences are solely responsible for the traits of an organism. b. Environmental influences can only responsible for the physical traits of an organism, such as height, weight, or color. c. Environmental influences can only be responsible for the behavioral traits of an organism . d. Environmental influences can affect gene activity. e. Separating genetic and environmental influences on organismal traits is easy.

10. Permanent, heritable changes in genetic information (DNA) are called a. evolution. b. defects. c. SNP. d. alleles. e. mutations.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 01_7e 11. A measurable or observable trait or characteristic is called a a. phenotype. b. genotype. c. single-nucleotide polymorphism. d. small interfering RNA. e. gene bank.

12. Which one of the following pairings between the subdiscipline of genetics and the phenomenon is INCORRECT? a. evolution—population genetics b. gene regulation—molecular genetics c. allelic frequency alteration—population genetics d. arrangement of genes on chromosome—transmission genetics e. chemical nature of the gene—transmission genetics 13. A change in allele frequency within a population over time leads to a. a genome. b. a phenotype. c. a genotype. d. mutations. e. evolution.

14. Which of the following scientists contributed significantly to the foundations of population genetics? a. James Watson b. Thomas Hunt Morgan c. Ronald Fisher d. Charles Darwin e. Frederick Sanger 15. Which of the following statements is FALSE concerning prokaryotic cells? a. They lack a nuclear membrane. b. They lack organelles such as chloroplasts. c. They are less complex than eukaryotic cells. d. They lack genetic information. e. They lack a true nucleus.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 01_7e 16. Which of the following examples of scientists and their contribution is matched INCORRECTLY? a. Watson and Crick—three-dimensional structure of DNA b. Mendel—principles of heredity using pea plants c. Gilbert and Sanger—DNA sequencing methods d. Morgan—polymerase chain reaction e. Sutton—genes on chromosomes as units of inheritance 17. Which of the following pairs is a part of a single nucleotide? a. nitrogenous base and sugar b. sugar and amino acid c. guanine and cytosine, two nitrogenous bases d. amino acid and nitrogenous base 18. Identify a TRUE statement from the following descriptions concerning genetics. a. The theory of pangenesis states that all living organisms are composed of cells. b. Bacteria and viruses are not useful in studying genes and inheritance because they are structurally and metabolically different from eukaryotic cells. c. Charles Darwin accurately described the laws of inheritance in his landmark book, On the Origin of Species. d. Many human traits, such as skin and hair color, are determined by more than a single gene. e. Evolution can occur without genetic changes in the population.

19. Which of the following species is considered a model genetic organism? a. the plant Linaria vulgaris b. the deer mouse Peromyscus maniculatus c. the worm Caenorhabditis elegans d. the frog Hyla chrysoscelis e. the chimpanzee Pan troglodytes 20. Which of the following would serve the LEAST well as a model for understanding basic mechanisms of inheritance? a. fruit flies b. humans c. yeast d. mice e. zebrafish

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 01_7e 21. Genetic information can be carried in which of the following biomolecules? a. proteins b. DNA but not RNA c. RNA but not DNA d. either DNA or RNA e. proteins but not RNA 22. What commonsense observation makes the theory of blending inheritance unlikely? a. An individual may inherit traits found in only one of his or her parents. b. Anatomical changes such as loss of a limb are not seen in the offspring of an individual. c. Evolution requires genetic change in populations. d. Some traits disappear in one generation and then reappear in the next generation. e. Offspring often look more like their parents than unrelated individuals.

23. Which of the following CORRECTLY describes the cell theory? a. Genetic information from different parts of the body travels to the reproductive organs. b. The cell is the compositional and functional unit of all life. c. Inside the germ cells, there exists a fully formed miniature adult that enlarges in the course of development. d. The genetic material itself blends, which cannot be separated out in future generations. e. Traits acquired in a person's lifetime become incorporated into the person's hereditary information, which will be passed on to his or her offspring.

24. Assume that a geneticist is doing a study with a wild mouse species. She captures 100 of these mice, takes a DNA sample from each, and sequences the same specific gene from each mouse. This gene has two alleles within this population. She then calculates the frequency of each of the two alleles from the sequencing results.

Which subdivisions of genetics would this study include? a. transmission and population genetics b. transmission and molecular genetics c. molecular genetics only d. molecular and population genetics e. transmission genetics only 25. The complete genetic makeup of any organism is referred to as a a. phylogeny. b. phenotype. c. genome. d. genotype. e. single-nucleotide polymorphism.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 01_7e 26. Which of the following combines molecular biology and computer science? a. single-nucleotide polymorphism b. microRNAs c. polymerase chain reaction d. bioinformatics e. eukaryotics 27. Which one of the following topics belongs to a different subdiscipline of genetics when compared with the rest? a. mechanism of gene regulation b. allele frequencies of a certain gene in different environments c. transcription d. chemical alternation of chromosomes e. mechanism of DNA replication 28. Which of the following scientists contributed significantly to the foundations of transmission genetics? a. James Watson b. Thomas Hunt Morgan c. John B. S. Haldane d. Charles Darwin e. Sewall Wright 29. Within cells, genes are located on structures called a. genomes. b. chromosomes. c. phenotypes. d. genotypes. e. alleles.

30. Choose the correct match between the scientists and the field of genetics to which they contributed. a. Haldane and Wright—transmission genetics b. Mendel—molecular genetics c. Gilbert and Sanger—population genetics d. Darwin—molecular genetics e. Morgan—transmission genetics

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 01_7e 31. Assume that a mutation occurs within a gene within an individual fruit fly. What will be the most likely series of consequences of this mutation? a. It will initially change the RNA sequence; then this change in genetic information will be transferred to the DNA sequence and finally result in a change in the protein made by the gene. b. It will initially change the DNA sequence; then this change in genetic information will be transferred to the RNA sequence and finally result in a change in the protein made by the gene. c. It will initially change the RNA sequence; then this change in genetic information will be transferred to the protein made by the gene and finally result in a change in the DNA. d. It will initially change the DNA sequence; then this change in genetic information will be transferred to the protein made by the gene and finally result in a change in the RNA. e. It will initially change the protein sequence made by the gene; then this change in genetic information will be transferred to DNA and finally result in a change in the RNA. 32. Albinism is rare in most human populations, occurring at a frequency of about 1 in 20,000 people. However, the trait occurs at a frequency of 1 in 200 in certain Hopi villages of Black Mesa in Arizona. In light of this example and others that you might be aware of, what can you conclude about particular alleles such as the allele for albinism? a. An allele that leads to an abnormal phenotype will be rare in most populations but common in Native American populations. b. An allele that leads to an abnormal phenotype will not be beneficial in any population. c. An allele that leads to an abnormal phenotype may be beneficial in some environments but harmful in others. d. An allele that leads to an abnormal phenotype will rise in frequency after many generations. e. An allele that leads to an abnormal phenotype will soon disappear from a population.

33. What commonsense observation makes the theory of preformationism unlikely? a. An individual may inherit traits found in both of his or her parents. b. Anatomical changes such as loss of a limb are not seen in the offspring of an individual. c. Evolution requires genetic change in populations. d. Alleles that result in abnormal phenotypes may be less common in some populations than in others. e. Offspring often look more like their parents than unrelated individuals.

34. Which of the following theories of inheritance is currently considered TRUE? a. germ-plasm theory b. pangenesis c. blending inheritance d. inheritance of acquired characteristics e. None of these theories is considered true based on new evidence.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 01_7e 35. What powerful piece of evidence suggests a genetic modification led to snakes developing without limbs? a. A few species of snakes, such as pythons, have vestigial leg bones. b. A few rare species of snakes actually develop with small limbs. c. Genes controlling leg development in mice can be introduced into snakes and cause them to develop limb-like structures. d. A genetic element called an enhancer believed to be responsible for suppression of limb development in snakes was isolated from a snake and introduced into mice. The resulting mice developed with truncated limbs. e. Sequencing of an enhancer element proved it was responsible for the suppression of limb development in snakes.

36. _____ is a change in allele frequency of a population over time. a. Blending inheritance b. Preformation c. Genome d. Evolution e. Phenotype 37. Among the model genetic organisms, Escherichia coli, a single-celled bacterium, is a prokaryote; Saccharomyces cerevisiae, one-celled yeast, is a eukaryote, as are Caenorhabditis elegans, a multicellular nematode worm, and Arabidopisis thaliana, a multicellular plant. Which of these organisms would NOT contain membrane-bound organelles? a. Escherichia coli b. Saccharomyces cerevisiae c. Escherichia coli and Saccharomyces cerevisiae d. Caenorhabditis elegans e. Caenorhabditis elegans and Arabidopisis thaliana 38. CRISPR/Cas9 is a powerful new method that allows a. DNA sequencing to be performed very quickly so that numerous individual genomes can be sequenced in a short period of time. b. precise editing of specific DNA sequences in living cells. c. the identification of genes involved in important medical conditions. d. the introduction of genes from one species into another species. e. the arrangement of genes on chromosomes.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 01_7e 39. How do DNA and RNA differ? a. DNA contains the nitrogenous base thymine while RNA contains the base uracil instead of thymine. b. DNA contains the nitrogenous base guanine while RNA contains the base cytosine instead of guanine. c. DNA is composed of repeating units called nucleotides while RNA is composed of repeating units called amino acids. d. DNA is composed of repeating units called amino acids while RNA is composed of repeating units called nucleotides. e. In DNA the nucleotides contain a sugar, a base, and a phosphate while in RNA the nucleotides contain no sugar.

40. A form of a gene that has a slightly different sequence than other forms of the same gene but encodes the same type of an RNA or protein is called a(n) a. locus. b. allele. c. homologous chromosome. d. heterozygote. e. homozygote.

41. The contribution Charles Darwin made to biology was to a. demonstrate the connection between Mendel's principles of inheritance and evolution. b. propose that evolution occurs by natural selection. c. develop the theory of evolution, based on earlier theories of population genetics. d. connect the fields of evolution and molecular genetics. e. determine the first DNA sequence for a free-living organism.

42. Which of the following sequences CORRECTLY shows the flow of genetic information during gene expression? a. RNA → DNA → protein b. protein → DNA → RNA c. DNA → RNA → protein d. DNA → protein → DNA e. None of the answers is correct.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 01_7e 43. Which of the following statements is TRUE? a. Each subdiscipline of genetics is very specific as to what is explored and does not overlap with the other subdisciplines. b. All phenotypes or traits are always determined by multiple genes. c. Albinism arises from the overexpression of the gene that controls the synthesis and storage of melanin. d. Humans make excellent model organisms because they have a variety of well-defined traits. e. None of the statements provided are true.

44. Which of the following statements is the CORRECT definition of meiosis? a. It is the method by which prokaryotic cells divide and produce daughter cells. b. It is the process by which the genetic information in DNA is transferred to RNA. c. It is the separation of chromosomes in the division of sex cells to produce gametes. d. It is the separation of chromosomes in the division of somatic cells in plants and animals. e. It is the process that produces multiple alleles of genes.

45. The experiments of Gregor Mendel can be placed into which subdivision of genetics? a. molecular genetics b. population genetics c. transmission genetics d. molecular genetics and transmission genetics e. population genetics and transmission genetics 46. The complete genetic makeup of an organism is referred to as its a. chromosome. b. alleles. c. locus. d. genome. e. phenotype.

47. Which of the following theories of inheritance is no longer accepted as TRUE? a. pangenesis b. blending inheritance c. inheritance of acquired characteristics d. preformationism e. None of the provided theories is currently considered true.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 01_7e 48. Which of the following statements is CORRECT? a. All genomes are encoded in DNA only. b. All genomes are encoded in nucleic acids. c. All genomes are encoded in proteins only. d. The genetic instructions are decoded completely differently in each organism. e. Molecular studies suggest life evolved from multiple primordial ancestors.

Indicate one or more answer choices that best complete the statement or answer the question.

49. Which of the following traits would make a species useful as a model genetic organism? (Select all that apply.) a. large number of progeny b. long generation time c. small size d. ability to be studied in a laboratory e. ability to be propagated inexpensively 50. The three-dimensional structure of DNA was first deciphered based on the work of which of the following individuals? (Select all that apply.) a. James Watson b. Francis Crick c. Maurice Wilkins d. Thomas Hunt Morgan e. Rosalind Franklin 51. The fruit fly Drosophila melanogaster is an important model system for studying inheritance in animals and genetic control of animal development, including humans. If researchers ultimately want to understand a biological process in humans, why might they want to study the process in fruit flies first? (Select all that apply.) a. Fruit flies are relatively easy to genetically manipulate and to isolate mutations. b. Fruit flies have short generation times and produce relatively large numbers of progeny. c. Fruit flies have simpler genomes than do humans. d. Fruit flies share all important physiological and developmental processes with humans. e. Fruit flies are small and easy to raise.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 01_7e 52. Albinism is rare in most human populations, occurring at a frequency of about 1 in 20,000 people. However, the trait occurs at a frequency of 1 in 200 in certain Hopi villages of Black Mesa in Arizona. Explain in terms of natural selection why the trait is so much more common among the Hopis of Black Mesa.

53. List and describe two significant events in the history of genetics that occurred during the twentieth century.

54. Albinism is rare in most human populations, occurring at a frequency of about 1 in 20,000 people. However, the trait occurs at a frequency of 1 in 200 in certain Hopi villages of Black Mesa in Arizona. Explain in terms of natural selection why albinism is so rare in most human populations.

55. Describe a discovery in genetics or an area of current research that you are concerned about that might have a negative impact on your life in the future. Explain why you think it might have a negative impact on you personally.

56. Many good ideas in science ultimately turn out to be incorrect, and this has happened several times in the history of genetics. In your own words, state one idea in the history of genetics that turned out to be incorrect.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 01_7e 57. Many good ideas in science ultimately turn out to be incorrect, and this has happened several times in the history of genetics. Summarize the evidence that ultimately caused an idea to be rejected by modern geneticists.

58. Why might bacteria and viruses be good model organisms for studying the basics of inheritance? Describe two advantages over studying genetics in mice, dogs, or humans.

59. Many good ideas in science ultimately turn out to be incorrect, and this has happened several times in the history of genetics. Why do you think a particular idea was widely accepted by scholars of that time? Include in your answer some evidence in favor of the idea, observations that seemed to support the idea, or other rationale for accepting the idea.

60. Write a paragraph explaining why genetics is considered a young science, even though people have been applying genetic principles for thousands of years.

61. The fruit fly Drosophila melanogaster is an important model system for studying inheritance in animals and genetic control of animal development, including humans. Evaluate fruit flies as a model system for human biology. What are their strengths and weaknesses as a model system?

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 01_7e 62. Describe one way in which discoveries in genetics will likely impact your life in the future.

63. Which features distinguish a prokaryotic cell from a eukaryotic cell?

64. Describe one way in which discoveries in genetics currently impact your daily life apart from this course.

65. What common features of heredity suggest that all life on Earth evolved from a common ancestor?

66. Though humans are the subject of intense genetic investigation, Homo sapiens are not generally considered a model genetic organism. Why not? Does this make sense?

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 01_7e Answer Key 1. b 2. a 3. c 4. d 5. e 6. e 7. a 8. a 9. d 10. e 11. a 12. e 13. e 14. c 15. d 16. d 17. a 18. d 19. c 20. b 21. d 22. d 23. b 24. d 25. c 26. d Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 01_7e 27. b 28. b 29. b 30. e 31. b 32. c 33. a 34. a 35. d 36. d 37. a 38. b 39. a 40. b 41. b 42. c 43. e 44. c 45. c 46. d 47. e 48. b 49. a, c, d, e 50. a, b, c, e 51. a, b, c, e

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 01_7e 52. Albinos occupy a privileged position among the Hopis of Black Mesa. In this culture, albinos are viewed as especially pretty, clean, and intelligent, and they often occupy positions of leadership. Albinos are celebrated in the villages as a sign of purity of Hopi blood in the community. Furthermore, albinos are often excused from normal male field labor because of their sensitivity to sunlight, causing them to be left behind in the village with the women during the daytime. This allows them extra mating opportunities compared to the other men of the village.

Therefore, the alleles that cause albinism are either selected for in this culture or at least not selected against as strongly as in other cultures, allowing the trait to occur at a much higher frequency.

53. 1900: Mendel's previously published work on pea plants, which stated basic principles of inheritance, was rediscovered.

1902: Sutton proposed that genes are located on chromosomes.

1910: Thomas Hunt Morgan began studies of transmission genetics, using fruit fly mutants.

1930s: Fisher, Haldane, and Wright outlined the founding principles of population genetics.

1940s: Organization of chromosomes and genes was studied using bacteria and viruses.

1940s–1950s: Evidence was accumulated for DNA as the genetic material; Watson and Crick described the DNA structure.

1966: The relationship between the chemical structure of DNA and the amino acid sequence of proteins was determined.

1973: The first recombinant DNA experiments were conducted.

1977: The Gilbert and Sanger methods for DNA sequencing were published.

1983: Mullis developed PCR. 1990: The first use of gene therapy was used in humans.

1990s: The Human Genome Project was started.

1995: The first genome of a free-living organism was sequenced (Haemophilus influenzae).

1996: The first genome of a eukaryote was sequenced (yeast).

2000–present: The human genome sequence was released.

54. In most populations, there is fairly strong selection against albinism because albinos don't produce melanin, causing their skin cells not to be protected from the damaging effects of sunlight. Also, the lack of melanin in their eyes causes them to have poor eyesight. Finally, in most cultures albinos are seen as abnormal, and they are not normally sought out for marriage and mating. Therefore, in most populations the alleles that cause albinism are selected against, and they decrease in frequency or are kept at a low level, causing the recessive trait to be rare.

55. Again, answers will vary, but an example is the possible abuse of genetic information about individuals that is becoming more available. Many people worry that results from genetic tests, for example, could be used to discriminate against individuals in the workplace and in the insurance marketplace. The Genetic Information Nondiscrimination Act offers limited protection against genetic discrimination, but the possibility of negative consequences from such tests remains.

Human genome editing is very controversial and could have unforeseen negative consequences not only to the edited individual's life but more broadly.

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 01_7e 56. Answers will vary but might include pangenesis, inheritance of acquired characteristics, preformationism, or blending inheritance, which are all described in Section 1.1. Pangenesis—The idea that information needed to encode each body structure is stored in that structure and transported to the reproductive organs and passed to the embryo at conception. Inheritance of acquired characteristics—The idea that traits acquired through use during one's lifetime can be passed to one's offspring. Preformationism—The idea that the sperm or egg carries a tiny preformed person whose development simply involves enlargement. Blending inheritance—The idea that the genetic material is a fluid that gets blended during sexual reproduction between a male and female, resulting in the production of traits in the offspring that are blended intermediates of those of the parents.

57. Answers will vary but should include specific evidence or observations that do not support the idea. Pangenesis— Observations of animals with body parts lost to injury producing normal offspring would not support pangenesis.

Inheritance of acquired characteristics— Experiments were conducted in which body parts were removed and normal offspring were produced, showing that the acquired characteristic was not inherited. Also, experiments in which offspring are raised in an environment different from that of their parents and do not develop their parents' traits would suggest that the environment influences development of these traits. Preformationism—Eventually better microscopes were produced that proved that gametes do not contain preformed people. Also, we eventually came to understand that both sperm and egg contribute genetic information during sexual reproduction.

Blending inheritance—Mendel showed that genes behave as particles that are not blended or changed during inheritance.

58. Bacteria and viruses have their genetic material (DNA) organized into genes, just like other organisms, so the basics of inheritance are the same in bacteria and viruses as in other organisms.

The genetic systems of bacteria and viruses are simpler when compared to higher eukaryotic organisms such as mice, dogs, or humans: they have fewer genes, fewer chromosomes, and less DNA. Bacteria and viruses reproduce more quickly than higher eukaryotic organisms: the generation time is shorter than for mice, dogs, or humans.

Bacteria and viruses are easy and less expensive to grow (take up less space, have less complicated nutritional needs) than vertebrates.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 01_7e 59. Answers will vary but should include specific evidence or observations that support the idea. Examples: Pangenesis—It is reasonable to assume that the information needed to build a structure must reside in that structure. It is less obvious that the information might also reside in other structures. Therefore, it is reasonable to envision the information being stored in each structure and transported to the reproductive structures before being passed to the next generation.

Inheritance of acquired characteristics—Observations to support this view would have been commonplace. For example, a man with a muscular physique would often have sons with muscular physiques. A talented musician often produced children with musical talent.

Preformationism—It would have been hard for people before the late 1800s to imagine how a complex organism could build itself from a single undifferentiated cell. Indeed, the problem has occupied developmental biologists for over 100 years. Preformationism is easier to understand. Add to that the poor optics of microscopes at that time, and it is easy understand how early biologists might have thought they could see a preformed person in a sperm or an egg, such as in Figure 1.11.

Blending inheritance—For example, a mating between a tall person and a short person producing a person of medium stature might have suggested blending inheritance.

60. Techniques for the observation of cells have been available only since the late 1500s, when the first microscopes were produced. The observation of chromosomes has been possible for only a century and a half. The widespread systematic study of genes and inheritance has been conducted only in the twentieth century, since the rediscovery of Mendel's work in 1900. The structure of DNA was determined only in the mid-twentieth century. Many molecular genetics techniques, like PCR, have been developed only in the last few decades. However, without understanding the nature of chromosomes and genes, plant and animal breeders have been applying the principles of inheritance for thousands of years to obtain desired characteristics in domesticated organisms.

61. Strengths—Fruit flies have proven to be an excellent model system for studying aspects of biology that they share with humans. Fruit flies are simpler in structure and physiology than humans and have a much simpler genome.

They are small and easy to raise, they have a short generation time, and they produce a large number of offspring.

Their chromosomes have been mapped and their genomes analyzed extensively. It is relatively easy to isolate and study mutants that are defective in specific processes of interest. These characteristics make them ideal for genetic studies of biological processes.

Weaknesses—Some aspects of fruit fly genetics and development are not shared with humans. Therefore, some features discovered in fruit flies will not apply directly to humans. Also, humans have many features that fruit flies lack. Fruit flies will not serve well as a model system for studying these features of human biology.

62. An example would be the use of genetic tests in medical practice. Genetic tests are already fairly commonplace and will become more common in the near future. In the future, most people will be offered one or more genetic tests as part of their ordinary medical care. They will need to understand the basis of the tests and their limitations, as well as how to interpret results and how to use the information provided.

Genome editing of humans, as with the CRISPR-modified babies born in 2019, may influence humankind more generally. Rules and regulations and their enforcement vary greatly through the world.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 01_7e 63. Prokaryotic cells lack a nuclear membrane and possess no true membrane-bound cell organelles, whereas eukaryotic cells possess a nucleus and membrane-bound organelles such as chloroplasts and mitochondria.

64. Answers will vary, but the best answers will include one or more specific discoveries in genetics and describe how they affect the student personally. Examples could come from those listed in Section 1.1 or from the student's background. This question and Questions 52 and 53 will work best if students are asked to consider ahead of time how these discoveries might impact their lives. For example, students might discuss the role of genetics and genetic technology in the Green Revolution of the 1950s and 1960s, which greatly expanded food production throughout the world, making food more efficient, more affordable, and more available to world populations. A much smaller part of the world's population works in agriculture, freeing up more people for work in other industries. They might also discuss the use of genetically modified crops in agriculture, including a significant proportion of corn and soybeans in the United States and other countries.

65. Despite the remarkable diversity of life on Earth, all genomes are encoded in nucleic acids. With few exceptions, the genetic code is common to all forms of life. Finally, the process by which genetic information is copied and decoded is remarkably similar for all forms of life.

66. Answers may vary. In some senses, the wealth of genetic information and studies concerning humans, the fact that the human genome has been sequenced, and the existence of experimental techniques to study humans provide similarities to model organisms. However, humans don't conform to true model organisms in terms of other experimentation advantages—short generation time, small size, inexpensiveness to propagate. Also, while natural, environmental, and substance-induced variations are studied in humans, experiments and studies must be set up within strong ethical guidelines and approvals not required for any other true model organism. The recognition of Homo sapiens as a distinct, separate form of life with an inherent dignity not common to other organisms prevents them from being classified as model organisms.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 02_7e Indicate the answer choice that best completes the statement or answers the question.

1. Which of the following statements is TRUE? a. Eubacteria are prokaryotes while the archaea are eukaryotes. b. Archaea are more closely related to eukaryotes than to eubacteria. c. Eukaryotes are more closely related to eubacteria than to archaea. d. Viruses are more closely related to prokaryotes than to eukaryotes. e. Eubacteria, archaea, and eukaryotes are all equally related.

2. The process of splitting the cytoplasm, which separates one cell into two, is termed a. cytokinesis. b. mitosis. c. anaphase. d. diakinesis. e. fusion.

3. To provide food for the developing embryo, a tissue called endosperm is produced through double fertilization. Endosperm has a ploidy of a. 1n. b. 2n. c. 3n. d. 4n. e. 5n.

4. Which chromosome in the following figure is MOST likely to be described as acrocentric?

a. (a) b. (b) c. (c) d. (d) e. Chromosomes (a) and (b) are both acrocentric.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 02_7e 5. In a typical flowering plant, a pollen grain that lands on a stigma grows a pollen tube to deliver _____ (how many?) sperm to the ovary. Fusion of a sperm with an egg produces a _____ n cell called a a. 1; 1; zygote. b. 2; 1; megasporocyte. c. 2; 2; zygote. d. 1; 2; microsporocyte. e. 1; 2; megasporocyte.

6. A diploid somatic cell from a rat has a total of 42 chromosomes (2n = 42). As in humans, sex chromosomes determine sex: XX in females and XY in males. What is the total number of chromosomes in a polar body cell from a rat? a. 21 b. 40 c. 41 d. 42 e. 84 7. Chromosome movement during anaphase is a result of a. disassembly of tubulin molecules by molecular motor proteins. b. kinetochore shortening causing chromosomes to pull apart. c. metaphasal plate splitting resulting in chromosomal disassembly. d. the cohesion protein attaching to the centromeres of sister chromatids. e. cilia movement inside the cellular structure.

8. Assume that the diploid or 2n number of chromosomes is 18 for a certain species of animal. How many DNA molecules will be found in metaphase II for this species? a. 9 b. 18 c. 36 d. 72 e. 24

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 02_7e 9. In tissue from the intestinal epithelium of a frog, the following proportions of cells were found at each stage of the cell cycle: Stage Interphase Prophase Prometaphase Metaphase Anaphase Telophase

Proportion of Cells 0.90 0.04 0.02 0.01 0.02 0.01

If the entire cell cycle in frog epithelium cells requires 20 hours for completion, what is the average duration of each stage? a. 18 hours for interphase, 0.4 hour for prophase, 0.2 hour for prometaphase, 0.2 hour for metaphase, 0.2 hour for anaphase, 0.4 hour for telophase b. 1.8 hours for interphase, 0.8 hour for prophase, 0.2 hour for prometaphase, 0.2 hour for metaphase, 0.2 hour for anaphase, 0.8 hour for telophase c. 18 hours for interphase, 0.8 hour for prophase, 0.4 hour for prometaphase, 0.2 hour for metaphase, 0.4 hour for anaphase, 0.2 hour for telophase d. 9 hours for interphase, 0.8 hour for prophase, 0.2 hour for prometaphase, 0.2 hour for metaphase, 0.6 hour for anaphase, 0.4 hour for telophase e. 18 hours for interphase, 0.8 hour for prophase, 0.6 hour for prometaphase, 0.2 hour for metaphase, 0.2 hour for anaphase, 0.8 hour for telophase 10. Which of the following statements is TRUE? a. Archaea resemble eukaryotes in cell structure. b. Evolutionarily, it is clear archaea are mostly closely related to bacteria. c. The evolutionary relationships among bacteria, archaea, and bacteria are well understood. d. While cell structure between archaea and bacteria is similar, certain genetic processes such as transcription are more similar between archaea and eukaryotes. e. While cell structure between archaea and eukaryotes is similar, certain genetic processes such as transcription are more similar between archaea and bacteria.

11. In eukaryotes, chromosomes do NOT contain a. ribosomes. b. chromatin. c. proteins. d. histones. e. DNA.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 02_7e 12. The centromere divides a chromosome into two sections or "arms." A chromosome is found to have two arms of equal lengths. Such a chromosome can be BEST described as a. telocentric. b. circular. c. acrocentric. d. metacentric. e. homologous.

13. A dividing eukaryotic cell is treated with a drug that inhibits the molecular motors associated with kinetochores. At which cell cycle stage would it stop? a. G1 b. S c. G2 d. M (metaphase) e. M (telophase) 14. Which of the following are NOT prokaryotes? a. eubacteria b. archaea c. viruses d. ancient bacteria 15. Assume that cells that are about to undergo meiosis are treated with a chemical that blocks crossing over but does not affect the cells in any other way, and four viable cells are produced by the two divisions of meiosis.

What will be the consequence of such a treatment? a. The four products of meiosis will be genetically identical. b. The four products of meiosis will all be genetically unique. c. All the chromosomes of two of the products of meiosis will have chromosomes that are paternal in origin, but the other two products will have chromosomes that are of both paternal and maternal origins. d. All the chromosomes of two of the products of meiosis will have chromosomes that are maternal in origin, but the other two products will have chromosomes that are of both paternal and maternal origins. e. Two of the products will be genetically identical but genetically different from the other two products, which will also be genetically identical.

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 02_7e 16. Which of the following does NOT occur during the G2 phase of the cell cycle? a. The G2/M checkpoint is reached. b. DNA replication and error checking are completed. c. The cell completes preparation for mitosis. d. The cell divides. e. All of these occur during the G2 phase of the cell cycle.

17. Prokaryotic chromosomes do NOT have telomeres because they a. do not go through mitosis. b. do not go through DNA replication. c. are in the cytoplasm. d. are circular. e. have no centromeres.

18. In prokaryotes, replication usually begins at a specific place on the chromosome called the a. binary fission site. b. origin of replication. c. origin of mitosis. d. anchoring site. e. kinetochore.

19. Humans have 23 pairs of chromosomes. Rarely, an egg is produced with 46 chromosomes instead of 23.

How might such an egg have originated? a. When the first polar body divides in meiosis II, all the chromatids go to one daughter cell. b. When the secondary oocyte divides in meiosis II, all the chromatids go to one daughter cell. c. When the second polar body divides in meiosis II, all the chromatids go to one daughter cell. d. When the primary oocyte divides in meiosis I, all the chromosomes go to the first polar body. e. When the secondary oocyte divides in meiosis I, all the chromatids go to the second polar body.

20. The cells of a mature pea plant have 14 chromosomes. In a pea plant ovary, how many chromosomes would the nucleus of a megaspore contain? a. 3 1/2 b. 7 c. 14 d. 21 e. 30

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 02_7e 21. Which of the following processes is unique to plants? a. meiosis b. double fertilization c. crossing over d. haploid gametes e. spermatogenesis 22. The cells of a mature pea plant have 14 chromosomes. How many chromosomes does a nucleus in the pea endosperm contain? a. 3 1/2 b. 7 c. 14 d. 21 e. 30 23. Which of the following statements is FALSE? a. Generally, chromosomes of prokaryotes are circular. b. Prokaryotes usually have a single molecule of DNA. c. Generally, chromosomes of eukaryotes are circular. d. Eukaryotes usually have multiple chromosomes. e. Eukaryote chromosomes are usually linear.

24. In order to be functional, a eukaryotic chromosome requires all of the following EXCEPT a. a centromere. b. origins of replication. c. a plasmid. d. telomeres.

25. If a healthy cell passes the G1/S checkpoint a. it will enter the G0 stage of the cell cycle. b. DNA will be replicated. c. it will not divide. d. it will proceed immediately to cytokinesis. e. it will die.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 02_7e 26. Suppose that a diploid cell contains eight chromosomes (2n = 8). How many different combinations in the gametes are possible? a. 2 b. 4 c. 8 d. 16 e. 64 27. A diploid somatic cell from a rat has a total of 42 chromosomes (2n = 42). As in humans, sex chromosomes determine sex: XX in females and XY in males. What is the total number of chromosomes present in the cell during metaphase I of meiosis? a. 21 b. 42 c. 84 d. 126 e. 168 28. In a flowering plant, the male part of the flower (the stamen) produces haploid microspores that divide by _____ to eventually produce sperm. a. mitosis b. meiosis c. gametogenesis d. spermatogenesis e. fertilization

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 02_7e 29. A "mistake" is happening during meiosis I in the following figure. Assume the second meiotic division is normal.

How many chromosomes would be expected in the four cellular products of this meiotic event?

a. All four cells would have four chromosomes. b. All four cells would have three chromosomes. c. Two cells would have three chromosomes, and two cells would have five chromosomes. d. Two cells would have six chromosome, and two cells would have 10 chromosomes. e. One cell would have three chromosomes, one cell would have five chromosomes, and two cells would have four chromosomes.

30. A geneticist observes 10 pairs of homologous chromosomes at metaphase I of meiosis in a newly discovered species of flowering plant. How many chromosomes should be found in a microsporocyte? a. 20 b. 10 c. 5 d. 40 e. 2 31. A chromosome with a centromere at the very end is called a. submetacentric. b. metacentric. c. acrocentric. d. acentric. e. telocentric.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 02_7e 32. The highly organized internal scaffolding of the nucleus is called the a. histone complex. b. spindle microtubules. c. nuclear cohesion. d. nuclear matrix. e. nuclear envelope.

33. Diploid cells are cells with _____ chromosomes. a. a single set of b. circular c. two sets of d. many sets of e. three sets of 34. A diploid somatic cell from a rat has a total of 42 chromosomes (2n = 42). As in humans, sex chromosomes determine sex: XX in females and XY in males. What is the total number of telomeres in a rat cell in G2? a. 21 b. 42 c. 84 d. 126 e. 168 35. Which of the following statements is FALSE? a. Errors in chromosome separation are rarely a problem for an organism. b. Errors in chromosome separation can result in a miscarriage. c. Errors in chromosome separation can result in cancer. d. Errors in chromosome separation can result in a child with severe handicaps. e. Errors in chromosome separation can cause numerous problems for an organism.

36. What might be the result if the breakdown of the shugoshin protein were premature? a. The cohesion protein would hold the chromosome arms together longer. b. The separation of homologous chromosomes would occur prematurely. c. The separation of sister chromatids would occur prematurely. d. Spindle fibers would not form. e. Sister chromatids would never separate.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 02_7e 37. Which of the following occurs during prometaphase? a. The chromosomes align in a single plane. b. DNA is replicated. c. Microtubules attach to the kinetochores. d. Mitotic spindles form. e. The two sister chromatids separate.

38. The attachment point on the chromosome for spindle microtubules is the a. telomere. b. centromere. c. origin of replication. d. sister chromatid. e. allele.

Indicate one or more answer choices that best complete the statement or answer the question.

39. Why is meiosis I also called reductional division? (Select all that apply.) a. The resulting cells are smaller than the original cell. b. The resulting cells have half the chromosome content of the original cell. c. The resulting cells have a quarter of the chromosome content of the original cell. d. The ploidy number is reduced by a factor of 2. e. Individual cell DNA content goes from 2n to 1n.

40. Somatic cancer cells often are unstable and divide inappropriately (divide when they should not be dividing).

In addition, such cells often contain losses of some chromosomes and extra copies of other chromosomes.

Defects in which of the following may be partially responsible for the aberrant behavior of cancer cells?

(Select all that apply.) a. spindle-assembly checkpoint b. G1/S checkpoint c. homologous chromosome pairing d. crossing over

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 02_7e 41. The cells illustrated here belong to a species with a diploid chromosome number of four. Each of the following cells is in which stage of mitosis or meiosis?

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 02_7e 42. Why are viruses considered to be neither prokaryotic nor eukaryotic?

43. Explain why mitosis does not produce genetic variation and how meiosis leads to the production of tremendous genetic variation.

44. (a) Draw a pair of acrocentric homologous chromosomes as they would appear in G2. Indicate centromeres with a small circle, and place the alleles A and a on each of the chromatids. (b) Draw the same chromosomes as they would appear in G1. Place the alleles A and a on each of the chromatids.

45. A diploid, eukaryotic cell in interphase has these two pairs of homologous chromosomes with the indicated arrangement of alleles:

Draw the chromosomes at the end of telophase of (a) mitosis and (b) meiosis II. Indicate placement of alleles on the chromosomes.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 02_7e 46. A diploid, eukaryotic cell in interphase has these two pairs of homologous chromosomes with the indicated arrangement of alleles:

Draw the chromosomes at the end of (a) prophase of mitosis and (b) prophase I (of meiosis I) with the most likely crossing-over events. Indicate placement of alleles on the chromosomes.

47. (a) Compare and contrast spermatogenesis and oogenesis in animals. For each process, be sure to include information about division of the nucleus, allocation of chromosomes to the various products, and division of the cytoplasm. (b) Why is the difference in cytoplasmic division between spermatogenesis and oogenesis important to reproduction, considering the different roles of sperm and eggs in reproduction?

48. (a) Describe the changing role of cohesin during the mitotic cell cycle. (b) Explain the importance of regulation of cohesin activity to normal cell division.

49. Microscopy to look at a cell's chromosomes is often performed when the cell is in mitotic metaphase. For example, karyotyping (extracting chromosomes from a single cell and photographing them to look for abnormalities) is performed on metaphase, rather than interphase, cells. Why?

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 02_7e 50. List two differences and two similarities between mitosis and meiosis.

51. Describe the difference between homologous chromosomes and sister chromatids.

52. Describe what is happening to chromosomes during the five substages of prophase I.

53. Describe the difference between the centromere and kinetochore.

54. Describe the difference between the sporophyte and gametophyte.

55. Write all possible genotypes of each of the cells resulting from mitosis and meiosis of a cell of the genotype shown below.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 02_7e 56. Describe the difference between meiosis I and meiosis II.

57. During prophase I of meiosis, crossing over is indicated by what microscopically visible structure?

58. What events during sexual reproduction are significant in contributing to genetic diversity?

59. What is one feature of meiosis that produces genetic variability in gametes? In two or three sentences, explain how this feature causes genetic uniqueness.

60. Why is mitosis important within the cell cycle?

61. B. subtilis is a type of bacterium. Under certain growth conditions, smc null mutants (i.e., completely lacking smc function) can develop with abnormal nucleoids and some have increased DNA content.* Based on your knowledge of SMC function, explain why these phenotypes might arise. *Reference: Britton R. A., Lin D. C., Grossman A. D. (1998) Characterization of a prokaryotic SMC protein involved in chromosome partitioning. Genes Dev. 12:1254–1259.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 02_7e 62. Describe the difference between G1 and G2 of the cell cycle.

63. Using the following choices, indicate the CORRECT phase(s) in parts a–e.

1. meiosis I prophase 2. meiosis I anaphase 3. meiosis II prophase 4. meiosis II anaphase 5. mitosis prophase 6. mitosis anaphase a. Chromosomes are in unseparated, sister-chromatid form at the end of phase(s) _____. b. Chromosomes condense during _____. c. Sister chromatids separate during _____. d. Chromosomes are randomly partitioned during _____, contributing to genetic diversity. e. Crossing over (genetic recombination) occurs in _____.

64. List and briefly describe three major cell cycle checkpoints. For each checkpoint, predict the consequences if the checkpoint fails to work properly.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 02_7e Answer Key 1. b 2. a 3. c 4. c 5. c 6. a 7. a 8. b 9. c 10. d 11. a 12. d 13. d 14. c 15. e 16. d 17. d 18. b 19. b 20. b 21. b 22. d 23. c 24. c 25. b 26. d Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 02_7e 27. b 28. a 29. c 30. a 31. e 32. d 33. c 34. e 35. a 36. c 37. c 38. b 39. b, d, e 40. a, b 41. a. meiosis I metaphase b. mitosis metaphase c. mitosis anaphase d. meiosis I anaphase e. meiosis II metaphase 42. Viruses do not possess the structure of a cell. Viruses are actually simple structures composed of an outer protein coat surrounding nucleic acid.

Viruses cannot reproduce outside of their host cells, suggesting that viruses evolved after cells were present and not before.

43. Mitosis produces cells that are genetically identical to the parent cell. Meiosis includes two distinct processes that contribute to the generation of genetic variation: Crossing over shuffles alleles on the same chromosome into new combinations, whereas the random distribution of pairs of homologous chromosomes, one member of each pair coming from the mother and the other from the father, shuffles alleles on different chromosomes into new combinations.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 02_7e

44.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 02_7e

45.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 02_7e

46.

47. (a) Division of the nucleus and allocation of the chromosomes to the products are essentially the same in both processes. Starting with a 2n germ cell, nuclear division is by meiosis I and II, and each product of meiosis contains one set of chromosomes (1n). The major difference is that division of the cytoplasm during meiosis I and II is equal in spermatogenesis and unequal in oogenesis. During oogenesis, meiosis I produces a large secondary oocyte with lots of cytoplasm and a polar body with very little cytoplasm. Meiosis II in the secondary oocyte produces a large ovum with lots of cytoplasm and a small second polar body. Therefore, only one large, functional egg is produced per primary oocyte, whereas four small, functional sperm are normally produced per primary spermatocyte. (b) The small size and other features of sperm structure suit them well to delivery of the haploid nucleus to the egg.

The large amount of cytoplasm in the egg suits it well to nourishing development of the embryo after fertilization.

48. (a) Cohesin keeps sister chromatids together after DNA replication during S phase through metaphase of mitosis.

The breakdown of cohesin allows the sister chromatids to separate from each other during anaphase. (b) Cohesin must be active beginning in S phase through metaphase in order to keep the sister chromatids together so that they can be properly aligned at the metaphase plate to ensure equal division of the genetic information to the two daughter cells. Cohesin must be inactivated or broken down in order to allow the sister chromatids to separate during anaphase so that each daughter cell will get one copy of the genes on each chromosome.

49. In metaphase, chromosomes are condensed and are more easily visualized.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 02_7e 50. Differences: (1) Mitosis occurs in somatic (nonsex) cells; meiosis occurs in sex cells to produce gametes.

(2) Meiosis involves chromosome pairing (of homologous chromosomes); mitosis does not.

(3) Mitosis produces nonsex cells; meiosis produces gametes directly or indirectly.

(4)Mitosis produces cells of the same ploidy; meiosis produces haploid cells from diploid cells.

(5) Meiosis has two consecutive divisions; mitosis has one.

(6) Mitosis produces two daughter cells; meiosis produces four daughter cells.

(7) Mitosis produces identical daughter cells; meiosis produces four different daughter cells.

Similarities: (1) Both involve the separation of replicated chromosomes during cell division.

(2) Both are processes to ensure that daughter cells in cell division receive a complete set of chromosomes.

(3) DNA replication must occur first.

(4) Cytokinesis usually occurs at the end of each.

51. Homologous chromosomes can have different alleles. Sister chromatids are duplicates and (except for errors in replication) are identical in sequence.

52. Leptotene—chromosomes contract and become visible Zygotene—chromosomes continue to condense and homologous chromosomes pair up and begin synapsis Pachytene—chromosomes become shorter and thicker; synaptonemal complex develops between homologous chromosomes Diplotene—centromeres of paired chromosomes move apart; the two homologs remain attached at each chiasma Diakinesis—centromeres move apart 53. A centromere is the physical location (DNA sequence) on a chromosome where the kinetochore and spindle microtubules attach. The kinetochore is composed of proteins that assemble on the centromere to provide a site for the spindle microtubules to attach.

54. The sporophyte is the diploid phase of a plant life cycle. The gametophyte is the haploid stage.

55. Mitosis: A/a B/b D/d or ABD/abd (diploid and heterozygous at all three loci) Meiosis: ABd, aBd, AbD, abD, Abd, aBD, ABD, abd (haploid at all three loci) 56. Homologs pair and segregate in meiosis I. Sister chromatids are paired and segregate in meiosis II. Crossing over occurs in meiosis I but not in meiosis II. 57. Chiasmata (chiasma) or the synaptonemal complex 58. (1) Crossing over changes allele combinations on chromosomes, so, after meiosis I, even sister chromatids are not genetically identical.

(2) Independent assortment of nonhomologous chromosomes ensures each gamete has a different combination of alleles for genes on nonhomologs.

(3) Two genetically unique gametes from each parent combine during fertilization to form a novel, genetically unique individual.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 02_7e 59. a. Independent assortment. In meiosis I—metaphase and anaphase—nonhomologous chromosomes distribute randomly. Alignment and separation of one pair of homologous chromosomes is independent of how a different pair separates. Different gametes that have different chromosomes can have different alleles for the same genes, so the gametes normally have different combinations of alleles. b. Crossing over. In meiosis I—prophase—portions of homologous chromosomes exchange, changing combinations of alleles of genes on a single chromosome, so not even sister chromatids are identical after crossing over. Each gamete has only one copy of each homolog, and each homolog now has a unique combination of alleles.

60. Mitosis is important because it results in two daughter cells that have identical nuclear chromosome complements so the daughter cells are genetically identical to each other and genetically identical to the parent cell from which they arose. The process of mitosis makes new cells and replaces cells that are worn out or damaged.

61. Answers should provide an overview and not be detailed. As described in Section 1.2, SMC (structural maintenance of chromosome) proteins encircle DNA during binary fission of prokaryotic cells to keep them untangled. So, SMC proteins are required for binary fission and chromosome segregation. Thus, smc null mutations will lead to defects associated with proper chromosome segregation, such as improper nucleoids and some cells having excess DNA content because excess DNA was segregated to one daughter cell.

62. G1 occurs before S phase and G2 occurs after S phase. During G1, cells grow in size; chromosomes are composed of a single chromatid. During G1, cells pass a critical checkpoint (the G1/S checkpoint), after which they are committed to undergoing cell division. During G2, the chromosomes are composed of two chromatids. There is another checkpoint during G2 that ensures cells are prepared for mitosis. Cells typically spend more time in G1 than in G2.

63. a. 1, 2, 3, 5 b. 1, 3, 5 c. 4, 6 d. 2 e. 1 64. (1) The G1/S checkpoint holds the cell in G1 until the cell has all of the enzymes necessary for replication of DNA. If the checkpoint failed, the cell would proceed into S without the necessary enzymes, causing the DNA not to be replicated properly or completely. This might cause the cell cycle to halt at the G2/M checkpoint. Alternatively, the cell might divide without the genetic material having been replicated, causing the daughter cells to receive incomplete genetic information. Both predictions are reasonable based on information in the chapter.

(2) The G2/M checkpoint is passed only if the cell's DNA is undamaged. If it fails to work properly, division would proceed in the presence of damaged DNA, possibly leading to mutations in the daughter cells and/or death of the daughter cells.

(3) The spindle-assembly checkpoint is during metaphase, and it ensures that each chromosome is aligned at the metaphase plate and attached to spindle fibers from opposite poles. This checkpoint depends on tension at the kinetochores of each chromosome. If the checkpoint fails, anaphase will occur even when the chromosomes are not aligned properly, allowing daughter cells to be produced with extra and/or missing chromosomes.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 03_7e Indicate the answer choice that best completes the statement or answers the question.

1. Gregor Mendel carried out a cross between two pea plants by taking pollen from a plant that was homozygous for round seeds and dusting the pollen onto the stigma of a plant homozygous for wrinkled seeds.

Which of the following would be the reciprocal cross that Mendel had carried out for this experiment? a. stigma of a plant homozygous for round seed pollinated with pollen from a plant homozygous for wrinkled-seed plant b. stigma of a plant homozygous for round seed pollinated with pollen from a plant heterozygous for wrinkled-seed plant c. stigma of a plant heterozygous for round seed pollinated with pollen from a plant homozygous for wrinkled-seed plant d. stigma of a plant homozygous for wrinkled seed pollinated with pollen from a plant homozygous for round-seed plant e. stigma of a plant homozygous for wrinkled seed pollinated with pollen from a plant homozygous for wrinkled-seed plant 2. Round seeds (R) are dominant to wrinkled seeds (r), and yellow seeds (Y) are dominant to green seeds (y). A true-breeding plant with round, yellow seeds is crossed to a true-breeding plant with wrinkled, green seeds.

What is the genotype of the F1 progeny? a. RRYY b. RrYY c. RRYy d. RrYy e. rryy 3. If two heterozygous Aa plants are crossed with each other, what will be the genotypic ratio found in the offspring? a. 1:1 b. 3:1 c. 1:1:1:1 d. 2:1 e. 1:2:1

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 03_7e 4. In dogs, black coat color (B) is dominant over brown (b), and solid coat color (S) is dominant over white spotted coat (s). A cross between a black, solid female and a black, solid male produces only puppies with black, solid coats. This same female was then mated with a brown, spotted male. Have of the offspring from this cross were black and solid, and half of the offspring were black and spotted. Which of the following could be the genotype of the black, solid male? a. BBSs b. BBss c. BbSS d. BbSs e. Bbss 5. In a cross between AaBbCcDdEe and AaBbccDdee, what proportion of the offspring would be expected to be A_bbC_ddE_? (A_ means AA or Aa.) a. 3/256 b. 3/32 c. 3/16 d. 3/8 e. 3/4 6. Round seeds (R) are dominant to wrinkled seeds (r), and yellow seeds (Y) are dominant to green seeds (y). A true-breeding pea plant with round, yellow seeds is crossed to a true-breeding plant with wrinkled, green seeds. The F1 progeny are allowed to self-fertilize. What is the probability of obtaining a round, yellow seed in the F2? a. 3/4 b. 1/16 c. 9/16 d. 3/16 e. 1/2 7. Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an AaBb zygote from a cross of AaBb × AABB? a. 1/4 b. 1/2 c. 1/16 d. 9/16 e. 1 (100%)

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 03_7e 8. Which of the following crosses would produce a 3:1 ratio of phenotypes in the next generation? a. AA × AA b. AA × aa c. Aa × Aa d. Aa × aa e. aa × aa 9. What important genetics concept is being shown in the diagram?

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 03_7e

a. Mendel's principle of independent assortment b. Mendel's principle of segregation c. The molecular nature of alleles Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 03_7e d. The chi-square goodness-of-fit test e. The results of a dihybrid cross

10. The figure shows that independent assortment results from the segregation of chromosomes at anaphase I of meiosis. As products of meiosis and no crossing over, what is the minimum number of meiotic events (an event consisting of the two produce the RY, Ry, rY, and ry gametes?

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 03_7e

a. one b. two Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 03_7e c. three d. four e. A correct answer can't be determined.

11. In Mendel's peas, purple flower color is dominant to white. From which of the following descriptions can you NOT infer the genotype completely? a. purple b. white c. pure-breeding purple d. pure-breeding white e. heterozygous 12. In dogs, black coat color (B) is dominant over brown (b), and solid coat color (S) is dominant over white spotted coat (s). A cross between a black, solid female and a black, solid male produces only puppies with black, solid coats. This same female was then mated with a brown, spotted male. Half of the offspring from this cross were black and solid, and half of the offspring were black and spotted. What is the genotype of the brown, spotted male? a. BBSS b. BbSS c. BBSs d. BbSs e. bbss 13. A couple has six daughters and is expecting a seventh child. What is the probability that this child will be a boy? a. 1/2 b. 1/4 c. 1/16 d. 1/64 e. 1/128 14. Which of the following was NOT one of Mendel's conclusions based on his monohybrid crosses? a. Genes are carried on chromosomes. b. Alleles exist in pairs. c. Alleles segregate equally into gametes. d. Alleles behave as particles during inheritance. e. One allele can mask the expression of the other allele.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 03_7e 15. In Mendel's peas, yellow seeds are dominant to green. A pure-breeding yellow plant is crossed with a purebreeding green plant. All of the offspring are yellow. If one of these yellow offspring is crossed with a green plant, what will be the expected proportion of plants with green seeds in the next generation? a. 0% b. 25% c. 50% d. 75% e. 100% 16. The ability to curl one's tongue into a U-shape is a genetic trait. Curlers always have at least one curler parent, but noncurlers can have one or both parents who are curlers. Using C and c to symbolize the alleles that control this trait, what is the genotype of a noncurler? a. CC b. Cc c. cc d. CC or Cc 17. Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an AABB zygote from a cross of AaBb × AaBb? a. 1/4 b. 1/2 c. 1/16 d. 9/16 e. 1 (100%)

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 03_7e 18. Honeybees have a haplo-diploid sex determination system where females develop from a fertilized egg (they are diploid, having one allele from the female queen and one allele from the male), and males develop from unfertilized eggs (they are haploid, having only one allele from the queen). Assuming that the female parent (queen) is heterozygous for a particular gene, what is the probability that a female offspring will inherit the recessive allele from her mother? What is the probability that a male offspring will inherit a recessive allele from his mother? a. The probability that a daughter will inherit a recessive allele from her mother is 50%; the probability that a son will inherit a recessive allele from his mother is 50%. b. The probability that a daughter will inherit a recessive allele from her mother is 50%; the probability that a son will inherit a recessive allele from his mother is 100%. c. The probability that a daughter will inherit a recessive allele from her mother is 100%; the probability that a son will inherit a recessive allele from his mother is 50%. d. The probability that a daughter will inherit a recessive allele from her mother is 100%; the probability that a son will inherit a recessive allele from his mother is 100%. e. The probability that a daughter will inherit a recessive allele from her mother is 0%; the probability that a son will inherit a recessive allele from his mother is 100%.

19. Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an AB gamete from an AABb individual? a. 1/4 b. 1/2 c. 1/16 d. 9/16 e. 1 (100%)

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 03_7e 20. While doing fieldwork in Madagascar, you discover a new dragonfly species that has either red (R) or clear (r) wings. Initial crosses indicate that R is dominant to r. You perform three crosses using three different sets of red-winged parents with unknown genotype and observe the following data: Cross

Phenotypes

1

72 red-winged, 24 clear-winged

2

12 red-winged

3

96 red-winged

Which cross is likely to have at least one parent with the genotype RR? a. cross 1 b. cross 2 c. cross 3 d. crosses 1 and 2 e. crosses 2 and 3 21. In a cross between pure-breeding tall plants with pure-breeding short plants performed by Mendel, all of the F1 are tall. When these plants are allowed to fertilize themselves, the F2 plants occur in a ratio of 3 tall to 1 short. Which of the following is NOT a valid conclusion from these results? a. The allele that produces the tall condition is dominant to the allele that produces the short condition. b. The difference between tall and short stature in these lines is controlled by a single gene pair. c. During production of gametes in F1 plants, the tall and short alleles segregate from each other equally into the gametes. d. The tall and short traits assort independently of each other in this cross. e. Fertilization occurs randomly between gametes carrying the tall and short alleles.

22. In a cross between AABbCcDD and AaBbccdd, what proportion of the offspring would be expected to be A_B_C_D_? (A_ means AA or Aa.) a. 3/256 b. 3/32 c. 3/16 d. 3/8 e. 3/4

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 03_7e 23. A phenotypically normal man has phenotypically normal parents, but he has a sister who has cystic fibrosis caused by a recessive mutant allele. What is the probability that the man is heterozygous for the cystic fibrosis allele? a. 1/4 b. 1/2 c. 3/4 d. 2/3 e. 1/3 24. If an organism of genotype Aa was used for a testcross, what was the genotype of the other individual used in the cross? a. AA b. Aa c. aa d. Either Aa or aa e. Either AA or Aa 25. If both husband and wife are known to be heterozygous for the autosomal recessive condition of albinism, what is the probability that among their four children, three will be normal and one will have albinism? a. 27/256 b. 3/4 c. 9/16 d. 13/32 e. 27/64

26. An organism produces the gametes shown. Which statement about the meiotic divisions that produced these gametes (a over) is TRUE?

a. After the end of meiosis I, two genetically distinct types of cells were present. b. The gametes shown could NOT have resulted from a single meiotic event with two divisions. c. Independent assortment takes place during meiosis II.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 03_7e 27. A space capsule crashes to earth with an alien life form aboard. Two creatures emerge from the capsule, one with green skin and one with yellow skin. The yellow creature soon gives birth to offspring fathered by the green creature, producing 12 green and 8 yellow offspring. Green skin in these diploid creatures is dominant to yellow skin. You are curious to find out if the number of offspring is significantly different from expected Mendelian ratios, so you perform a chi-square test. What is the chi-square value for this cross? a. 0.2 b. 0.4 c. 0.8 d. 1.2 e. 1.6 28. This question is based off on question 23.

If both husband and wife are known to be heterozygous for the autosomal recessive condition of albinism, what is the probability that among their four children, two will be normal and two will have albinism? a. 27/256 b. 27/128 c. 27/64 d. 13/32 e. 3/4 29. Which of the following statements is TRUE? a. The genotype is the physical appearance of a trait. b. Alleles, genes, and loci are different names for the same thing. c. The phenotype of a dominant allele is never seen in the F1 progeny of a monohybrid cross. d. A testcross can be used to determine whether an individual is homozygous or heterozygous for a dominant allele.

30. This is based on question 45.

Assume the genes A–E represent five different genes that control five different traits. In a cross between AaBbCcDdEe and AaBbccDdee, what proportion of the offspring would be expected to show the dominant phenotype for every trait? a. 27/256 b. 27/128 c. 27/64 d. 13/32 e. 3/4

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 03_7e 31. In dogs, short fur is dominant to long fur. A dog with short fur is crossed with a dog with long fur. In a litter of four, all of the puppies have short fur. What is the BEST conclusion? a. The black poodle is definitely homozygous. b. The black poodle is probably homozygous. c. The black poodle is definitely heterozygous. d. The black poodle is probably heterozygous. e. The genotype of the black poodle cannot be inferred with this information.

32. A geneticist is conducting an experiment by making a testcross. She expects the offspring of the testcross to result in a 1:2:1 ratio. She wants to see if her data for the three phenotypic classes could be reasonably assumed to have deviated from the expected values by chance. How many degree(s) of freedom should she use when evaluating the chi-square goodness-of-fit test? a. 1 b. 2 c. 3 d. 4 e. 5 33. In a cross between AaBbCc and AaBbcc, what proportion of the offspring would be expected to be A_bbcc? (A_ means AA or Aa.) a. 3/256 b. 3/32 c. 3/16 d. 3/8 e. 3/4 34. Round seeds (R) are dominant to wrinkled seeds (r), and yellow seeds (Y) are dominant to green seeds (y). A true-breeding pea plant with round, yellow seeds is crossed to a true-breeding plant with wrinkled, green seeds. The F1 progeny are allowed to self-fertilize. What is the probability of obtaining a yellow seed in the F2? a. 3/4 b. 1/16 c. 9/16 d. 3/16 e. 1/2

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 03_7e 35. Freckles are caused by a dominant allele. A man has freckles, but one of his parents does not have freckles.

The man has fathered a child with a woman that does not have freckles. What is the probability that their child has freckles? a. 1/4 b. 1/3 c. 1/2 d. 2/3 e. 3/4 36. Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an offspring with the AB phenotype from a cross of aabb × AABB? a. 1/4 b. 1/2 c. 1/16 d. 9/16 e. 1 (100%) 37. Round seeds (R) are dominant to wrinkled seeds (r), and yellow seeds (Y) are dominant to green seeds (y). A true-breeding pea plant with round and yellow seeds is crossed to a true-breeding plant with wrinkled and green seeds. The F1 progeny are allowed to self-fertilize. What is the probability of obtaining a wrinkled, green seed in the F2? a. 3/4 b. 1/16 c. 9/16 d. 3/16 e. 1/2 38. Round seeds (R) are dominant to wrinkled seeds (r), and yellow seeds (Y) are dominant to green seeds (y). A plant of unknown genotype is testcrossed to a true-breeding plant with wrinkled, green seeds. The offspring produced were 53 round and yellow, 49 round and green, 44 wrinkled and yellow, and 51 wrinkled and green. What is the likely genotype of the parent in question? a. RRYY b. RrYY c. RRYy d. RrYy e. rryy

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 03_7e 39. Why was the pea plant an ideal plant for Mendel to use? a. generation time that is several years b. simple traits that are easy to identify c. low numbers of offspring produced d. expensive and time-consuming to grow e. All of the answers are correct.

40. If an AaBbccDd individual is crossed to an AaBbCcDD individual, what is the probability that the first offspring will be of AabbccDd genotype? a. 3/64 b. 1/64 c. 1/8 d. 1/32 e. 1/16 41. In horses, black coat color is dominant to chestnut brown. Suppose that a black horse is mated with a chestnut brown one and the offspring are twin foals, one black and one chestnut brown. What can you conclude about the genotype of the black parent? a. The genotype must be BB. b. The genotype must be bb. c. The genotype must be Bb. d. The genotype could be either BB or Bb. e. The genotype cannot be determined from these data.

42. Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an AB gamete from an AaBb individual? a. 1/4 b. 1/2 c. 1/16 d. 9/16 e. 1 (100%) 43. Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an offspring with the AB phenotype from a cross of AaBb × AaBb? a. 1/4 b. 1/2 c. 1/16 d. 9/16 e. 1 (100%) Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 03_7e 44. In dogs, black coat color (B) is dominant over brown (b), and solid coat color (S) is dominant over white spotted coat (s). A cross between a black, solid female and a black, solid male produces only puppies with black, solid coats. This same female was then mated with a brown, spotted male. Half of the offspring from this cross were black and solid, and half of the offspring were black and spotted. What is the genotype of the female? a. BBSS b. BbSS c. BBSs d. BbSs e. bbss 45. Freckles are caused by a dominant allele. A man has freckles, but one of his parents does not have freckles.

What is the man's genotype? a. homozygous dominant b. homozygous recessive c. heterozygous d. heterologous e. homologous 46. Which of the following crosses would produce a 1:1 ratio of phenotypes in the next generation? a. AA × AA b. AA × aa c. Aa × Aa d. Aa × aa e. aa × aa 47. In animals, the inability to make the pigment melanin results in albinism, a recessive condition. Two unaffected parents, who have decided to have three children, have a first child that has albinism (genotype aa). What is the probability that the second and third children will also have albinism? a. 1/4 b. 1/2 c. 1/16 d. 9/16 e. 1 (100%)

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 03_7e 48. Genes come in different versions called a. alleles. b. loci. c. genotypes. d. chromosomes. e. genomes.

Indicate one or more answer choices that best complete the statement or answer the question.

49. A cross was made and a chi-square test was done to see if the results were consistent with the expected 1:1:1:1 ratio from this cross. Using p = 0.05 as the cutoff value, which of the following calculated chi-square values would be consistent with the deviations from the expected values being due to chance? (Select all that apply.) a. 5.961 b. 0.536 c. 6.444 d. 2.384 e. 1.376 50. Which statement(s) about probabilities is/are TRUE? (Select all that apply.) a. The probability of a woman giving birth to three girls in a row is 1/2 + 1/2 + 1/2. b. The chi-square test is used to determine if observed outcomes are consistent with expected outcomes. c. The probability of two or more independent events occurring together is calculated by adding their independent probabilities. d. Branched diagrams are used for determining probabilities of various phenotypes or genotypes for genetic crosses involving more than one gene pair. e. The probability of rolling an even number on a single roll of a six-sided die is 1/6 + 1/6 + 1/6.

51. What is the difference between a backcross and a testcross?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 03_7e 52. In peas, tall (T) is dominant to short (t). A homozygous tall plant is crossed with a short plant. The F1 plants are self-fertilized to produce the F2. Both tall and short plants appear in the F2. If the tall F2 plants are selffertilized, what types of offspring and proportions will be produced?

53. A man has either an AaBB or AABb genotype with equal probability. Assume these genes assort independently. What is the overall probability that the man will produce an Ab gamete?

54. You have learned that purple flowers are dominant to white in Mendel's peas. When walking the grounds of Mendel's monastery, you come across a stray purple pea plant. You suspect that it is descended from Mendel's experimental plants, but you have no idea of its exact heritage. Propose two ways that you could determine the plant's genotype with respect to the flower color. Assume that you have any other pea plants that you might want to use in your analysis. Provide expected results and interpretations of possible results for your experiments. Which of the two ways would be easier and why?

55. What conclusions did Mendel make from his monohybrid crosses?

56. In deer mice, red eyes (r) are recessive to normal black eyes (R). Two mice with black eyes are crossed.

They produce two offspring, one with red eyes and one with black eyes. Give the genotypes of parents and offspring of this cross.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 03_7e 57. A man has either an AaBB or AABb genotype with equal probability. Assume these genes assort independently. What is the probability of the man producing an AB gamete?

In the introduction to this chapter, 'The Genetics of Blond Hair in the South Pacific," you learned that a variant (allele) of the TYRP1 gene leads to blond hair if an individual has the genotype TT. 58. Is the allele dominant or recessive? Explain why.

59. Two pea plants with purple flowers are crossed. Among the offspring, 63 have purple flowers, and 17 have white flowers. With a chi-square test, compare the observed numbers with a 3:1 ratio and determine if the difference between observed and expected could be a result of chance.

60. The following cross produced 125 progeny: aaBbCcDd × AaBbccDD. Assume all the genes assort independently and that the uppercase letters represent dominant alleles. a. How many offspring are expected to express the phenotype ABCD? b. How many offspring are expected to have the genotype aaBBccDd? c. How many offspring are expected to have the genotype AaBbCcDd?

In the introduction to this chapter, 'The Genetics of Blond Hair in the South Pacific," you learned that a variant (allele) of the TYRP1 gene leads to blond hair if an individual has the genotype TT. 61. In Chapter 1 (Figure 1.7), model genetic organisms were discussed. How did knowledge of a model organism aid the scientists in this case?

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 03_7e 62. Let us assume that hairy toes (hh) and brittle ear wax (ww) are both recessive traits in humans. In families of three children where both parents are heterozygotes at both loci (they have smooth toes and sticky ear wax) what is the probability that the family will consist of one hairy-toed, brittle ear-waxed boy and two smoothtoed, sticky ear-waxed girls?

63. Compare and contrast Mendel's principle of segregation and the principle of independent assortment.

64. Sex in mammals is determined by the X and Y sex chromosomes: Males are XY and females are XX. How do you explain the 50–50 sex ratio in mammalian progeny?

65. Using the diploid cell shown here (at interphase), illustrate/describe Mendel's principles of segregation and independent assortment.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 03_7e 66. In a plant species, you notice that purple and yellow leaf colors, as well as hairy and smooth stems, segregate.

In this species, yellow (P) is dominant to purple (p), and hairy (S) is dominant to smooth (s). You cross a plant with purple leaves and hairy stems to a plant with yellow leaves and hairy stems and generate the following progeny: Progeny Class 1 2 3 4 Total:

Phenotypes 68 yellow leaves, hairy stems 66 purple leaves, hairy stems 22 purple leaves, smooth stems 25 yellow leaves, smooth stems 181

a. Indicate the most likely genotypes for each parent. b. Propose a hypothesis to explain the progeny results. Based on this hypothesis, what ratios are expected for each of the four classes of progeny? c. Using the chi-square method, test your hypothesis and indicate whether you accept or reject it.

67. Albinism is a recessive condition resulting from the inability to produce the dark pigment melanin in skin and hair. A man and woman with normal skin pigmentation want to have two children. The man has one parent with albinism; the woman has parents with normal pigmentation but a brother with albinism. a. What is the probability that at least one child will be have albinism? b. What is the probability of both children having normal pigmentation?

In the introduction to this chapter, 'The Genetics of Blond Hair in the South Pacific," you learned that a variant (allele) of the TYRP1 gene leads to blond hair if an individual has the genotype TT. 68. The scientists who examined the genetics of the blond hair trait in the Solomon Islands determined the blond hair trait was associated with the short of chromosome 9. Describe the technique they used and why it was successful.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 03_7e 69. In peas, tall (T) is dominant to short (t). A homozygous tall plant is crossed with a short plant. The F1 plants are self-fertilized to produce the F2. Both tall and short plants appear in the F2. If the short F2 plants are selffertilized, what types of offspring and proportions will be produced?

70. For a particular plant, red flowers (A) are dominant over yellow flowers (a). An initial cross was made between a plant that was true-breeding for red flowers and another plant true-breeding for yellow flowers. F1 progeny, all having red flowers, were allowed to self- pollinate and form seeds, which were then planted to generate F2 progeny. Pollen from all the resulting F2 plants was pooled and used to fertilize true-breeding yellow plants. What proportion of the progeny resulting from this cross would be expected to have yellow flowers?

71. How did Sutton's chromosome theory of inheritance link Mendel's work with a more mechanistic understanding of heredity?

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 03_7e Answer Key 1. a 2. d 3. e 4. c 5. a 6. c 7. a 8. c 9. b 10. b 11. a 12. e 13. a 14. a 15. c 16. c 17. c 18. a 19. b 20. e 21. d 22. d 23. d 24. c 25. e 26. b Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 03_7e 27. c 28. b 29. d 30. a 31. b 32. b 33. b 34. a 35. c 36. e 37. b 38. d 39. b 40. d 41. c 42. a 43. d 44. c 45. c 46. d 47. c 48. a 49. a, b, c, d, e 50. b, d, e

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 03_7e 51. A testcross is used to determine genotypes of individuals with a dominant phenotype that may be heterozygous or homozygous for a dominant allele. The unknown genotypes are revealed by crossing the dominant individual to a "tester" that is known to be homozygous for the recessive allele in question.

A backcross is the mating of F1 progeny back to one of their parents or to an individual with a genotype identical to one of the parents. A backcross can also be a testcross if the original parent is homozygous for the recessive allele.

52. One-third of the tall F2 plants are TT, and 2/3 are Tt. The TT plants when self-fertilized will produce all tall plants.

The Tt plants when self-fertilized will produce 3 tall to 1 short. If all F2 plants produce an equal number of offspring, then 1/3 of the offspring will be from TT plants and will be tall. Three-quarters of the plants from Tt will be tall, but since only 2/3 of the F2 plants are Tt, this represents 2/3 × 3/4 = 6/12 = 1/2 of the total. Therefore, 1/3 + 1/2 = 5/6 will be tall. The remaining 1/6 of the offspring will be short.

53. Only the AABb genotype can produce Ab gametes, and the expected probability of Ab gametes produced will be 1/2. Because the parent has an equal probability (50%) of having either genotype (AaBB or AABb), the final probability of the parent producing an Ab gamete is (1/2) × (1/2) = 1/4 = 0.25 = 25%.

54. The easiest way would be to allow the plant to fertilize itself. If it produces only purple offspring, it must be PP. If it produces 3 purple : 1 white, it must be Pp. This is easy because you don't have to do any manipulations.

Another way would be to do a testcross of the unknown purple plant with a white tester (pp). If the plant is PP, then the offspring of this testcross would be all purple. If the unknown plant is Pp, then the offspring will be 1 purple : 1 white. This is harder because you have to set up the crosses.

55. (1) Progeny inherit genetic factors from both parents.

(2) Each individual possesses two factors (alleles) that control the appearance of each phenotypic trait.

(3) The two alleles in each individual separate (segregate) during gametogenesis and are randomly distributed with equal probability of being distributed into the gametes.

(4) From a cross between two true-breeding (homozygous) parents expressing different phenotypes for a given trait, traits that appeared unchanged in the F1 heterozygous offspring were dominant, and traits that disappeared in F1 heterozygous offspring were recessive.

56. For red eyes (recessive) to be expressed in the progeny (i.e., to segregate in the progeny), both parents must be heterozygotes (Rr). Rr × Rr ® 1 (RR), 2 (Rr), 1 (rr) = 3:1 (black eyes to red eyes). Note that even with only two progeny produced, you have determined the genotypes of the parents because you observe segregation of the recessive allele (red eyes).

57. For genotype AaBB, the probability of generating an AB gamete is 1/2. For genotype AABb, the probability of generating an AB gamete is 1/2. Again, because both genotypes have equal probabilities, the final probability of the parent generating an AB gamete is (1/2) × (1/2) + (1/2) × (1/2) = 2/4 = 0.5 = 50%.

58. Recessive because two copies of the allele are required for the blond trait to be displayed.

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 03_7e 59. Plugging the numbers into the chi-square equation: (63 – 60)2/60 + (17 – 20)2/20 = 0.15 + 0.45 = 0.6 (the chisquare statistic). Degrees of freedom = the number of classes (phenotypes) – 1 = 2 – 1 = 1. Looking at the chisquare table for one degree of freedom and using the standard critical value of 5%, we do not reject the hypothesis and thus conclude that the differences between observed and expected progeny numbers are a result of chance.

We can further conclude that both purple parents are heterozygous at the single gene locus controlling flower color (at least for purple and white).

60. a. Figure out the likelihood of a dominant phenotype for each of the four independent monohybrid cross components (e.g., aa × Aa = 1/2 Aa, 1/2 aa, etc.). Therefore, A_ = 1/2 (62 or 63); B_ = 3/4 (92 or 93); C_ = 1/2 (62 or 63); D_ = 1 (125). Thus, for phenotype ABCD: (1/2)(3/4)(1/2)(1) = 3/16 of the progeny, which represents 23 or 24 offspring. b. Simplify this quadric hybrid cross by breaking it down into four independent monohybrid cross components (e.g., aa × Aa = 1/2 Aa, 1/2 aa, etc.). This strategy can be used for any multihybrid cross involving genes that assort independently. Thus, for aaBBccDd: (1/2)(1/4)(1/2)(1/2) = 1/32. 1/32 × (125) = 3.9 = 3 or 4 offspring. c. For AaBbCcDd: (1/2)(1/2)(1/2)(1/2) = 1/16. 1/16 × (125) = 7 or 8 offspring.

61. The GWAS located the genetic variant associated with blond hair to the short arm of chromosome 9. However, that did not pinpoint the specific gene responsible. Scientists wanted to narrow down candidate gene(s) on that region of chromosome 9, and they looked to possible functional significance; that is, they searched for candidate genes whose predicted/known gene products might reasonably be predicted to have something to do with hair color. It was the comparison of the human gene sequence of TYRP1 to its homolog in mice that led the scientists to examine that gene further. In mice, a model organism TYRP1 gene was known to encode an enzyme involved in melanin production and pigmentation.

62. Three traits are segregating in these crosses—toe hair, ear wax, and sex; we assume that these three loci all assort independently. The cross is Hh Ww XX by Hh Ww XY. The probability of getting one hh ww XY and two H_ W_ XX in any order is the product of the probabilities: The probability of getting the hairy-toed, brittle ear-waxed boy (hh ww XY) is 1/4 × 1/4 × 1/2 = 1/32 and the probability of getting a smooth-toed, sticky ear-waxed girl (H_ W_ XX) is 3/4 × 3/4 × 1/2 = 9/32. The probability of getting a second smooth-toed, sticky ear-waxed girl (H_ W_ XX) is the same: 9/32. There are three possible orders for the three children (boy first, second, or third); thus the overall probability is 3 × 1/32 × 9/32 × 9/32 = 243/32,768.

63. The principle of segregation involves segregation (separation) of the alleles of a gene pair as the paired homologs on which they reside separate during anaphase I of meiosis. This principle can be demonstrated using a single pair of homologous chromosomes.

The principle of independent assortment involves the random assortment of alleles from different homologs into separate daughter cells during meiosis I. This principle can only be demonstrated using two or more pairs of homologous chromosomes.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 03_7e 64. Sex ratio segregates like any other trait. By forming a Punnett square, you see that an XX mother and XY father form offspring in a 50–50 ratio of males and females.

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 03_7e 65. (1) Principle of segregation: The separation of paired homologs distributes the alleles contained on each homolog to different gametes.

(2) Principle of independent assortment: Homologs are randomly assorted along the metaphase plate and are subsequently distributed (in complete sets) to gametes.

66. a. Using P for the leaf color locus and S for the stem type locus: PpSs × ppSs. b. A possible hypothesis is that each trait is controlled by independently assorting single gene pairs and that one allele in each pair exhibits complete dominance over the other. Class 1 = 3/8, class 2 = 3/8, class 3 = 1/8, class 4 = 1/8. c. (68 – 68)2/68 + (66 – 68)2/68 + (22 – 23)2/23 + (25 – 23)2/23 = 0.276. This number (the chi-square statistic) is small compared to the critical value; therefore you confidently accept your hypothesis.

Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 03_7e 67. If we use the allele symbols A (normal) and a (albino), the man's genotype must be Aa since he has normal pigmentation but one of his parents is aa. The woman has a brother with albinism, which means both her parents must be carriers (Aa). However, the woman (who does not have albinism) could have either an AA or Aa genotype. In the woman's case, the aa (albino) genotype must be excluded as a possibility. Therefore, the probability of the woman being AA is 1/3, and the probability that she is Aa is 2/3.

First, we should calculate the probability of the couple having a child with albinism each time a child is born. If the woman is Aa, then the mating is Aa × Aa, and P(albino) = 1/4. However, there is only a 2/3 chance that she is Aa. So overall for this mating, P(albino) = P(man is Aa) × P(woman is Aa) × 1/4 = 1 × 2/3 × 1/4 = 2/12 = 1/6.

Conversely, the probability that a child will be normal P(normal) = 1 – P(albino) = 5/6.

If the couple plans to have two children, there are four possible outcomes, which are given in the following table along with each probability. The overall probability of each outcome is calculated using the multiplication rule.

Child 1 Normal (5/6) Normal (5/6) Albino (1/6) Albino (1/6)

Child 2 Normal (5/6) Albino (1/6) Normal (5/6) Albino (1/6)

Probability 25/36 5/36 5/36 1/36

a. The probability that at least one child will have albinism corresponds to the last three outcomes of the table.

Since it can happen in any of three different ways, the three probabilities should be added to get the final probability. P(at least one albino) = 11/36 = 0.306. b. The probability of both children having normal pigmentation is given in the table. P(both are normal) = 25/36 = 0.694.

68. They used a genome-wide associated study (GWAS). By collecting saliva and hair samples from a significant portion of people on the islands and searching a statistically significant correlation between the blond hair trait and a genetic variant, they located the putative gene responsible for the trait to the short arm of chromosome 9.

Presumably people with black hair did not show the variation associated with blond hair.

Answers about why the approach was successful may vary. Points that can be brought up include the fact that hair colour was an obvious trait to follow, so categorizing saliva and hair samples from individuals with black versus blond hair should have been straightforward and unambiguous. The Solomon Islands had a large enough population with enough blond-haired individuals to produce a large enough group to conduct statistical analysis.

However, the Solomon Islands are geographically isolated and have a sufficiently isolated population to predict that a single genetic variation associated with blond hair might reasonably be found through a GWAS. 69. Self-fertilizing the short F2 progeny (tt) will only yield short progeny.

Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 03_7e 70. The F2 progeny will have the genotypes AA:Aa:aa in a 1:2:1 ratio (out of four plants, there will be one AA plant, two Aa plants, and one aa plant). When pooled, 50% of the gametes produced by these plants will contain only a alleles. Of the gametes produced by the true-breeding yellow plant, 100% will contain a alleles. This means that 50% of the F3 progeny will be homozygous recessive (aa) with yellow flowers.

71. Sutton documented the fact that each homologous pair of chromosomes consists of one maternal chromosome and one paternal chromosome. Showing that these pairs segregate independently into gametes in meiosis, he concluded that this process is the biological basis for Mendel's principles of heredity.

Copyright Macmillan Learning. Powered by Cognero.

Page 30

Name:

Class:

Date:

Chap 04_7e Indicate the answer choice that best completes the statement or answers the question.

1. The sex determination system used by Drosophila is called a. the X:A sex determination system. b. the ZZ–ZW sex determination system. c. the XX–XO sex determination system. d. the XX–XY sex determination system. e. genic sex detemination.

2. In humans, occasionally a baby is found that has the XY chromosomal karyotype but is phenotypically female.

Which of the following statements might be a CORRECT explanation for at least some of these unusual cases? a. A mutation has occurred in the SRY gene, making it inactive. b. An extra piece of autosomal chromosome 15 is probably present in the genome but is too small to be detected. c. A small piece of the X chromosome is missing but is too small to be detected. d. A small piece of the X chromosome is duplicated but too small to be detected. e. The ratio of number of X chromosomes to number of sets of chromosomes is incorrect.

3. A woman is phenotypically normal, but her father had the sex-linked recessive condition of red–green color blindness. If she has children with a man with normal vision, what is the probability that their first child will have normal vision and their second child will be color blind? a. 1/16 b. 3/8 c. 3/16 d. 3/6 e. 8/27 4. In species of birds, males are the homogametic sex and females the heterogametic sex. Which of the following statements is TRUE in this system of sex determination? a. The gender of the offspring is determined by the female parent. b. Male offspring have a ZW chromosome constitution. c. The gender of the offspring is determined by the male parent. d. Female offspring have a ZZ chromosome constitution. e. Female and male offspring have the same chromosome constitution.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 04_7e 5. If a female Drosophila that is heterozygous for a recessive X-linked mutation is crossed to a wild-type male, what proportion of female progeny will have the mutant phenotype? a. 100% b. 0% c. 33% d. 25% e. 50% 6. Women known to be heterozygous or carriers for the sex-linked recessive condition of hemophilia A were studied to determine the time required for their blood to clot. It was found that the time required for their blood to clot varied from individual to individual. The values obtained ranged from normal clotting at one extreme to clinical hemophilia at the other extreme. What is the MOST probable correct explanation for these findings? a. Some women had only one X chromosome and it is inactive. b. Some women had three copies of the X chromosome, which allowed them to make extra amounts of gene products for their X-linked genes. c. The women with normal clotting times probably had a mother with hemophilia while those with abnormal clotting times probably had fathers with hemophilia. d. Random X inactivation probably results in individuals with different proportions of cells in their bodies expressing the normal allele at the hemophilia locus. e. In women with abnormal clotting times, there was probably an interaction between an allele of a gene on the X chromosome and an allele of an autosomal gene that reduced the expression of the X-linked gene.

7. In Drosophila, flies that are XXY but have a normal diploid set of autosomes have been found. Such XXY chromosome constitutions have also been found in humans. The sex of such individuals is expected to be a. male in both Drosophila and humans. b. female in both Drosophila and humans. c. male in Drosophila but female in humans. d. female in Drosophila but male in humans. e. intersex in Drosophila and male in humans.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 04_7e 8. Women with Turner syndrome (XO) and normal women (XX) are clearly different phenotypically. In addition, the vast majority of XO conceptions abort before birth. However, both XO and XX women have one active X chromosome since the X in XO women remains active and one might expect that they would therefore have similar phenotypes. What is the MOST reasonable explanation for their different phenotypes? a. XO women do not have a copy of the SRY gene. b. Some genes remain active on the inactive X chromosome, and XX women will have two copies of these genes expressed and XO women only one copy. c. In XO women, the single X chromosome has no partner to pair with during mitosis so that each cell division is delayed by pairing problems with the single X not finding a pairing partner. d. XO women are missing a copy of the Xist gene so that they are forced to develop partway along the male pathway during embryogenesis. e. XO women have problems during development because mitosis is abnormal.

9. A eukaryotic diploid cell from an organism with the XX–XO sex determination system has two pairs of autosomes and one X chromosome, shown here.

What is the probability of a gamete from this individual that has the following genotype: alleles A and b, chromosome X? a. 1/2 b. 1/4 c. 1/6 d. 1/8 e. 1/16

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 04_7e 10. What is the role of the SRY gene in humans? a. It initiates the X inactivation process in females. b. It is located on the X chromosome and causes the X to pair with the Y chromosome during male meiosis. c. It is located on the Y chromosome and initiates the developmental pathway toward the male phenotype. d. It is located on an autosomal chromosome and represses expression of autosomal genes in order to balance their expression level with genes on the X chromosome. e. None of the answers is correct.

11. Species in which individuals have only male or only female reproductive structures are called a. hermaphrodites. b. diploids. c. dioecious. d. homogametic. e. monoecious.

12. Which of the following human genotypes is associated with Klinefelter syndrome? a. XXY b. XYY c. XXX d. XO e. XX 13. Which of the following chromosome constitutions would never lead to a viable human baby being born? a. XXX b. XYY c. XO (O = the absence of a second chromosome) d. YY e. XXY

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 04_7e 14. In Drosophila yellow body color is caused by a sex-linked recessive allele and brown eye color is an autosomal recessive trait. A phenotypically normal female fly, heterozygous for both genes, is crossed to a male with normal body color and brown eyes. What proportion of their offspring should have yellow body color and brown eyes? a. 1/16 b. 1/8 c. 1/4 d. 3/16 e. 3/8 15. Human males with XY chromosomes are _____ and produce two different kinds of gametes, whereas females with XX chromosomes are _____ and produce only one kind. a. homogametic; heterogametic b. dioecious; monoecious c. heterogametic; homogametic d. monoecious; dioecious e. monoecious; heterogametic 16. In which of the following individuals would you expect to find two Barr bodies in their somatic cells? a. XX b. XO c. XXY d. XXYY e. XXX 17. In which of the following phenotypic females do testes develop? a. XY with a deletion that removes the SRY gene b. XO c. XY with the X-linked recessive condition of androgen-insensitivity syndrome d. XX e. XXX

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 04_7e 18. A eukaryotic diploid cell from an organism with the XX–XO sex determination system has two pairs of autosomes and one X chromosome, shown here.

A diploid cell from this individual begins to go through meiosis. After the completion of meiosis I, it becomes two cells. One of these two cells now undergoes meiosis II. Which of the following is a possible normal combination of chromosomes in one of the subsequent two cells after the completion of meiosis II? a. one chromosome with the A allele, one chromosome with the B allele, and two X chromosomes b. one chromosome with the A allele, one chromosome with the a allele, one with B allele, one with b allele, and two X chromosomes c. one chromosome with the A allele, one chromosome with the a allele d. one chromosome with the b allele, one chromosome with the B allele, one X chromosome e. one chromosome with the a allele, one chromosome with the B allele, and two X chromosomes 19. Normal males (XY) and Klinefelter males (XXY) both possess only one active X chromosome. Nonetheless, there are clearly phenotypic differences between the two. What is the MOST reasonable explanation as to why such differences exist? a. The Y chromosome has higher gene expression levels when two X chromosomes are present compared to one X. b. The Y chromosome has lower gene expression levels when two X chromosomes are present compared to one X. c. Some genes remain active on inactive X chromosomes so XXY males would have higher expression levels for these genes compared to XY males. d. XXY males exhibit a higher rate of problems during mitotic divisions than XY males. e. XXY males don't have a copy of the SRY gene.

20. A female with androgen-insensitivity syndrome, a sex-linked recessive condition, has a. two X chromosomes, both carrying mutant alleles in the gene that makes the androgen receptor. b. a pair of ovaries that overproduce estrogen. c. an XXX chromosome constitution that causes her not to produce testosterone. d. a pair of internal testes that produce testosterone. e. an inactive SRY gene.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 04_7e 21. You discover a new mutation in Drosophila that causes an abnormal wing shape. When a male fly with this mutation is crossed to a homozygous normal female, the F1 are all wild-type phenotypically. The F2 flies consist of 1/2 phenotypically normal females, 1/4 phenotypically normal males, and 1/4 males with abnormal wings. What is the MOST likely explanation for the new mutant allele? a. It is X linked and recessive. b. It is X linked and dominant. c. It is autosomal and recessive. d. It is autosomal and dominant. e. It is Y linked.

22. A eukaryotic diploid cell from an organism with the ZZ–ZW sex determination system has two pairs of autosomes and a pair of sex chromosomes, Z and W, shown here.

From what type of individual is this cell? a. male b. female c. hermaphrodite d. monoecious e. intersex 23. Assume that an XX mouse is homozygous for a mutation in the Xist gene that causes the gene to not be expressed. What would be the expected consequences of such a situation? a. The mouse would have two Barr bodies instead of one. b. The mouse would develop as a male. c. The mouse would have a Y chromosome. d. Both X chromosomes would be active. e. Nondisjunction would occur at each meiotic event.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 04_7e 24. A Barr body is a(n) a. gene on the X chromosome that is responsible for female development. b. patch of cells that has a phenotype different from surrounding cells because of variable X inactivation. c. inactivated X chromosome, visible in the nucleus of a cell that is normally from a female mammal. d. extra X chromosome in a cell that is the result of nondisjunction. e. extra Y chromosome in a cell that is the result of nondisjunction.

25. A woman has normal vision although her maternal grandfather (her mother's father) had red–green color blindness, a sex-linked recessive trait. Her maternal grandmother and the woman's own father are assumed to not possess a copy of the mutant allele. The woman marries a man with normal vision, although his father was color blind. What is the probability that the first child of this couple will be color blind? a. 1/2 b. 1/4 c. 1/8 d. 1/16 e. 1/12 26. A eukaryotic diploid cell from an organism with the ZZ–ZW sex determination system has two pairs of autosomes and a pair of sex chromosomes, Z and W, shown here.

Assume A and B are dominant alleles. If this individual were crossed to an individual of genotype Aa Bb, what is the probability of an offspring being a female with an A__B__ genotype? a. 1/8 b. 1/16 c. 9/16 d. 9/32 e. 3/32

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 04_7e 27. What is the expected outcome for a human embryo with the XXXY chromosome constitution? a. It would likely develop into a female who will not respond to the hormone testosterone. b. It would likely develop into a sterile male with reduced testes. c. It will always abort early in development before birth. d. It would likely develop into a tall female who may be slightly cognitively impaired. e. It would likely develop into a fertile man with a completely normal male phenotype.

28. A male fruit fly carries the allele for yellow body color on his X chromosome. How would this male's genotype for this body color gene be described? a. holandric b. heterozygous c. homozygous d. homogametic e. hemizygous 29. Joan is phenotypically normal but had a child with the autosomal recessive disease cystic fibrosis (CF) from a previous marriage. Joan's father has hemophilia A, a sex-linked recessive condition where the blood fails to clot properly. Her father has survived due to recent treatment advances. Joan now intends to marry Bill, who is also phenotypically normal but who has a sister, Jill, with CF. Bill's parents are phenotypically normal, and there is no history of hemophilia A in his family. Assume that Joan and Bill do marry and have a child. What is the probability that this child will have CF but will not have hemophilia A? (Hint: This problem requires that you utilize concepts from Chapter 3 as well as Chapter 4.) a. 1/8 b. 1/12 c. 1/24 d. 3/32 e. 5/32 30. Species in which an individual organism has both male and female reproductive structures are called a. monoecious. b. haploid. c. diploid. d. dioecious. e. heterogametic.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 04_7e 31. The Lyon hypothesis helps us to understand which phenomenon in mammals? a. X-linked inheritance b. evolution of the Y chromosome c. dosage compensation between males and females d. development of male and female secondary sexual characteristics e. sex determination 32. In a germ-line cell from a female grasshopper (XX–XO sex determination system), when do the homologous X chromosomes segregate? a. during mitosis b. during meiosis I, anaphase c. during meiosis II, anaphase d. They do not segregate; gametes contain a copy of X and a copy of Y. e. just before meiosis begins 33. X and Y chromosomes are not homologs, but in meiosis they do pair and segregate in XY organisms to create 50% haploid gametes with an X chromosome and 50% haploid gametes with a Y chromosome. How is pairing achieved? a. Since all other homologous chromosomes pair, the remaining two chromosomes pair by default. b. Pairing proteins are capable of binding to different genes on the X and Y chromosomes, which allows them to pair. c. Pseudoautosomal regions that are homologous exist at the tips of both the X and Y chromosomes, and they allow pairing. d. They don't actually pair. Random segregation generally ensures the X and Y chromosomes separate. e. None of the statements is correct.

34. An XXY chromosome constitution produces _____ development in humans and _____ development in fruit flies. a. female; female b. male; male c. female; male d. male; female e. male, intersex

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 04_7e 35. Familial vitamin-D-resistant rickets is an X-linked dominant condition in humans. If a man is afflicted with this condition and his wife is normal, it is expected that among their children, all the daughters would be affected and all the sons would be normal. In families where the husband is affected and the wife is normal, this is almost always the outcome among their children when such families have been studied. Very rarely an unexpected result occurs in such families where a boy is born with the disorder. If the chromosomes of such unusual boys are examined, what might be expected to be found? a. Some of the boys are XYY. b. Some of the boys are XY but have lost the SRY gene from their Y chromosome. c. Some of the boys are YY. d. Some of the boys are XXY. e. Some of the boys are XXX. 36. What is the apparent purpose for X inactivation in humans and other mammals? a. It allows for the levels of expression of genes on the X chromosome to be similar in males and females. b. It allows for the levels of expression of genes on the autosomes to be similar to the levels of genes on the X chromosome. c. It suppresses the expression of genes on the Y chromosome in males. d. It reduces the amount of nondisjunction during meiosis in females. e. It enhances the level of pairing between the two X chromosomes during meiosis in females.

37. A woman is phenotypically normal, but her father had the sex-linked recessive condition of red–green color blindness. If she marries a man with normal vision, what is the probability that their two children will both have normal vision? a. 4/9 b. 1/16 c. 9/16 d. 3/8 e. 3/4 38. Which of the following statements about the sex-linked recessive trait of red–green color blindness in humans is FALSE? a. A phenotypically normal daughter can have a color-blind father. b. A phenotypically normal daughter can have a color-blind mother. c. A color-blind daughter can have a normal father. d. A color-blind daughter can have a phenotypically normal mother. e. A color-blind son could have a color-blind father.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 04_7e 39. In humans there is a genetic disorder that results from a dominant mutation present in a gene located in the pseudoautosomal region of the Y chromosome and on the X chromosome. Which of the following statements is CORRECT? a. All affected men marrying normal women will have no affected daughters. b. All affected women marrying normal men will have affected daughters and no affected sons. c. All affected men marrying normal women will have affected daughters, but all the sons will be normal. d. All affected women marrying normal men will have only normal sons and daughters. e. None of the statements is correct.

40. Human females with XY chromosomes and a phenotype that includes the absence of a uterus and ovaries and the presence of internal testes are likely to have which of the following mutations? a. a mutation in the SRY gene b. a mutation in the androgen receptor gene c. a deletion that removes much of the Y chromosome d. They likely do not carry a mutation but may have been premature babies. e. a defect in Xist gene involved in X inactivation 41. James is a 42-year-old man with hemophilia, a sex-linked recessive condition. His daughter, Susan, who has normal blood clotting, is married to Fred, who also has hemophilia. Susan and Fred are expecting their first child, and an ultrasound shows that the fetus is male. What is the approximate probability that their new son will have hemophilia? a. 1 b. 3/4 c. 2/3 d. 1/2 e. 1/4 42. During the evolution of the human Y chromosome, all of the following are assumed to have occured EXCEPT a. the original chromosome was an autosome that eventually evolved into the Y chromosome. b. one of the early events in the evolution of the Y chromosome was the acquisition or evolution of a gene somewhat similar to the current human SRY gene. c. many of the genes on the original ancestral chromosome suffered mutations and became inactive during the evolution of the Y chromosome. d. many of the genes on the early X chromosome that were responsible for critical cellular functions got moved to the evolving Y chromosome. e. several palindromic regions evolved or were acquired and are now present on the Y chromosome.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 04_7e 43. In humans, occasionally a baby is found that has the XX chromosomal karyotype but is phenotypically male.

Which of the following statements might be a CORRECT explanation for at least some of these unusual cases? a. A mutation has occurred in the SRY gene, making it inactive. b. An extra pierce of autosomal chromosome 15 is probably present in the genome but is too small to be detected. c. A small piece of autosomal chromosome 15 is missing but is too small to be detected. d. A piece of chromosomal material containing an active SRY gene is found attached to one of the X chromosomes. e. The ratio of number of X chromosomes to number of sets of chromosomes is incorrect.

44. What is the sex chromosome constitution of a male duck-billed platypus? a. XX b. XY c. XO d. ZZ e. XXXXXYYYYY 45. A male has Klinefelter syndrome with the chromosome constitution of XXXY. How many Barr bodies should be found in the nuclei of cells of this individual? a. 0 b. 1 c. 2 d. 3 e. 4 46. In mice, what else is required besides the absence of Sry (the mouse homology of SRY) to bring about ovary and female feature development? a. a third X chromosome b. pro-Mullerian hormone c. anti-testosterone hormone d. chicken ovalbumin upstream promoter transcription factor (COUP-TFII) e. All of these options are correct.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 04_7e 47. A eukaryotic diploid cell from an organism with the ZZ–ZW sex determination system has two pairs of autosomes and a pair of sex chromosomes, Z and W, shown here.

What is the probability of a gamete from this individual that has the following genotype: alleles A and b, chromosome Z? a. 1/2 b. 1/4 c. 1/6 d. 1/8 e. 1/16 48. In a germ-line cell from a human male that is dividing, when do the X and Y chromosomes segregate? a. during mitosis b. during meiosis I, anaphase c. during meiosis II, anaphase d. They do not segregate; gametes contain a copy of X and a copy of Y. e. just before meiosis begins 49. If a male bird that is heterozygous for a recessive Z-linked mutation is crossed to a wild-type female, what proportion of the progeny will be mutant males? a. 0% b. 100% c. 75% d. 50% e. 25%

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 04_7e 50. Red–green color blindness is X-linked recessive. A woman with normal color vision has a father who is color blind. The woman has a child with a man with normal color vision. Which phenotype is NOT expected? a. a color-blind female b. a color-blind male c. a noncolor-blind female d. a noncolor-blind male e. a color-blind male or a color-blind female 51. A new mutation in Drosophila is found called peach, which causes a peach-like body color. A peach male is crossed to a homozygous wild-type female, and the offspring consists of peach females and normal males. If a heterozygous peach female is crossed to a normal male, the offspring consists of 1/4 peach females, 1/4 normal females, 1/4 peach males, and 1/4 normal males. What is the MOST likely mode of inheritance for peach? a. It is X linked and recessive. b. It is X linked and dominant. c. It is autosomal and recessive. d. It is autosomal and dominant. e. It is Y linked.

52. How is dosage compensation accomplished in Drosophila? a. One of the two X chromosomes is inactivated in females. b. Drosophila has no obvious mechanism of dosage compensation. c. The Y chromosome has an increased level of gene expression. d. The activity of genes on the male X chromosome is doubled but not that of genes on the X chromosome of females. e. The activity of genes on half of the autosomes is doubled and the activity of genes on the X chromosomes of both males and females is decreased.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 04_7e 53. How many Barr bodies (condensed X chromosomes) would you predict in this boy's cells?

a. one per somatic cell b. two per somatic cell c. three per somatic cell d. none e. either two or three depending on the tissue type 54. A eukaryotic diploid cell from an organism with the ZZ–ZW sex determination system has two pairs of autosomes and a pair of sex chromosomes, Z and W, shown here.

A diploid cell from this individual begins to go through meiosis. After the completion of meiosis I, it becomes two cells. One of these two cells now undergoes meiosis II. Which of the following is a possible normal combination of chromosomes in one of the subsequent two cells after the completion of meiosis II? a. one chromosome with the A allele, one with the B allele, one Z, one W b. one chromosome with the A allele, one with the a allele, one with the B allele, one with the b allele, one Z, one W c. a pair of chromosomes with A alleles, a pair of chromosomes with B alleles, a pair of Z chromosomes d. one chromosome with an a allele, one chromosome with a B allele, one W e. one chromosome with the B allele, one with the b allele, one Z Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 04_7e 55. With the XX–XO sex determination system, generally: a. female offspring have one X chromosome, and it is inherited from their father. b. male offspring have one X chromosome, and it is inherited from their mother. c. male offspring have one X chromosome, and it is inherited from their father. d. female offspring have one X chromosome, and it is inherited from their mother. e. male offspring have two X chromosomes, one inherited from each parent.

56. In which of the following organisms is gender/sex determined by the temperature during embryonic development? a. humans b. mice c. fruit flies d. many snakes and birds e. many turtles and alligators 57. Which of the following statements about X inactivation in mammalian females is FALSE? a. Females that are heterozygous for an X-linked gene have patches of cells that express one allele and patches of cells that express the other. b. Some genes on the inactive X continue to be expressed after the chromosome is inactivated. c. X inactivation is random as to which X is inactivated and takes place early in embryonic development. d. Inactivation is thought to be initiated by expression of the Xist gene on the X that will remain active. e. Once an X chromosome first becomes inactivated in a cell, that same X will remain inactivated in somatic cells that are descendants of this cell.

58. Miniature wings in Drosophila are due to an X-linked allele (Xm) that is recessive to the wild-type allele for normal long wings (X+). Sepia eyes are produced by an autosomal allele (se), which is recessive to the wildtype allele for red eyes (se+). A female that is homozygous for normal wings and has sepia eyes is crossed with a male that has miniature wings and is homozygous for red eyes. The F1 offspring are intercrossed to produce the F2 generation. What proportion of the F2 females is expected to have miniature wings and sepia eyes? a. 1/4 b. 3/16 c. 1/2 d. 1/8 e. 0

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 04_7e 59. Which statement BEST summarizes our current understanding of the origin of the Y chromosome? a. The Y chromosome is thought to have arisen spontaneously in an ancestor of mammals millions of years ago. b. The Y chromosome is thought to have arisen as a fusion of two autosomes. c. The Y chromosome is thought to have arisen as a broken fragment of the X chromosome. d. The Y chromosome is thought to have been derived along with the X chromosome from a pair of autosomes. e. The Y chromosome arose from the fusion of the X chromosome with one of the autosomes.

60. A eukaryotic diploid cell from an organism with the XX–XO sex determination system has two pairs of autosomes and one X chromosome, shown here.

From what type of individual is this cell? a. male b. female c. hermaphrodite d. monoecious e. intersex 61. A phenotypically normal man marries a phenotypically normal woman whose father was color blind. They have a color-blind daughter with Turner syndrome (XO). What is the BEST explanation for this result? a. nondisjunction at meiosis II in the mother b. nondisjunction at meiosis I in the mother c. nondisjunction at meiosis I or meiosis II in the father d. mitotic nondisjunction in the first division of the zygote e. two sperm fertilizing the same egg simultaneously

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 04_7e 62. What level of male hormones would you expect to find in a person with androgen-insensitivity syndrome?

63. Explain the development of female external anatomy in individuals with androgen-insensitivity syndrome.

64. Red–green color blindness is an X-linked recessive condition. Juliet has a bit of difficulty passing the red– green color distinction test when she tries to get her driver's license. Her husband is not color blind, and neither is her son, Henry, nor her daughter, Roxanne. Roxanne has a son who is color blind. What is Juliet's genotype for the color-blindness allele? How would you explain her partial color blindness?

65. Predict the sexual phenotype of a person who is XY but whose Y chromosome carries a deletion of the SRY gene. Explain your prediction.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 04_7e 66. Red–green color blindness is X-linked recessive. A woman with normal color vision has a father who is red– green color blind. The woman has four sons, none of whom are colorblind. In this family there are no instances of chromosome loss or gain such as occurs due to nondisjunction in meiosis. Following are three explanations for why none of the sons are color blind. For each, state if color blindness is possible or not possible; then give the reason for your choice. (a) Explanation 1: None of the sons are color blind because the mother does not carry the color-blindness allele. (b) Explanation 2: None of the sons are color blind because none of them inherited the color-blindness allele from the mother. (c) Explanation 3: None of the sons are color blind because the mother inactivated the X chromosome with the recessive color-blindness allele, and that is the one each son inherited.

67. Compare and contrast the patterns of inheritance expected for Y-linked and X-linked recessive inheritance in humans.

68. While doing summer fieldwork on a remote Indonesian island, you discover a new genus of lizard closely related to Komodo dragons. You attempt to discover what sex determination system it uses by performing a series of controlled crosses on the island, using an isolated pair of lizards. Initially, all your crosses yield only males (in significant numbers). As fall begins and you prepare to leave the island, you find that your last cross yielded only females (in significant numbers). Suggest a mode of sex determination that explains these data.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 04_7e 69. A man and a woman are trying to have children but are unsuccessful. The man's autosomes appear normal, but his sex chromosomes, shown in the following diagram, are not. The diagram also shows a normal male's sex chromosomes for reference. Can you tell if the mutation came from the man's mother or the man's father?

Explain how you can tell.

70. In a few sentences, describe what effect removal of the Xist gene would have. Include a description of what Xist encodes and why a mutation would have this effect.

71. List three dosage compensation strategies for equalizing the amount of sex chromosome gene products.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 04_7e 72. A man and a woman are trying to have children but are unsuccessful. The man's autosomes appear normal, but his sex chromosomes, shown in the following diagram, are not. The diagram also shows a normal male's sex chromosomes for reference. In two to three sentences, explain the man's situation, including the type of chromosome mutation he carries, the specific regions of specific chromosomes involved, and why he is male.

73. You are trying to develop a new species of newt as an experimental model system. You know that in other species of newt, green (G) is dominant to brown (g) skin color and is determined by a sex-linked gene. You cross brown males to green females and see that in the F1 all the males are green and all the females are brown. Which is the heterogametic sex in your species of newt?

74. Describe the inheritance of each possible combination of your answer to question 65, including the parent and meiotic stage in which an unusual event occurred.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 04_7e 75. You cross a female rat with pink toe pads (T) and pointy ears (Xe) to a male rat with black toe pads (t) and round ears (XE). The t and e alleles are both recessive, and the ear-shaped gene is X linked, whereas the toe pad color gene is autosomal. The F1 progeny all have pink toe pads. What is the genotype of parental generation? What is the genotype of the F1 progeny? If the F1 are crossed to produce F2 progeny, what proportion of the F2 will be black-padded, pointy-eared males?

76. Marsupials, like cats, achieve dosage compensation by X inactivation. You are working in a lab that has discovered a mutation on the X chromosome in marsupials in the same gene that causes the tortoiseshell fur color phenotype in cats. You cross an X+Y black-furred male with an XOXO orange-furred female. You expect that the X+XO female progeny will have tortoiseshell fur (like cats). Surprisingly, you find that all the females (n = 25) have solid orange fur. Offer a hypothesis to explain these results and describe a genetic test to support your hypothesis.

77. Suppose that an apparently female athlete fails a gender test and is not allowed to compete in her event. The gender test is based on examination of cheek cells for the presence of one or more Barr bodies. Later, it is discovered that the athlete has androgen-insensitivity syndrome. Evaluate the decision of the officials to exclude the athlete from athletic competition as a female in light of your knowledge of androgen-insensitivity syndrome. Do you think the athlete should be allowed to compete as a female?

78. Explain the genders of human and diploid Drosophila XXY individuals.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 04_7e 79. The boy in questions 65 and 66 has an X-linked recessive condition that is not seen in either parent. With this additional information, what can you conclude about the allelic composition of his parents and how he got this condition?

80. This boy's parents' X chromosomes can be distinguished by polymorphism differences. One of his mother's X chromosomes has the polymorphism allele 7 and the other has the allele 14; his father's X has the allele 5.

Assuming each parent contributed at least one sex chromosome, what polymorphism combinations are possible in the boy's X chromosomes?

81. Explain how dosage balance is achieved between X-linked genes and autosomal genes in mammals.

82. Explain the genders of human and diploid Drosophila XO individuals.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 04_7e 83. Compare and contrast the terms "sex determination" and "sexual development."

84. Suppose that an apparently female athlete fails a gender test and is not allowed to compete in her event. The gender test is based on examination of cheek cells for the presence of one or more Barr bodies. Later, it is discovered that the athlete has androgen-insensitivity syndrome. What is the chromosome constitution of a person with androgen-insensitivity syndrome?

85. List three different mechanisms for generating sexes in dioecious species.

86. Calvin Bridges crossed white-eyed females to red-eyed males and found rare red-eyed males and white-eyed females in the progeny. Explain how he used the Drosophila sex determination system and nondisjunction to demonstrate that the gene for red/white eye color is on the X chromosome.

87. How might an XY person with a deletion of the SRY gene be distinguished from a person with androgeninsensitivity syndrome?

88. Characterize the gonads of a person with androgen-insensitivity syndrome as testes, ovaries, or intersex.

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 04_7e 89. Suppose that an apparently female athlete fails a gender test and is not allowed to compete in her event. The gender test is based on examination of cheek cells for the presence of one or more Barr bodies. Later, it is discovered that the athlete has androgen-insensitivity syndrome. Explain why the athlete failed the gender test.

What did the technician see in the test, and how was it interpreted?

90. Some organisms have multiple X and Y chromosomes and even different numbers of X and Y chromosomes.

You have discovered such a species. Females have eight X chromosomes, whereas males have four X and two Y. Describe the X and Y constitution of the gametes produced by this species—both male and female— that allows these chromosome numbers to be stably maintained.

91. List five different sex determination systems and a representative organism for each.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 04_7e Answer Key 1. a 2. a 3. c 4. a 5. b 6. d 7. d 8. b 9. d 10. c 11. c 12. a 13. d 14. b 15. c 16. e 17. c 18. c 19. c 20. d 21. a 22. b 23. d 24. c 25. c 26. d Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 04_7e 27. b 28. e 29. a 30. a 31. c 32. b 33. c 34. d 35. d 36. a 37. c 38. c 39. e 40. b 41. d 42. d 43. d 44. e 45. c 46. d 47. d 48. b 49. a 50. a 51. b 52. d 53. a 54. d Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 04_7e 55. b 56. e 57. d 58. e 59. d 60. a 61. c 62. Based on the presence of normal testes, one would expect to find normal levels of male hormones in these individuals.

63. These individuals lack the androgen receptor on the cells of their bodies. Therefore, the cells cannot respond to the presence of male hormones, and the external anatomy develops according to the "default" pathway, which is female.

64. Juliet is heterozygous, so every diploid cell in her eyes contains both X chromosomes, one with the defective red– green color vision allele and one with the normal color vision allele. She is not completely color blind because some of her cells are inactivating the X chromosome with the recessive, defective red–green color vision allele and expressing the normal color vision allele, allowing her to see color. She has some difficulty with color vision because some of her cells are inactivating the X with the normal red–green color vision allele and expressing the red–green color-blind phenotype. Her eyes are actually a mosaic of cells with two different phenotypes.

65. The SRY gene is the primary determinant of maleness in humans. If it is deleted, the gonads will not be induced to differentiate as testes, and the individual would likely follow the female developmental pathway. (Actually, this condition exists as Swyer syndrome. It turns out that the gonads do not develop into functional ovaries, indicating that genes other than the SRY also have roles in sexual differentiation. However, these individuals develop as apparently normal females until the lack of female hormones from the ovaries causes puberty not to be induced without hormone therapy.)

66. (a) Not possible. If there are no aneuploidies, the mother must have inherited the X with the recessive colorblindness allele from her father. (b) Likely. She is heterozygous, so each son has a 50% chance of inheriting the X chromosome with the dominant allele for normal color vision. (c) Not possible. X inactivation is reversed when an X chromosome is passed to offspring.

67. Y-linked inheritance is easy to recognize because the trait is passed from a father to all of his sons and none of his daughters and continues to pass from fathers to sons in every generation. The trait does not affect females. An Xlinked recessive trait is more common in males but can affect females as well. A son inherits the trait from his mother, who is typically a phenotypically normal carrier. Often, the trait will also be present in the mother's father or brothers or other male relatives. An affected female normally occurs from an affected father and a carrier mother.

Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 04_7e 68. The crosses yielded all males or all females from the same parents. Male and female progeny were correlated with climatic conditions (summer versus fall). Environmental sex determination that is dependent on temperature is a likely explanation.

69. He inherited the translocation chromosome from his father because his mother could not have carried the SRYcontaining chromosome.

70. Xist encodes an RNA that coats an X chromosome to inactivate it. Null mutation would eliminate Xist RNA, and no X chromosome would ever be inactivated. Lack of Xist RNA would lead to improper dosage compensation in a female but have no effect on dosage compensation in a male.

71. (1) Inactivation of one sex chromosome in the homogametic sex (2) Halving the activity of genes on both sex chromosomes in the homogametic sex (3) Increasing the activity of genes on the sex chromosome in the heterogametic sex 72. He has an X with a translocation, meaning part of one chromosome has been moved to another. The translocated part is from Y and carries SRY, which determines maleness. He is missing the rest of Y, including the genes required for male fertility.

73. Because the F1 females have the recessive brown phenotype, they must be hemizygous (i.e., they inherited a brown allele from their father and no allele from their mother). Therefore, the females of this species are the heterogametic sex. We use the ZZ–ZW nomenclature for species with heterogametic females. Therefore, your F1 females and males are Zg W and ZGZg , respectively.

74.

Polymorphisms seen Inherited from father Inherited from mother Nondisjunction in

Boy’s genotype X5 X7Y 5,7 X5 Y X7 father, meiosis I

X5 X14Y 5,14 X5 Y X14 father, meiosis I

X7 X14Y 7,14 Y X7 X14 mother, meiosis I

X7 X7Y X14 X14Y 7 14 Y Y X7 X7 X14 X14 mother, meiosis mother, II meiosis II

Nondisjunction in meiosis I: Homologs fail to segregate (e.g., X5 and Y or X7 and X14).

Nondisjunction in meiosis II: Sister chromatids fail to separate (e.g., X7 and X7, or X14 and X14, or both).

75. The parental generation was T/T; Xe/Xe and t/t; XE/Y. The F1 were T/t; XEXe and T/t; XeY. Black-padded, pointy-eared males are t/t and Xe/Xe. One-quarter of the F2 progeny will be t/t, 1/4 will be XeY. Therefore, 1/16 of the progeny will be black-padded, pointy-eared males.

Copyright Macmillan Learning. Powered by Cognero.

Page 30

Name:

Class:

Date:

Chap 04_7e 76. It appears that the X+ chromosome from the male was inactivated in every female offspring. So perhaps in marsupials, unlike in cats, X inactivation is not random. Instead, in marsupials only the paternal X chromosome is inactivated. If this is true (and, in fact, it is), then it may be tested genetically. Crossing an orange-furred XOY male to a X+X+ black-furred female should produce only black-furred female progeny even though their genotype is X+XO, the same as the orange-furred female progeny from the first cross.

77. The decision to ban the athlete very likely was unfair. Other than the presence of testes and the absence of internal female reproductive structures, she is anatomically female. Although she has male hormones in her system, her muscle and other cells do not respond to the presence of the male hormones. Very likely, she does not gain a competitive advantage over other females by the presence of the Y chromosome, testes, and male hormones.

78. An XXY human is male. The SRY on the Y chromosome determines maleness in spite of the two X chromosomes.

An XXY Drosophila is female. Drosophila is normally diploid, so there are two of each autosome. If there are two X chromosomes, the X:A ratio is 1.0, which is a female.

79. His mother must be a carrier of the allele for the condition because she passed it to the son without showing it herself. The father could not have been a carrier because he did not show the condition. If the son has the recessive condition, the recessive allele must be on both his X chromosomes. Therefore, the son’s two X chromosomes are the same and were inherited from his mother. The nondisjunction occurred in the mother’s germline, in meiosis II, when sister chromatids failed to separate.

80. (1) 7, 7 (fingerprint would show just 7) (2) 14, 14 (fingerprint would show 14) (3) 7, 14 (4) 7, 5 (5) 5, 14 81. In both males and females only one X chromosome is fully active, potentially creating an imbalance in gene expression between genes on the X chromosome and autosomes. The imbalance is prevented by upregulation of genes on the X chromosome.

82. An XO human is female. In the absence of SRY from a Y chromosome, the person will develop as a female, in spite of having only one X chromosome. An XO Drosophila is male. Drosophila is normally diploid, so there are two of each autosome. If there is a single X chromosome, the X:A ratio is 1/2, or 0.5, which is a male.

Copyright Macmillan Learning. Powered by Cognero.

Page 31

Name:

Class:

Date:

Chap 04_7e 83. Sex determination is a mechanism by which a program of sexual development is established. It may depend on the presence or absence of sex chromosome(s), a sex chromosome: autosome ratio, genotypes at several non-sexchromosome loci, and/or on environmental influences. Sex determination is "upstream" or early in starting the program of sexual development of an individual. Atypical composition of sex-determining genes can result in a nonwild-type sex determination mechanism.

Sexual development, that is, the development of sexual characteristics, involves many genes that are "downstream" of the sex determination mechanism. In a wild-type situation, those genes are responsible for the program of sexual development triggered by the sex determination mechanism; for example, XX humans develop as females and XY humans develop as males. However, alterations in one or more of the genes that bring about sexual development can result in a sexual phenotype was not anticipated by the sex-determination mechanism.

84. The chromosome constitution is XY. 85. (1) Chromosomal sex determination (XX–XO, XX–XY, or ZZ–ZW) (2) Genetic sex determination (3) Environmental sex determination 86. Bridges looked at the chromosomes of the rare flies under the microscope and showed that the rare white-eyed females had two X chromosomes and one Y, and the rare red-eyed males had only one sex chromosome. The rare white-eyed females were XwXwY; they were white-eyed because they contained only recessive w alleles, and female because their X:A ratio = 2:2 = 1.0. The rare red-eyed males were X+O; they were red-eyed because of the dominant + allele, and male because their X:A ratio was 1:2 = 0.5. This demonstrated that genes that confer phenotypes were located on X chromosomes.

87. A person with androgen-insensitivity syndrome would have testes and a level of male hormones characteristic of males. An XY person with a deletion of SRY would not have testes or high levels of male hormones.

88. Because of the presence of a normal Y chromosome, the gonads develop as testes.

89. Because the athlete is XY, the technician would have seen no Barr bodies in her cells. The absence of Barr bodies is normally characteristic of males, so the test was interpreted to indicate that the athlete is genetically male.

90. Females must produce one type of gamete, with four X chromosomes, whereas males produce two types of gametes, with four X chromosomes or with two Y chromosomes. Fertilization using a four-X male gamete produces an eight-X female, whereas fertilization with a two-Y gamete produces a male.

91. (1) XX–XO; grasshoppers (2) XX–XY; humans (3) ZZ–ZW; birds (4) X:A; Drosophila (5) Environmental mollusks; many turtles, crocodiles, and alligators

Copyright Macmillan Learning. Powered by Cognero.

Page 32

Name:

Class:

Date:

Chap 05_7e Indicate the answer choice that best completes the statement or answers the question.

1. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A pure-breeding pink plant is mated with a pure-breeding white plant. All of the F1 are blue. When the blue F1 plants are selfed, the F2 plants occur in the ratio 9 blue : 3 pink : 4 white. The genotype that produces white is ww. The presence of one W allele allows pink or blue color to occur. The alleles at the hypostatic locus are B (blue) and b (pink).

What is the genotype of the F1 plants? a. bb WW b. bb Ww c. Bb Ww d. Bb ww e. BB ww 2. Leber hereditary optic neuropathy (LHON) is a human disease that exhibits cytoplasmic inheritance. It is characterized by rapid loss of vision in both eyes, resulting from the death of cells in the optic nerve. A teenager loses vision in both eyes and is later diagnosed with LHON. How did this individual MOST likely inherit the mutant DNA responsible for this condition? a. a nuclear gene from the father b. a nuclear gene from the mother c. a mitochondrial gene from the father d. a mitochondrial gene from the mother e. Any of these answers is possible.

3. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > cch > ch > c. Indicate the phenotypic ratios expected of rabbits with the cross Cch × cc. a. 1 full color : 1 chinchilla b. 1 full color : 1 Himalayan c. 1 chinchilla : 1 Himalayan d. 3 full color : 1 chinchilla e. 2 full color :1 Himalayan : 1 albino

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 05_7e 4. You are studying body color in an African spider and have found that it is controlled by a single gene with four alleles: B (brown), br (red), bg (green), and by (yellow). B is dominant to all the other alleles, and by is recessive to all the other alleles. The bg allele is dominant to by but recessive to br. You cross a pure-breeding brown spider with a pure-breeding green spider. Predict the phenotype of the progeny. a. half brown, half green b. three-fourths brown, one-fourth green c. all brown d. all green e. all yellow 5. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > cch > ch > c. Indicate the phenotypic ratios expected of rabbits with the cross Ccch × Cch . a. 1 full color : 1 chinchilla b. 1 full color : 1 Himalayan c. 1 chinchilla : 1 Himalayan d. 3 full color : 1 chinchilla e. 2 full color : 1 Himalayan : 1 albino 6. A mother with blood type B has a child with blood type O. Give all possible blood types for the father of this child. a. O b. B, AB c. A, AB d. A, B, O e. A, B, AB, O 7. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A pure-breeding pink plant is mated with a pure-breeding white plant. All of the F1 are blue. When the blue F1 plants are selfed, the F2 plants occur in the ratio 9 blue : 3 pink : 4 white. The genotype that produces white is ww. The presence of one W allele allows pink or blue color to occur. The alleles at the hypostatic locus are B (blue) and b (pink).

What is the genotype of the pink parent? a. bb WW b. bb Ww c. Bb Ww d. Bb ww e. BB ww Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 05_7e 8. In purple people eaters, purple is dominant to white. A true-breeding white mutant is mated with a different true-breeding white mutant. All of the F1 are purple. When the purple F1 offspring mate with each other, their offspring occur in the ratio of 9 purple : 7 white. Which phenomenon explains the purple F1 offspring? a. recessive epistasis b. dominant epistasis c. complementation d. mutation e. suppression 9. In the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but the B and b alleles can only be expressed if the animal has an mm genotype at a third gene locus.

Animals that are M_ are yellow regardless of which allele from the B locus is present. A mating between a solid yellow-tailed male yawncat and a solid brown-tailed female yawncat produces 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. What is the MOST likely genotype of the male parent? a. MM BB RR b. MM Bb RR c. Mm Bb RR d. Mm BB Rr e. Mm Bb Rr 10. Suppose that the "fabulous" phenotype is controlled by two genes, A and B, as shown in the following diagram. One copy of allele A produces enough enzyme 1 to convert "plain" to "smashing." Allele a produces no enzyme 1. One copy of allele B produces enough enzyme 2 to convert "smashing" to "fabulous." Allele b produces no enzyme 2. The A and B genes are both autosomal and assort independently.

What will be the phenotype(s) of the F1 offspring of a true-breeding "fabulous" father and a true-breeding "plain" mother (aa bb)? a. all "plain" b. all "smashing" c. all "fabulous" d. "plain" females and "fabulous" males e. "fabulous" females and "smashing" males

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 05_7e 11. In the endangered African watchamakallit, the offspring of a true-breeding black parent and a true-breeding white parent are all gray. When the gray offspring are crossed among themselves, their offspring occur in a ratio of 1 black : 2 gray : 1 white. Upon close examination of the coats, each hair of a gray animal is gray.

What is the mode of inheritance? a. one gene pair with black dominant to white b. one gene pair with codominance c. one gene pair with incomplete dominance d. two gene pairs with recessive epistasis e. two gene pairs with duplicate genes 12. In the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but the B and b alleles can only be expressed if the animal has an mm genotype at a third gene locus.

Animals that are M_ are yellow regardless of which allele from the B locus is present. A mating between a solid yellow-tailed male yawncat and a solid brown-tailed female yawncat produces 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. What is the probability of the next offspring from these same two parents having a spotted brown tail? a. 1/2 b. 3/16 c. 1/4 d. 1/16 e. 9/16 13. Male-limited precocious puberty results from a rare autosomal allele (P) that is dominant over the allele for normal puberty (p) and is expressed only in males. A male and female that both went through normal puberty have two sons. The first son undergoes precocious puberty, but the second undergoes normal puberty. What is the genotype of the father? a. PP b. Pp c. pp d. PP or Pp e. Pp or pp

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 05_7e 14. Suppose that extra fingers and toes are caused by a recessive trait, but it appears in only 60% of homozygous recessive individuals. Two heterozygotes conceive a child. What is the probability that this child will have extra fingers and toes? a. 0.05 b. 0.10 c. 0.15 d. 0.25 e. 0.33 15. Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink, 3/16 black, and 1/16 white. What kind of gene interaction is this? a. recessive epistasis b. dominant epistasis c. duplicate recessive epistasis d. duplicate dominant epistasis e. dominant and recessive epistasis 16. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. Two heterozygous birds are mated. What fraction of the total offspring is expected to exhibit cock feathering? a. 0 b. 1/8 c. 1/4 d. 1/2 e. 3/4 17. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > cch > ch > c. Indicate the phenotypic ratios expected of rabbits with the cross Cch × ch c. a. 1 full color : 1 chinchilla b. 1 full color : 1 Himalayan c. 1 chinchilla : 1 Himalayan d. 3 full color : 1 chinchilla e. 2 full color : 1 Himalayan : 1 albino

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 05_7e 18. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A pure-breeding pink plant is mated with a pure-breeding white plant. All of the F1 are blue. When the blue F1 plants are selfed, the F2 plants occur in the ratio 9 blue : 3 pink : 4 white. The genotype that produces white is ww. The presence of one W allele allows pink or blue color to occur. The alleles at the hypostatic locus are B (blue) and b (pink).

What is the genotype of the white parent? a. bb WW b. bb Ww c. Bb Ww d. Bb ww e. BB ww 19. Polydactyly is the condition of having extra fingers or toes. Some polydactylous persons possess extra fingers or toes that are fully functional, whereas others possess only a small tag of extra skin. This is an example of: a. variable expressivity. b. complete dominance. c. independent assortment. d. complementation. e. cytoplasmic inheritance.

20. What phenomenon describes a genetic trait that is expressed more strongly or earlier in development with each generation? a. epigenetics b. maternally determined progeny phenotypes c. epistasis d. anticipation e. genomic imprinting 21. Which of the following is a characteristic exhibited by cytoplasmically inherited traits? a. present in both males and females b. usually inherited from one parent, typically the maternal parent c. reciprocal crosses that give different results d. extensive phenotypic variation exhibited, even within a single family e. All of the answers are correct.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 05_7e 22. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. Two heterozygous birds are mated. What fraction of the female offspring is expected to exhibit cock feathering? a. 0 b. 1/8 c. 1/4 d. 1/2 e. 3/4 23. You are studying body color in an African spider and have found that it is controlled by a single gene with four alleles: B (brown), br (red), bg (green), and by (yellow). B is dominant to all the other alleles, and by is recessive to all the other alleles. The bg allele is dominant to by but recessive to br. You cross a pure-breeding brown spider with a pure-breeding green spider to produce F1 progeny. You then cross one of these F1 progeny to a br/by spider to produce F2 progeny. What would be the expected percentage of red progeny in the F2? a. 0% b. 25% c. 33% d. 50% e. 100% 24. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A pure-breeding pink plant is mated with a pure-breeding white plant. All of the F1 are blue. When the blue F1 plants are selfed, the F2 plants occur in the ratio 9 blue : 3 pink : 4 white. How many gene pairs control the flower color phenotype? a. 0 b. 1 c. 2 d. 3 e. 4

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 05_7e 25. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. Two heterozygous birds are mated. What fraction of the male offspring is expected to exhibit cock feathering? a. 0 b. 1/8 c. 1/4 d. 1/2 e. 3/4 26. You are studying a gene that controls ossicone (horn) length in giraffes. The wild-type long-ossicone allele (L) is dominant to the mutant short-ossicone (l) allele. However, the L allele is only 60% penetrant. You cross two heterozygous giraffes. What proportion of offspring would you expect to exhibit the long-ossicone phenotype?

Assume the penetrance of L applies equally to both homozygotes and heterozygotes. a. 0.40 b. 0.45 c. 0.55 d. 0.60 e. 0.75 27. The SRY gene is located on the Y chromosome. This single gene encodes a protein called a transcription factor that binds to DNA and stimulates the transcription of other genes that lead to the development of male sex characteristics, including physical, biochemical, and behavioral phenotypes. What concept in genetics BEST describes this example? a. dominance b. discontinuous characteristic c. polygenic characteristic d. phenocopy e. pleiotropy 28. In a certain species of plant, flowers occur in three colors: blue, pink, and white. A pure-breeding pink plant is mated with a pure-breeding white plant. All of the F1 are blue. When the blue F1 plants are selfed, the F2 plants occur in the ratio 9 blue : 3 pink : 4 white. What is the name for this type of interaction? a. recessive epistasis b. dominant epistasis c. duplicate recessive epistasis d. duplicate dominant epistasis e. dominant and recessive epistasis

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 05_7e 29. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > cch > ch > c. Indicate the phenotypic ratios expected of rabbits with the cross cchch × ch c. a. 1 full color : 1 chinchilla b. 1 full color : 1 Himalayan c. 1 chinchilla : 1 Himalayan d. 3 full color : 1 chinchilla e. 2 full color : 1 Himalayan : 1 albino 30. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Walnut crossed with single produces 1 walnut, 1 rose, 1 pea, and 1 single offspring. a. RR PP × rr pp b. RR Pp × rr pp c. Rr PP × rr pp d. Rr Pp × rr pp e. Rr pp × rr pp 31. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Pea crossed with single produces 1 single offspring. a. rr PP × rr pp b. RR Pp × rr pp c. Rr PP × rr pp d. Rr Pp × rr pp e. rr Pp × rr pp

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 05_7e 32. A mother with blood type A has a child with blood type A. Give all possible blood types for the father of this child. a. O b. B, AB c. A, AB d. A, B, O e. A, B, AB, O 33. The himalayan allele in rabbits produces dark fur at the extremities of the body—on the nose, ears, and feet.

The dark pigment develops, however, only when a rabbit is reared at a temperature of 25°C or lower; if a Himalayan rabbit is reared at 30°C, no dark patches develop. What does this exemplify? a. dominance b. discontinuous characteristic c. genetic imprinting d. phenocopy e. temperature-sensitive allele 34. Suppose that the "fabulous" phenotype is controlled by two genes, A and B, as shown in the following diagram. One copy of allele A produces enough enzyme 1 to convert "plain" to "smashing." Allele a produces no enzyme 1. One copy of allele B produces enough enzyme 2 to convert "smashing" to "fabulous." Allele b produces no enzyme 2. The A and B genes are both autosomal and assort independently.

What will be the phenotype(s) of the F2 offspring of a true-breeding "fabulous" father and a true-breeding "plain" mother (aa bb)? a. 9 "fabulous" : 7 "plain" b. 13 "fabulous" : 3 "plain" c. 9 "fabulous" : 3 "smashing" : 4 "plain" d. 12 "plain" : 3 "fabulous" : 1 "smashing" e. 15 "fabulous" : 1 "smashing"

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 05_7e 35. Interactions among the human ABO blood group alleles involve _____ and _____. a. codominance; complete dominance b. codominance; incomplete dominance c. complete dominance; incomplete dominance d. epistasis; complementation e. continuous variation; environmental variation 36. In the Mexican hairless breed of dog, the hairless condition is produced by the heterozygous genotype (Hh).

Normal dogs are homozygous recessive (hh). Puppies homozygous for the H allele are usually born dead. If the average litter size at weaning (after nursing) is six puppies in matings between hairless dogs, what would be the average expected number (not a ratio) of hairless and normal offspring at weaning from matings between hairless and normal dogs? a. six normal dogs and two hairless dogs b. two normal dogs and one hairless dog c. five normal dogs and one hairless dog d. three normal dogs and three hairless dogs e. four normal dogs and four hairless dogs 37. Coat color is determined by two loci, A and B, in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink, 3/16 black, and 1/16 white.

What is the genotype of the black progeny? a. A_ B_ b. A_ BB c. aa B_ d. aa bb e. A_ B_ and A_ bb 38. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Rose crossed with pea produces 20 walnut offspring. a. RR pp × rr PP b. Rr pp × rr Pp c. Rr pp × rr PP d. RR pp × rr Pp e. Rr pp × Rr Pp Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 05_7e 39. The very rare Bombay blood phenotype in humans (first discovered in Bombay, India) results in blood type O because of the lack of both the A and B antigens in individuals who are of hh genotype. This genotype results in blood type O regardless of the genotype at the unlinked I locus. If two parents are both of IA IB Hh genotype (type AB blood), what is the probability that their first child will also have type blood B? a. 3/8 b. 2/3 c. 1/16 d. 1/4 e. 3/16 40. Explain the term "maternal effect" with respect to genetics and biochemistry. a. Dummy Answer 41. The bicoid mutation (bcd– ) in fruit flies is inherited as a maternal effect recessive allele. What is the expected ratio of phenotypes in the offspring of a cross between a bcd+/bcd– female and a bcd+/bcd– male? a. 1 normal : 1 mutant b. 3 normal : 1 mutant c. 3 mutant : 1 normal d. all normal e. all mutant 42. In C. elegans, the gene fem-1 influences male development. Homozygous fem-1 mutants develop as females if their mothers are also homozygous fem-1 mutants. However, homozygous fem-1 mutants with heterozygous fem-1 mothers [fem-1/ fem-1(+)] show some male development. Suggest what might be happening. a. Dummy Answer 43. Which organelle in an animal cell, in addition to the nucleus, contains genes? a. lysosome b. ribosome c. mitochondrion d. Golgi body e. vesicle 44. Crossing two yellow mice results in 2/3 yellow offspring and 1/3 nonyellow offspring. What percentage of offspring would you expect to be nonyellow if you crossed two nonyellow mice? a. 25% b. 33% c. 66% d. 75% e. 100% Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 05_7e 45. Achondroplasia is a common cause of dwarfism in humans. All individuals with achondroplasia are thought to be heterozygous at the locus that controls this trait. When two individuals with achondroplasia mate, the offspring occur in a ratio of 2 achondroplasia : 1 normal. What is the MOST likely explanation for these observations? a. Achondroplasia is incompletely dominant to the normal condition. b. Achondroplasia is codominant to the normal condition. c. The allele that causes achondroplasia is a dominant lethal allele. d. The allele that causes achondroplasia is a recessive lethal allele. e. The allele that causes achondroplasia is a late-onset lethal allele.

46. The phenomenon in which a gene's expression is determined by its parental origin is called: a. sex-influenced. b. sex-limited. c. genomic imprinting. d. maternal effect. e. paternal effect.

47. The red kernel color in wheat is caused by the presence of at least one dominant allele in each of two independently segregating gene pairs (e.g., R_B_). Wheat plants with rrbb genotypes have white kernels, and plants with genotypes R_bb and rrB_ have yellow kernels. You cross a plant homozygous for red kernels (RR BB) with a plant homozygous for white kernels. What are the relative proportions of the phenotypic classes expected in the F2 progeny after selfing the F1 progeny? a. 9 red; 3 white; 4 yellow b. 9 red; 6 yellow; 1 white c. 9 yellow; 4 red; 1 white d. 9 red; 3 yellow; 4 white e. 12 red; 3 yellow; 1 white 48. In order to determine if mutations from different organisms in the same species that exhibit the same phenotype are allelic, which test would you perform? a. testcross b. epistasis test c. complementation test d. allelic series test e. biochemical test

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 05_7e 49. Hair color is determined in Labrador retrievers by alleles at the B and E loci. A dominant allele B encodes black pigment, whereas a recessive allele b encodes brown pigment. Alleles at a second locus affect the deposition of the pigment in the shaft of the hair; dominant allele E allows dark pigment (black or brown) to be deposited, whereas recessive allele e prevents the deposition of dark pigment, causing the hair to be yellow. A black female Labrador retriever was mated with a yellow male. Half of the puppies were black, and half were yellow. If the genotype of the black female parent was Bb Ee, then what was the genotype of the other parent? a. bb ee b. bb EE c. Bb ee d. BB ee e. BB EE 50. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. Two birds heterozygous for cock feathering are mated. What are the phenotypes of the parents? a. male with cock feathering, female with hen feathering b. male with hen feathering, female with cock feathering c. male with cock feathering, female with cock feathering d. male with hen feathering, female with hen feathering e. cannot be determined from the information given 51. Deafness is often inherited in humans as an autosomal recessive trait. Assume that this is the case here. Two severely deaf people meet and marry. They have four children, and all of them have normal hearing. What is the MOST reasonable explanation for this outcome? a. The mutant allele for deafness shows variable expressivity, and the normal children may have some hearing loss that is difficult to detect. b. Complementation has occurred in the children, indicating that the deafness mutations in the parents involved different genes. c. Deafness in this family shows a genetic maternal effect with the condition being determined by the genotype of the mother. d. Deafness in this family is most likely caused by an epigenetic change such as the addition of methyl groups to the DNA. e. The mutant allele for deafness shows anticipation and becomes more weakly expressed in succeeding generations.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 05_7e 52. In humans, blood types A and B are codominant to each other, and each is dominant to O. What blood types are possible among the offspring of a couple of blood types AB and A? a. A, B, AB, and O b. A, B, and AB only c. A and B only d. A, B, and O only e. A and AB only 53. Two loci control body color in beetles. In a cross between a black beetle and a white beetle you obtain a ratio of 9 black to 7 white beetles. What kind of gene interaction is this? a. recessive epistasis b. dominant epistasis c. duplicate recessive epistasis d. duplicate dominant epistasis e. dominant and recessive epistasis 54. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Rose crossed with walnut produces 6 walnut, 6 rose, 2 single, and 2 pea offspring. a. RR pp × rr PP b. Rr pp × rr Pp c. Rr pp × rr PP d. RR pp × rr Pp e. Rr pp × Rr Pp 55. A mother with blood type AB has a child with blood type B. Give all possible blood types for the father of this child. a. O b. B, AB c. A, AB d. A, B, O e. A, B, AB, O

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 05_7e 56. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Rose crossed with single produces 10 rose and 11 single offspring. a. RR PP × rr pp b. RR Pp × rr pp c. Rr PP × rr pp d. Rr Pp × rr pp e. Rr pp × rr pp 57. A deletion of a small region on the long arm of chromosome 15 causes a developmental disorder in children called Prader–Willi syndrome when the deletion is inherited from the father. However, the deletion of this same region of chromosome 15 can also be inherited from the mother, but this inheritance results in a completely different set of symptoms, called Angelman syndrome. What type of genetic phenomenon does this represent? a. sex-influenced b. genomic imprinting c. cytoplasmic inheritance d. maternal effect e. paternal effect 58. Hair color is determined in Labrador retrievers by alleles at the B and E loci. A dominant allele B encodes black pigment, whereas a recessive allele b encodes brown pigment. Alleles at a second locus affect the deposition of the pigment in the shaft of the hair; dominant allele E allows dark pigment (black or brown) to be deposited, whereas recessive allele e prevents the deposition of dark pigment, causing the hair to be yellow.

What type of gene interaction does this represent? a. recessive epistasis b. dominant epistasis c. duplicate recessive epistasis d. duplicate dominant epistasis e. dominant and recessive epistasis 59. What is a "paternal effect" in terms of genetics? How do you think it might be brought about biochemically? a. Dummy Answer

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 05_7e 60. A mother with blood type A has a child with blood type AB. Give all possible blood types for the father of this child. a. O b. B, AB c. A, AB d. A, B, O e. A, B, AB, O 61. In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > cch > ch > c. Indicate the phenotypic ratios expected of rabbits with the cross Cc × ch c. a. 1 full color : 1 chinchilla b. 1 full color : 1 Himalayan c. 1 chinchilla : 1 Himalayan d. 3 full color : 1 chinchilla e. 2 full color : 1 Himalayan : 1 albino 62. Male-limited precocious puberty results from a rare autosomal allele (P) that is dominant over the allele for normal puberty (p) and is expressed only in males. A male and female that both went through normal puberty have two sons. The first son undergoes precocious puberty, but the second undergoes normal puberty. What is the genotype of the mother? a. PP b. Pp c. pp d. PP or Pp e. Pp or pp

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 05_7e 63. In the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but the B and b alleles can only be expressed if the animal has an mm genotype at a third gene locus.

Animals that are M_ are yellow regardless of which allele from the B locus is present. A mating between a solid yellow-tailed male yawncat and a solid brown-tailed female yawncat produces 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. What is the MOST likely genotype of the female parent? a. mm bb rr b. Mm bb rr c. mm Bb rr d. mm bb Rr e. mm Bb Rr 64. Coat color is determined by two loci, A and B, in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink, 3/16 black, and 1/16 white.

What is the genotype of the pink progeny? a. A_ B_ b. A_ bb c. aa B_ d. aa bb e. A_ B_ and A_ bb 65. Coat color is determined by two loci, A and B, in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink, 3/16 black, and 1/16 white.

What is the genotype of the white progeny? a. A_ B_ b. A_ bb c. aa B_ d. aa bb e. A_ B_ and A_ bb 66. In domestic chickens, some males display a plumage pattern called cock feathering. Other males and all females display a pattern called hen feathering. Cock feathering is an autosomal recessive trait that is exhibited in males only. What type of inheritance is exhibited by this trait? a. sex-linked b. sex-limited c. sex-influenced d. autosomal recessive e. autosomal dominant

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 05_7e 67. A 24-year-old woman with myotonic dystrophy, a genetic disorder caused by a dominant mutant allele, has significant muscle weakness and wasting. She has a 1-year-old son with very poor muscle tone, severe weakness, and significant developmental delay. Her affected 55-year-old father has minor cataracts but no muscle weakness. Which of the following terms BEST describes this situation? a. codominance b. complementation c. multiple alleles d. compound heterozygote e. anticipation 68. The presence of a beard on some goats is determined by an autosomal gene that is dominant in males and recessive in females. Heterozygous males are bearded, while heterozygous females are beardless. What type of inheritance is exhibited by this trait? a. sex-linked b. sex-limited c. sex-influenced d. autosomal recessive e. autosomal dominant 69. In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Rose crossed with single produces 31 rose offspring. a. RR PP × rr pp b. RR pp × rr pp c. Rr PP × rr pp d. Rr Pp × rr pp e. Rr pp × rr pp

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 05_7e 70. A man is heterozygous for a rare trait that is assumed to be caused by a dominant autosomal mutant allele. He marries a woman who is phenotypically normal. They have a phenotypically normal daughter who marries a phenotypically normal man. Unexpectedly their first child, a girl, has the same abnormal trait as her grandfather. Which of the following BEST explains this situation? a. The trait is really due a mutation in a cytoplasmic gene and can be only inherited from the mother. b. This is a sex-influenced trait and is expressed more often in females than in males. c. This is an example of genomic imprinting, and the mutant allele is only expressed if it is inherited from the father. d. The mutant allele shows incomplete dominance when heterozygous with the normal allele. e. The mutant allele shows reduced penetrance, and the normal daughter carries the mutant allele.

71. You are studying a gene that controls ossicone (horn) length in giraffes. The wild-type long-ossicone allele (L) is dominant to the mutant short-ossicone (l) allele. However, the L allele is only 60% penetrant. You cross two heterozygous giraffes. What proportion of offspring would you expect to exhibit the short-ossicone phenotype? Assume the penetrance of L applies equally to both homozygotes and heterozygotes. a. 0.25 b. 0.40 c. 0.45 d. 0.55 e. 0.60 72. The R locus determines flower color in a new plant species. Plants that are genotype RR have red flowers, and plants that are rr have white flowers. However, Rr plants have pink flowers. What type of inheritance does this demonstrate for flower color in these plants? a. complete dominance b. incomplete dominance c. codominance d. complementation e. lethal alleles 73. The cross of CcDd × CcDd results in a 12:3:1 phenotypic ratio. What is the expected phenotypic ratio when a CcDd × ccdd testcross is made involving the same two genes? (They are assorting independently.) a. 1:1:1:1 b. 2:1:1 c. 3:1 d. 9:3:3:1 e. 1:1

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 05_7e 74. Multifactorial traits are influenced by _____ and _____. a. dominance; codominance b. epistasis; pleiotropy c. age; sex d. genetic imprinting; reduced penetrance e. multiple genes; environment 75. Assume that a polygenic character in Drosophila is produced through the activity of five genes, each with two alleles. Two individuals who are both heterozygous for all five genes are crossed. How many different genotypes are expected in the offspring? a. 9 b. 27 c. 81 d. 243 e. 729 76. A mother of blood type A gives birth to a child with blood type O. Which of the following could NOT be the blood type of the father? a. A b. B c. O d. AB e. Any of these is a possible blood type of the father.

77. You are studying body color in an African spider and have found that it is controlled by a single gene with four alleles: B (brown), br (red), bg (green), and by (yellow). B is dominant to all the other alleles, and by is recessive to all the other alleles. The bg allele is dominant to by but recessive to br. You cross a pure-breeding brown spider with a pure-breeding green spider to produce F1 progeny. You then cross one of these F1 progeny to a br/by spider to produce F2 progeny. What would be the expected percentage of yellow progeny in the F2? a. 0% b. 25% c. 33% d. 50% e. 100%

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 05_7e 78. You are studying body color in an African spider and have found that it is controlled by a single gene with four alleles: B (brown), br (red), bg (green), and by (yellow). B is dominant to all the other alleles, and by is recessive to all the other alleles. The bg allele is dominant to by but recessive to br. You cross a pure-breeding brown spider with a pure-breeding green spider. Predict the genotype of the progeny. a. B/bg b. Br/bg c. br/by d. by /bg e. B/by Indicate one or more answer choices that best complete the statement or answer the question.

79. Huntington's disease tends to strike earlier and lead to a more rapid progression of symptoms as it moves from generation to generation. Also, it tends to strike earlier and progress more rapidly when it is passed from the father to his children rather than from the mother. Which genetic phenomenon (or phenomena) is/are likely operating here? (Select all that apply.) a. incomplete penetrance b. sex-limited inheritance c. genetic anticipation d. parental imprinting e. epistasis 80. Explain the differences between incomplete dominance and continuous variation.

81. A homozygous strain of corn that produces yellow kernels is crossed with another homozygous strain that produces purple kernels. When the F1 are interbred, 280 of the F2 are yellow and 70 are purple. a. If kernel color is controlled by a single gene pair with yellow dominant to purple, what would be the expected ratio of yellow to purple in the F2? b. Do the observed data differ significantly from that expected in (a)? Explain your answer. c. Provide an alternative explanation for the inheritance of kernel color and evaluate it by comparing the observed data to that expected from your alternative hypothesis.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 05_7e 82. How do incomplete penetrance and variable expressivity differ?

83. With regard to the ear shape phenotypes described in the previous question, how could you test the relative importance of environmental and genetic factors?

84. You are studying a coat color gene (B, brown) in Mexican bats. You have isolated a recessive allele (b) that causes yellow coat color, but you suspect that the phenotype may be sensitive to environmental conditions. To test your hypothesis, you examine the segregation ratio of phenotypes in F1 progeny from a cross between two heterozygotes. You do this once at normal laboratory temperatures (28°C) and once at temperatures closer to their native habitat (34°C) and record the following data:

28°C 34°C

Brown 153 170

Yellow 47 30

a. What ratio do you expect in each experiment if temperature does not affect the phenotype? b. What test can you use to determine if the ratio you observed is significantly different from the expected ratio? c. Using that statistical test, is either observed ratio more different from the expected ratio than one would expect from chance alone? If so, suggest a biological explanation.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 05_7e 85. Assume that a human gene is imprinted. Normal individuals have one active copy of the gene, the copy inherited from their fathers. The copy inherited from their mothers is normally inactive.

In one large family a mutation in this gene occurs, and this mutation results in part of the gene being deleted.

The defective gene is never active regardless of the parent from which it is inherited. Individuals who carry the mutation are heterozygous and also have a normal copy of the gene whose activity depends upon which parent it was inherited from. Individuals who have no active copy of the gene have several medical problems but live long enough to have children of their own. This family includes two phenotypically normal parents who have two affected children, a son and a daughter. Which parent, father or mother, did the affected children inherit this mutation from? Why?

86. Two mice of the same species have different ear shapes. You find that one mouse, having ears of normal shape, was caught in a field in Kenya. The other mouse, with curled ears, was caught in the frozen tundra of Greenland. You have determined that both mice have identical genotypes at the gene loci controlling ear shape. How would you explain the differences in ear shape?

87. Discuss the difference between "cytoplasmic inheritance" and "genetic maternal effect."

88. What is a dominant epistatic gene?

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 05_7e 89. A yeast geneticist isolates two different haploid mutant yeast strains, strain A and strain B, which cannot grow unless the amino acid leucine is added to the growth media. Wild-type yeast strains can make their own leucine and do not require that it be added to the growth media. The geneticist discovers that each mutant yeast strain contains a single recessive mutation that leads to the observed leucine-requiring phenotype. When she crosses the two mutant strains together, she observes that the resulting diploid can grow without leucine added to the growth media. Explain the allelic relationship between the mutations in these two strains.

90. A geneticist is examining a culture of fruit flies and discovers a single female with strange spots on her legs. The new mutation is named melanotic. When a female melanotic fly is crossed with a normal male, the following progeny are produced: 123 normal females, 125 melanotic females, and 124 normal males. In subsequent crosses with normal males, melanotic females are frequently obtained but never any melanotic males. Provide a possible explanation for the inheritance of the melanotic mutation. (Hint: The cross produces twice as many female progeny as male progeny.)

91. Cloning is a procedure by which exact genetic duplicates are made. Using cloning techniques, you have produced 10 cloned cows. However, each of the individual calves have fur color that looks very different from the others. Explain how this may have occurred.

92. In some plant species, a single pair of alleles is involved in both flower color and stem color. For example, a plant with red flowers may also have red stems, whereas white-flowered varieties of the same species have green stems. How would you explain this observation?

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 05_7e 93. You observe continuous variation in tail length in a wild population of rats. How would you determine whether this variation is an example of variable expressivity or polygenic inheritance?

94. How do incomplete dominance and codominance differ?

95. In the model organism C. elegans, a nematode worm, sex is determined through a negative regulatory pathway in which gene activity in one step of the pathway acts to inhibit the activity of the gene in the step immediately downstream. Wildtype C. elegans with an XX genetic makeup develop with female soma whereas wildtype XO C. elegans develop with male soma. The gene fem-1 promotes male somatic development; XX and XO fem-1 null homozygotes develop female soma. ("Null" refers to an allele that completely eliminates gene activity.) The gene tra-1 promotes female somatic development; XX and XO tra1 null homozygotes develop male soma. Double fem-1 null and tra-1 null XX and XO mutants develop male soma. Which gene is epistatic and why?

96. Explain how a phenotype like the height in a tree species can be due to the influence of both genes and environment.

97. Describe the differences between dominance, codominance, and incomplete dominance.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 05_7e 98. List at least four phenomena that can alter expected Mendelian phenotypic ratios in genetic crosses.

99. How does epistasis differ from Mendel's principle of dominance?

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 05_7e Answer Key 1. c 2. d 3. b 4. c 5. d 6. d 7. a 8. c 9. e 10. c 11. c 12. d 13. c 14. c 15. b 16. b 17. b 18. e 19. a 20. d 21. e 22. a 23. b 24. c 25. c 26. b Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 05_7e 27. e 28. a 29. c 30. d 31. e 32. e 33. e 34. c 35. a 36. e 37. c 38. a 39. e 40. a 41. d 42. a 43. c 44. e 45. d 46. c 47. b 48. c 49. d 50. d 51. b 52. b 53. c 54. e Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 05_7e 55. e 56. e 57. b 58. a 59. a 60. b 61. e 62. b 63. d 64. e 65. d 66. b 67. e 68. c 69. b 70. e 71. d 72. b 73. b 74. e 75. d 76. d 77. a 78. a 79. c, d

Copyright Macmillan Learning. Powered by Cognero.

Page 30

Name:

Class:

Date:

Chap 05_7e 80. (1) In incomplete dominance, there will be three distinct phenotypes because the phenotype of a heterozygote is intermediate in appearance between the phenotypes of the two homozygotes. Incomplete dominance involves a single gene locus.

(2) Continuous variation refers to phenotypic variation exhibited by quantitative traits that are overlapping and distributed from one extreme to another. The continuous variation of quantitative traits is usually controlled by several genes whose alleles have an additive effect on the phenotype.

81. a. Because there are only two progeny classes, the simplest explanation is monohybrid inheritance with an expected ratio of 3 yellow : 1 purple. b. A chi-square test of the observed numbers using an expected 3:1 ratio suggests rejection of this hypothesis (χ2 = 4.7, 0.025 < p < 0.05). c. So, the next hypothesis to test is dihybrid inheritance. However, the progeny clearly don't segregate in a classic Mendelian dihybrid ratio (9:3:3:1)—there are only two phenotypic classes. Therefore, some kind of epistasis is likely. There are three epistatic ratios with two phenotypic classes to test: 9:7, 15:1, and 13:3. Dividing each phenotypic class by 16 suggests that the 13:3 ratio is the closest. The 13:3 ratio is standard for the kind of epistasis called "dominant and recessive" interaction. In this particular kind of epistasis, only two F2 phenotypes are generated, because a dominant genotype (e.g., A_) present at one locus and the recessive genotype at the other locus (bb) produce identical phenotypes in a 13:3 ratio (e.g., A_ B_, A_ bb, and aa bb produce one phenotype, and aa B_ produces another phenotype). To further substantiate that this is the correct ratio, a chi-square test can be done. In fact, among the alternatives only the 13:3 ratio and accompanying genetic hypothesis should not be rejected (χ2 = 0.35 with 0.9 < p < 0.5).

82. (1) If some individuals in a population don't express a trait, even though they have the corresponding genotype, the trait is said to exhibit incomplete penetrance in that population. When using the term penetrance, therefore, think of populations. Polydactyly (extra fingers and toes) exhibits incomplete penetrance in human populations.

(2) A trait exhibiting variable expressivity is not expressed at the same degree among all individuals expressing it.

Male pattern baldness in humans is an example of a trait that exhibits variable expressivity.

Note: For incomplete penetrance, not everyone with the genotype will express the phenotype. For variable expressivity, everyone with the genotype expresses the phenotype to some degree. Of course, some traits may (and often do) exhibit both incomplete penetrance and variable expressivity.

83. Develop true-breeding strains of mice for normal and curled ears in their original habitats. Then rear and observe one pair of each (one control pair and one test pair) true-breeding strain in each habitat in Kenya and Greenland, mating them with only their fellow Greenland or Kenyan siblings and raising all offspring under the same conditions as originally present for the native species (e.g., the original location outside). If the phenotypes of the experimental and control offspring reared in Kenya are all the same (normal) and the phenotypes of both sets of mice in Greenland are all the same (curled), then the two-ear phenotypes are caused by environment. For example, they may result from temperature-sensitive alleles. Note that strict temperature sensitivity could be tested under laboratory conditions without the need to transport mice to different countries. If the phenotypes of the two sets of mice at any one geographic location are not all the same, then there is a genetic component involved in ear shape.

Note that there may be both genetic and environmental components.

Copyright Macmillan Learning. Powered by Cognero.

Page 31

Name:

Class:

Date:

Chap 05_7e 84. a. This is a simple monohybrid cross with brown dominant to yellow, so expect 3 brown : 1 yellow. b. The chi-square test. c. The chi-square test for the treatment at 34°C yields a value of χ2 = 10.67, indicating a significant difference from the expected ratio of 3:1. This suggests that elevated temperatures reduce the penetrance of the yellow phenotype.

The chi-square test for the treatment at 28°C yields a value of 0.08, indicating that the data fit the expected ratio of 3:1.

85. The mutation must have been inherited from the father since he normally provides the active copy of the gene to his children. If a child receives the inactive mutant copy of the gene from the father, the child will be affected since he or she will not have a functional copy of the gene. The mother's copy of the gene is normally turned off when inherited by her children. It would not matter if this copy was mutated or not when coming from the mother since it is not active anyway. None of her children would be affected if the mutant gene came from the mother.

86. Because both mice have the same genotype at the relevant loci controlling ear shape, there is most likely an effect of environment on phenotype. Because Kenya and Greenland have quite different climates (but also different food sources, humidity, sunlight intensities, etc.), it is possible that the very different temperature ranges within each region resulted in differential expression of identical genotypes for ear shape in each mouse—for example, differential expression of temperature-sensitive allele(s) involved in ear development. However, the phenotypic differences could also be a result of differences in diet, light conditions, exposure to chemicals, nutrition, or a range of other nongenetic factors.

87. (1) In cytoplasmic inheritance, the genes controlling a given trait are inherited exclusively from the mother (through cytoplasmic organelles such as mitochondria) and can be expressed in both male and female progeny.

(2) In genetic maternal effect, each individual's phenotype is determined by the genotype of the mother. Typically, the offspring's phenotype is determined by factors loaded into the oocyte and encoded by the mother's genome.

So while genes related to the trait are inherited from both parents (not so for cytoplasmic inheritance), in a given generation, phenotype is determined exclusively by the mother's, not the offspring's, genotype.

88. In dominant epistasis a dominant allele, if present, determines the phenotype of a given trait regardless of which alleles are present at the hypostatic locus.

89. The mutations in strains A and B are NOT allelic because complementation was observed. Strain A contains a mutation at gene A, which is recessive (a), and strain B contains a mutation at a separate genetic locus, gene B, which is also recessive. Strain A contains a wild-type B gene, and strain B contains a wild-type A gene. These wild-type genes complement the corresponding mutant alleles in the diploid.

Copyright Macmillan Learning. Powered by Cognero.

Page 32

Name:

Class:

Date:

Chap 05_7e 90. These observations can be explained by a single X-linked locus with two segregating alleles. The skewed sex ratio (2 female : 1 male) in the F2 suggests a recessive lethal allele on the X chromosome that kills males that carry the lethal allele in one copy on their one X chromosome. The phenotype of the female parent also suggests that the allele is dominant for the melanotic trait. We will represent the mutant allele as M and the normal allele as +.

91. A given phenotype arises from a genotype that develops within a particular environment. How the phenotype develops is determined by the effects of genes and environmental factors, and the balance between these influences varies from character to character. Since we are told that the calves are genetically identical, there must be environmental variation that explains the phenotypic differences. Even within the "constant" environment of a cow's womb, there is environmental variation!

92. This phenomenon, called pleiotropy, is the condition where a single gene affects multiple, apparently unrelated, phenotypic traits. In many other cases of pleiotropy, a single gene affects more than two phenotypic traits. For example, a mutant white-eye gene in Drosophila (fruit fly) also affects the structure and color of internal organs, causes reduced fertility, and decreases life expectancy. Another example involves sickle-cell anemia in humans (caused by a single nucleotide change in a hemoglobin gene), which has adverse effects on different organs and tissues.

93. Take male and female rats from each phenotypic extreme (shortest and longest tails). Interbreed short with short and long with long under controlled laboratory conditions for several generations. If this is polygenic inheritance, then you will be able to develop different homozygous lines for short and long tails. But if, after several generations, each line continues to produce progeny classes exhibiting significant variance in tail length, you could assume variable expressivity is the primary basis for the variation because the genotypes for each extreme line are (theoretically) homozygous and isogenic. Therefore, variances in tail length observed within each line cannot be the result of variable polygenic genotypes.

94. Incompletely dominant traits show an intermediate phenotype in the heterozygote, while codominant traits show both phenotypes in the heterozygote (e.g., AB alleles of blood type). Incompletely dominant traits often result from one allele being inactive and the other allele being active. Heterozygotes thus produce half as much gene product as individuals homozygous for the normal allele and may have an intermediate phenotype compared to the two homozygotes. With codominance, both alleles are usually active and produce functional gene products.

95. tra-1 is epistatic because null mutations in the tra-1 gene mask the effects of null mutations in the fem-1 gene. That is, regardless of the state of fem-1 activity (on or off), tra-1 null mutants develop male somas.

Copyright Macmillan Learning. Powered by Cognero.

Page 33

Name:

Class:

Date:

Chap 05_7e 96. The height reached by a tree at maturity is a phenotype that is strongly influenced by environmental factors, such as the availability of water, sunlight, and nutrients. Nevertheless, the tree's genotype still imposes some limits on its height: an oak tree will never grow to be 300 meters tall no matter how much sunlight, water, and fertilizer are provided.

97. (1) Dominance is the condition in which one allele of a gene pair completely masks or inhibits phenotypic expression of the other allele.

(2) Codominance is the condition in which the complete expression of both alleles of a given gene pair is observed in heterozygotes; that is, the expression of neither allele influences the expression of the other.

(3) Incomplete (or partial) dominance is the condition in which one allele is active and normally expressed and the other allele is often inactive and doesn't produce a functional product. Heterozygotes exhibit phenotypes that are intermediate between those of the two homozygotes.

98. (1) Linkage (2) Epistasis (3) X-linked genes (4) Lethal recessive alleles (5) Environmental effects (6) Continuous traits (7) Variable expressivity 99. Phenotypic expression is often the result of products produced by multistep metabolic pathways involving several different genes; each gene encodes an enzyme that regulates a specific biochemical step or event. Epistasis refers to the interaction among two or more genes that control a common pathway. For example, a mutation in any single gene contributing to a metabolic pathway can affect the expression of other genes in the pathway and, of course, the final phenotype, depending on which biochemical step that gene controls.

Epistasis thus involves interaction among alleles located at different gene loci. This is in contrast to dominance, which involves interaction between alleles located at the same gene locus.

Copyright Macmillan Learning. Powered by Cognero.

Page 34

Name:

Class:

Date:

Chap 06_7e Indicate the answer choice that best completes the statement or answers the question.

1. What is/are the possible inheritance pattern(s) for the characteristic in Pedigree 2? Assume no new mutations and complete penetrance.

a. autosomal recessive only b. autosomal dominant only c. X-linked recessive only d. X-linked dominant only e. All of these are possible.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 06_7e 2. Below are concordance rates for five different human disorders studied in monozygotic (MZ) and dizygotic (DZ) twins.

One of the above disorders is in part caused by significant genetic effects. In addition to a significant genetic contribution toward the disorder, there are also significant environmental effects that contribute toward the disorder phenotype. To which of the above disorders would this description MOST likely apply? a. A b. B c. C d. D e. E 3. This form of prenatal testing is most commonly performed between the 15th week and 18th week of pregnancy. a. chorionic villus sampling b. preimplantation genetic analysis c. amniocentesis d. heterozygote screening e. presymptomatic screening

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 06_7e 4. Both members of an expecting couple are carriers of Tay–Sachs disease, which is an autosomal recessive condition that causes deterioration of nervous function within several months after birth and usually death by age 4. The couple wants to avoid having a child with Tay–Sachs disease, and a DNA test is available to detect the mutant allele. It is early in the pregnancy, and the couple is willing to abort the fetus if it is homozygous for the allele that causes Tay–Sachs disease. However, the couple wishes to have the abortion performed as early as possible if one is to be done. Which would be the MOST appropriate test to determine the genotype of the fetus? a. amniocentesis b. newborn screening c. preimplantation genetic diagnosis d. heterozygote screening e. chorionic villus sampling 5. The pedigree below shows the segregation of an autosomal recessive trait. If III-3 and III-4 marry and have a child, what is the probability that this child will show the trait?

a. 1/4 b. 1/6 c. 1/8 d. 1/2 e. 2/3 6. A trait is caused by a rare dominant autosomal allele with full penetrance. Practically all individuals with the trait are heterozygous and result from matings between affected and normal individuals. What would be the expected concordance values for monozygotic and dizygotic twins? a. 100% for both types of twins b. 100% for monozygotic twins and 90% for dizygotic twins c. about 90% for both types of twins d. 100% for monozygotic twins and 50% for dizygotic twins e. 75% for monozygotic twins and 25% for dizygotic twins Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 06_7e 7. Which of the following is NOT a typical characteristic of human traits that follow an autosomal recessive inheritance pattern? a. They often "skip" generations. b. They appear equally in males and females. c. Parents of affected children are often phenotypically normal themselves. d. When affected individuals marry phenotypically normal individuals, their children are often phenotypically normal. e. All of these options are characteristic of autosomal recessive inheritance.

8. The pedigree below shows the segregation of an autosomal recessive trait in humans. Unless there is evidence to the contrary, assume that individuals who marry into the family do not carry the recessive allele. If IV-1 and IV-2 marry, what is the probability that their first child will have this trait?

a. 1/12 b. 1/16 c. 1/24 d. 1/36 e. 1/8 9. This form of prenatal testing is most commonly performed between the 10th week and 12th week of pregnancy and involves the insertion of a soft plastic into the vagina to obtain cells. a. chorionic villus sampling b. preimplantation genetic analysis c. amniocentesis d. heterozygote screening e. presymptomatic screening

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 06_7e 10. The pedigree below shows the segregation of an autosomal recessive trait. What is the probability that the child of III-1 and III-2 will be affected? Assume that individuals from outside the families are homozygous normal.

a. 1/9 b. 1/64 c. 1/36 d. 1/12 e. 1/16 11. Preimplantation genetic testing involves: a. testing a newborn infant for a genetic disorder immediately after birth. b. testing a single cell of an early embryo for a genetic disorder before the embryo is considered for implantation into the mother's uterus. c. testing multiple sperm from a man who is heterozygous for a serious genetic disorder and then using only sperm that have normal alleles for in vitro fertilizations. d. removing cells from a fetus, testing them for a genetic disorder, and then allowing abnormal fetuses to be aborted. e. testing primary oocytes (eggs) for genetic disorders before using them for in vitro fertilizations.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 06_7e 12. Which of the following scenarios would be the most useful as an adoption study for distinguishing genetic versus environmental effects on characteristics? a. A child is adopted by her grandparents. b. A child is adopted by her aunt (her mother’s adopted sister) and the aunt’s husband. The aunt and husband have three biological children. c. A child is adopted through an adoption agency to unrelated parents. The adoptive parents have no other children. d. A child is adopted through an adoption agency to unrelated parents. The adoptive parents have three biological children. e. A child is adopted through an adoption parent agency to unrelated parents. The adoptive parents have two other adopted children and three biological children.

13. Which of the following is NOT normally used to study the inheritance of human traits or disorders? a. pedigree analysis b. twin studies c. adoption studies d. All of these are used in the study of human traits or disorders. e. None of these is used in the study of human traits or disorders.

14. Could the characteristics followed in the pedigree be caused by an X-linked recessive allele?

a. Yes, all individuals fit the X-linked recessive inheritance pattern. b. No, the offspring of I-1 and I-2 contradict an X-linked recessive inheritance. c. No, the offspring of I-3 and I-4 contradict an X-linked recessive inheritance. d. No, the offspring of II-3 and II-4 contradict an X-linked recessive inheritance. e. Possibly, but I-4 would have to be a heterozygous carrier if there is X-linked inheritance.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 06_7e 15. What is the MOST likely mode of inheritance for the trait segregating in the pedigree below?

a. X-linked recessive b. autosomal recessive c. could be either X-linked recessive or autosomal recessive d. X-linked dominant e. autosomal dominant 16. Which of the following is NOT correctly identified as an advantage of using amniocentesis? a. It can be used to detect chromosome abnormalities prenatally in the fetus. b. In some cases, it can be used to obtain fetal DNA so that tests can be done to determine if the fetus may have a particular genetic disorder. c. It can normally be done as early as the sixth week of pregnancy. d. It has a slightly lower risk of complications than chorionic villus sampling. e. All of these are advantages of the use of amniocentesis.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 06_7e 17. Phenylketonuria (PKU) is an autosomal recessive condition that can lead to mental retardation. It is caused by an enzyme deficiency that interferes with phenylalanine metabolism, causing a by-product to accumulate to levels that are toxic to brain development. Intellectual disability due to PKU can be prevented by a special diet that strictly restricts intake of phenylalanine, but the diet must be started soon after birth to be effective. Which of the following would be the MOST appropriate and cost-efficient way of identifying individuals who need the special diet? a. heterozygote screening b. newborn screening c. presymptomatic testing d. prenatal diagnosis e. amniocentesis 18. If the phenotype followed in Pedigree 3 is autosomal recessive, then what is the genotype of I-1? Assume no new mutations and complete penetrance.

a. homozygous dominant b. heterozygous c. homozygous recessive d. hemizygous dominant e. hemizygous recessive

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 06_7e 19. The ability to roll the tongue is caused by a dominant allele. A woman is a "roller," but one of her parents is not. The woman is expecting a child with a man who is a "nonroller." What is the probability that their first child will be a "roller"? a. 1/4 b. 0 c. 3/4 d. 1/2 e. 1 20. Heterozygous genetic screening is used mainly to: a. detect genetic disorders in newborn infants. b. detect adult members of a particular population who may be heterozygous carriers for recessive disorders. c. detect in healthy adults the presence of a mutant allele that may predispose them to some serious health problem later in life. d. detect fetuses who may be heterozygous carriers for recessive disorders and may eventually be at risk of having children of their own with these disorders. e. detect adults who may be heterozygous for serious autosomal dominant disorders.

21. If monozygotic and dizygotic twins have the same concordance value for a trait, which of the following is TRUE? a. Genetic factors are more important than environmental factors for the trait. b. The trait is probably influenced by numerous genes. c. The trait will more likely appear in males than in females. d. The trait will be passed on from fathers to their daughters but not to their sons. e. The trait is entirely due to environmental factors.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 06_7e 22. If the characteristic followed in the pedigree is X-linked recessive allele, what is III-1's genotype?

a. hemizygous for a dominant allele b. hemizygous for a recessive allele c. definitely heterozygous d. definitely homozygous dominant e. either heterozygous or homozygous dominant 23. A trait is caused by a rare recessive autosomal allele with full penetrance. Practically all individuals with the trait result from matings between normal individuals. What would be the expected concordance values for monozygotic and dizygotic twins? a. 100% for both types of twins b. 100% for monozygotic twins and 50% for dizygotic twins c. about 50% for both types of twins d. 100% for monozygotic twins and 75% for dizygotic twins e. 100% for monozygotic twins and 25% for dizygotic twins

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 06_7e 24. What is the MOST likely mode of inheritance in the pedigree below?

a. autosomal recessive b. autosomal dominant c. X-linked recessive d. X-linked dominant e. Y-linked 25. In pedigree analysis, the proband is: a. the individual having the trait or disorder from whom the pedigree is initiated. b. the medical geneticist who analyzes the pedigree to find the mode of inheritance for the disorder. c. the parents of the first child in the family to show the trait or disorder. d. one of the grandparents or great grandparents who are in the first generation of the pedigree. e. the most common software package that geneticists use to analyze pedigrees.

26. Most pedigrees showing the hypothetical human trait show the following characteristics: - Only males are affected.

- Affected fathers always pass the trait to sons.

What is the MOST likely mode of inheritance for this disorder? a. autosomal recessive b. autosomal dominant c. X-linked recessive d. X-linked dominant e. Y-linked

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 06_7e 27. June has two brothers with Becker muscular dystrophy (BMD), an X-linked recessive condition that allows affected males to survive into adulthood. Her parents are phenotypically normal. She marries Sheldon, who also has BMD. June and Sheldon have a daughter. What is the probability that this daughter will have BMD? a. near 0 b. 1/4 c. 1/2 d. 3/4 e. 1/8 28. If the characteristic followed in the pedigree is autosomal recessive, what is III-1's genotype?

a. either homozygous dominant or heterozygous b. definitely heterozygous c. definitely homozygous dominant d. must be homozygous recessive e. homozygous dominant since he is a male, but a female would be heterozygous 29. Which term refers to mating between closely related people? a. consanguinity b. probanding c. congenital d. concordance e. discordance

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 06_7e 30. In an effort to identify the influence of genetic factors on both Type 1 diabetes and Type II diabetes, researchers calculated concordance rates for monozygotic twins. Concordance rates of 30–50% have been found for Type I diabetes with concordance rates of 80% for Type II. For both kinds of diabetes, dizygotic concordance rates were about 15%. What does this information suggest concerning the relative effect of genetic and environmental factors for each type of diabetes? a. Genetic factors have little or no role in the occurrence of Type I diabetes. b. Type II diabetes is primarily determined by nongenetic factors. c. Genetic influences exert a larger role in Type II diabetes than in Type I diabetes. d. Type II diabetes appears to be caused by a single unidentified gene. e. Environmental factors have no role in the occurrence of either type of diabetes.

31. Suppose that a harmful autosomal recessive condition is found at a particularly high frequency within a certain population. Public health officials would like to reduce the incidence of the condition in the population using genetic testing and genetic counseling. Which of the following programs would be MOST appropriate and most effective in achieving the desired outcome? a. presymptomatic genetic testing b. heterozygote screening c. newborn screening d. preimplantation genetic diagnosis e. amniocentesis 32. If the pedigree below is for an autosomal recessive characteristic, which individuals are definitely heterozygous?

a. I-1, I-2, II-2, II-4, II-5 b. I-1, I-2, I-4, III-1 c. I-1, I-2, II-4, II-5, III-1 d. II-2, II-4, II-5, II-3 e. II-2, II-4, II-5, III-1

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 06_7e 33. In one study with both monozygotic and dizygotic twins, the concordance value for a neurological disorder was 100% for monozygotic twins. Which of the following statements is CORRECT? a. The concordance value for dizygotic twins will be also 100%. b. There will be no phenotypic variation in susceptibility for the disorder for all the pairs of twins in the study. c. The concordance value for dizygotic twins will be 25%. d. For dizygotic twins, all the variation in susceptibility for the disorder will be due to environmental factors. e. Most of the variation in susceptibility for the disorder for monozygotic twins is likely due to genetic factors, assuming that the concordance value for dizygotic twins is significantly less than 100%.

34. Which of the following conditions is MOST commonly screened for in maternal blood tests? a. phenylketonuria and chromosome abnormalities b. Tay–Sachs disease and neural-tube defects c. cancer and multiple sclerosis d. chromosome abnormalities and neural-tube defects e. None of the answers is correct.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 06_7e 35. What is/are the possible inheritance pattern(s) for the characteristic in Pedigree 1? Assume no new mutations and complete penetrance.

a. autosomal recessive only b. autosomal dominant only c. X-linked recessive only d. X-linked dominant only e. All of these are possible.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 06_7e 36. The pedigree below shows the segregation of an autosomal dominant trait. If IV-1 marries IV-5, what is the probability that their first child will be affected? Assume that the homozygous mutant genotype is viable and has the same phenotype as the heterozygote.

a. 1/4 b. 1/2 c. 3/4 d. 3/8 e. 1/8 37. Which description of a Y-linked trait in humans is CORRECT? a. All the sons of an affected father will be affected with the trait. b. Half the sons of a mother whose father was affected with the trait will be affected. c. Half the sons of an affected father will not be affected with the trait, and the other half will be infertile. d. All the daughters of an affected father will be phenotypically normal, but half of their own sons will be affected with the trait. e. The parents of an affected man likely were both phenotypically normal.

38. To establish a successful and cost-effective screening program for detecting adult heterozygous carriers of an autosomal recessive disease, which of the following is NOT important? a. Detection of the disorder is possible in the fetus. b. The disease is clinically significant. c. A high-risk population can be identified. d. Genetic counseling is provided with the testing. e. An effective treatment is available for the disorder.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 06_7e 39. The pedigree below shows a rare autosomal recessive trait segregating. What is the probability of the first child of a marriage between III-3 and III-7 being affected?

a. 1/12 b. 1/4 c. 1/8 d. 1/16 e. 1/24 40. Lucy is 16 weeks pregnant and undergoes maternal serum screening to measure her maternal alpha fetoprotein levels (AFP). Her AFP level is several times higher than normal levels. For which of the following disorders should Lucy be offered additional testing? a. trisomy 21, a chromosome abnormality b. sickle-cell disease, an autosomal recessive disorder c. spina bifida, a neural-tube defect d. hemophilia A, a sex-linked recessive disorder e. Lucy does not need additional testing for any disorder.

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 06_7e 41. Assuming that it is not a Y-linked trait, suggest the MOST likely mode of inheritance for the rare trait shown in the pedigree below.

a. autosomal recessive b. autosomal dominant c. X-linked recessive d. X-linked dominant e. mitochondrial 42. "Vanishing twin syndrome" occurs when one developing baby in a set of multiple pregnancies is lost early during development. Imagine a mother shows twin pregnancies and subsequently experiences vanishing twin syndrome. Her other pregnancy goes to term, and a healthy child is born. Later, that child is found to be a genetic mosaic or chimera. What most likely occurred? a. The two initial pregnancies resulted from monozygotic twins, and one twin was lost spontaneously. b. The two initial pregnancies resulted from monozygotic twins, and one embryo absorbed the other. c. The two initial pregnancies resulted from dizygotic twins, and one embryo was lost spontaneously. d. The two initial pregnancies resulted from dizygotic twins, and one embryo absorbed the other. e. None of these options are correct.

43. Most pedigrees showing a hypothetical human trait show the following characteristics: - If a phenotypically normal woman with an affected father has children with an unaffected man, half of the sons and none of the daughters are affected.

- Affected females always have an affected father and an affected maternal grandfather.

- The trait is never passed from father to son.

What is the MOST likely mode of inheritance for this disorder? a. autosomal recessive b. autosomal dominant c. X-linked recessive d. X-linked dominant e. Y-linked Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 06_7e 44. In pedigree analysis, consanguinity refers to: a. mating between two heterozygous carrier parents. b. the realization that phenotypes are often more closely related between children and grandparents than between children and parents. c. mating between two closely related parents. d. a situation where the children of two parents are adopted. e. a situation where only one individual in the entire pedigree is affected with the trait or disorder.

45. The pedigree below shows the sex-linked recessive inheritance of Nance–Horan syndrome, a rare Mendelian disorder in which affected persons have cataracts and abnormally shaped teeth. What is the probability that the first child of IV-6 and V-3 would have Nance–Horan syndrome?

a. 1/4 b. 1/2 c. 1/8 d. 3/16 e. 3/4 46. Heterozygote screening normally involves: a. testing healthy individuals to see if they possess mutant alleles that may make them ill later in life. b. testing newborn infants to see if they have a genetic disorder so that they can be treated immediately. c. examining fetal cells to see if they have a serious genetic disorder so that the pregnancy can be terminated if the parents so desire. d. testing to determine if two parents are related to each other. e. testing adult members of a particular population to identify heterozygous carriers for a recessive disorder.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 06_7e 47. The ability to roll the tongue is caused by a dominant allele. A woman is a "roller," but one of her parents is not. What is the woman's genotype? a. homozygous dominant b. heterozygous c. homozygous recessive d. either homozygous recessive or homozygous dominant e. It cannot be determined from this information.

48. Which of the following statements is NOT true concerning newborn screening? a. It is normally done soon after the birth of a child. b. It is particularly important to test for serious conditions where there is no treatment available. c. Most newborn testing is done by collecting a small amount of an infant's blood and then using it for analysis to detect specific genetic disorders. d. The genetic disorder phenylketonuria is one of the conditions usually tested for with newborn screening. e. All of the statements are true.

49. Could the characteristic followed in the pedigree below be caused by an autosomal dominant disease? Why or why not?

a. Yes, all individuals fit the autosomal dominant inheritance pattern. b. No, the offspring of I-1 and I-2 contradict an autosomal dominant inheritance. c. No, the offspring of I-3 and I-4 contradict an autosomal dominant inheritance. d. No, the offspring of II-3 and II-4 contradict an autosomal dominant inheritance. e. Yes, the offspring of I-1and I-2 are consistent with an autosomal dominant inheritance pattern.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 06_7e 50. Most pedigrees showing a hypothetical human trait show the following characteristics: - Males and females are equally affected.

- Two unaffected parents can have an affected child.

- In families in which the parents are unaffected but the children are affected, one-fourth of the children are affected.

What is the MOST likely mode of inheritance for this disorder? a. autosomal recessive b. autosomal dominant c. X-linked recessive d. X-linked dominant e. Y-linked

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 06_7e 51. If the phenotype followed in Pedigree 3 is X-linked recessive, then what is the genotype of II-2? Assume no new mutations and complete penetrance.

a. homozygous dominant b. heterozygous c. homozygous recessive d. hemizygous dominant e. hemizygous recessive 52. Imagine that a human characteristic is determined by genotype only, with no environmental influence. Based on the amount of shared genetic information in MZ (monozygotic) and DZ (dizygotic) twins, what would you expect the concordance values to be in these two sets of twins? a. The concordance values would be near 100% for both types of twins. b. The concordance values would be near 50% for both types of twins. c. The concordance values should be close to 100% for MZ twins and a concordance value much less for DZ twins. d. The concordance values should be close to 50% for MZ twins and about 25% for DZ twins. e. The concordance values should be close to 25% for MZ twins and about 75% for DZ twins.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 06_7e 53. Explain what the following data say about the genetic and environmental influence on each characteristic. How did you interpret these data? (MZ = monozygotic twins and DZ = dizygotic twins.)

Trait Right- or left-handedness Manic depression

Concordance in MZ Concordance in DZ 80 75 80 20

a. Handedness and manic depression both have a strong genetic basis and little or no environmental basis. b. Handedness and manic depression are both strongly affected by environmental factors with little or no genetic basis. c. Handedness and manic depression are strongly affected by both environmental factors and genetic factors. d. Handedness is strongly affected by environmental factors with little or no genetic basis, while manic depression is significantly affected by genetic factors. e. Handedness is significantly affected by both environmental and genetic factors, while manic depression is significantly affected by environmental factors with little genetic basis.

54. Most pedigrees showing the hypothetical human trait show the following characteristics: - Females are affected twice as frequently as males.

- Affected fathers may have affected daughters but never affected sons.

- Half the children of affected mothers and normal fathers are affected.

What is the MOST likely mode of inheritance for this disorder? a. autosomal recessive b. autosomal dominant c. X-linked recessive d. X-linked dominant e. Y-linked 55. Which statement is INCORRECT concerning an X-linked recessive trait in humans? a. An affected man often has phenotypically normal parents. b. All the sons of an affected woman will be expected to be affected. c. An affected woman almost always has an affected mother. d. An affected man usually has a mother who carries the recessive allele. e. A phenotypically normal woman whose father was affected is likely to be heterozygous for the condition.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 06_7e 56. Which of the following is NOT a characteristic of X-linked recessive traits in humans? a. More males than females are affected. b. Approximately one-half of the sons of a female carrier are affected. c. They cannot be passed from father to son. d. Phenotypically normal daughters of affected men are always carriers. e. Affected daughters always have an affected mother.

57. Two parents are phenotypically normal, but one of their four biological children has a typical autosomal recessive trait. The other three children are phenotypically normal. It is very likely that: a. the affected child is a girl. b. the affected child is a boy. c. the trait was expressed by one of the grandparents of the children. d. the parents are both heterozygous for the trait. e. if the affected child eventually marries a phenotypically normal spouse, all of their children will have the trait.

Indicate one or more answer choices that best complete the statement or answer the question.

58. Assume that the concordance values for a particular disease are 63% for monozygotic twins and 36% for dizygotic twins. Which of the following are TRUE? (Select all that apply.) a. Environmental factors are important in susceptibility to this disease. b. Genetic factors are important in susceptibility to the disease with monozygotic twins but not with dizygotic twins. c. Genetic factors are not important in susceptibility to this disease. d. Genetic factors are important in susceptibility to this disease. e. Environmental factors are important in susceptibility to this disease with dizygotic twins but not with monozygotic twins.

59. Which of the following statements concerning chorionic villus sampling are TRUE? (Select all that apply.) a. A karyotype can be made from fetal cells. b. It is usually performed between the 10th and 12th weeks of pregnancy. c. It has fewer complications than does amniocentesis if parents choose to abort the pregnancy. d. Biochemical analyses can be performed on fetal cells. e. The fetal cells obtained usually do not need to be cultured.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 06_7e 60. In what way(s) are humans useful for genetic study? (Select all that apply.) a. Specific crosses are easily carried out. b. Familial records with phenotypic data are often available. c. Generation times are short. d. Brood sizes are large. e. Many well characterized traits are available for study.

61. Which of the following are typical reasons why individuals or couples may seek genetic counseling? (Select all that apply.) a. A husband and wife are closely related, and they wish to have children. b. A husband is 15 years older than his wife, and they wish to have children. c. A person knows of a genetic disease in the family. d. A couple has given birth to a child with a genetic disease. e. A couple experiences difficulties in achieving a successful pregnancy.

62. Tony, who is not diseased, has a sister with cystic fibrosis (CF). Neither of his parents has CF. Tony is expecting a child with Tina. Tina's family history is unknown. If the frequency of heterozygotes in the general population is 1/50, what is the probability that Tony and Tina's child will have CF? Explain each factor in your calculation.

63. Today, more than 1000 genetic tests are available to doctors and patients for use in diagnosis and treatment of health conditions, and more will become available in the near future. Genetic testing is becoming more common in medical practice, and genetic testing is expected to become commonplace. Explain the complications that are associated with genetic testing and why specialized genetic counselors are needed in medical practice.

64. Describe at least two reasons why testing for a genetic condition might be advantageous.

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 06_7e 65. Is it possible that the characteristic in the pedigree below could be X-linked dominant? If so, what is the genotype of each individual? If not, explain why not, giving specific genotypes. Use alleles B and b.

66. Tony, who is not diseased, has a sister with cystic fibrosis (CF). Neither of his parents has CF. Tony is expecting a child with Tina. Tina's family history is unknown. What is the probability that Tony is heterozygous for the CF gene? Explain your answer.

67. How have the eras of genomics, proteomics, and large-scale data collection, analysis, and presentation in databases helped people with genetic syndromes or diseases?

68. Assume a pedigree is well-established for several generations for a particular trait. What factors may complicate the genetic analysis of the pedigree? (Hint: Consider potential complicating factors raised in Section 6.2 and Chapter 5, "Extensions and Modifications of Basic Principles.")

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 06_7e 69. Is it possible that the characteristic in the pedigree below could be autosomal dominant? If so, what is the genotype of each individual? If not, explain why not, giving specific genotypes. Use alleles D and d.

70. Is it possible that the characteristic in the pedigree below could be autosomal dominant? If so, what is the genotype of each individual? If not, explain why not, giving specific genotypes. Use alleles D and d.

71. The American College of Medical Genetics recommends mandatory screening of newborn infants for 35 conditions, and many states have adopted this recommendation into law. a. Explain the benefits of mandatory newborn screening for medical conditions. b. Why might some people be concerned about or even opposed to mandatory screening of newborns? c. Do you support mandatory screening of newborn infants for genetic disorders? Defend your position with facts and arguments.

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 06_7e 72. Is it possible that the characteristic in the pedigree below could be X-linked dominant? If so, what is the genotype of each individual? If not, explain why not, giving specific genotypes. Use alleles B and b.

73. Is it possible that the characteristic in the pedigree below could be X-linked recessive? If so, what is the genotype of each individual? If not, explain why not, giving specific genotypes. Use alleles A and a.

74. Draw a four-generation pedigree, following colorblindness, using the following information: In generation I, neither parent is colorblind. In generation II, a son (II-1) is colorblind, and a daughter (II-2) is not. II-2 and a man with normal color vision (II-3) have a daughter (III-1) who is not colorblind. III-1 and a man with normal color vision (III-2) have a daughter (IV-1) who is not colorblind. Write in all the genotypes that are known.

Use symbols Xc and Y to follow the sex chromosomes and sex-linked genes.

Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 06_7e 75. Researchers studying genetic determination of childhood asthma noted 65% concordance for monozygotic twin pairs in which at least one twin has asthma and 37% concordance for dizygotic twin pairs in which at least one twin has asthma. a. Explain the meaning of 65% concordance for asthma in monozygotic twins. b. Interpret the difference in concordance for asthma between monozygotic and dizygotic twins. c. If genes do influence childhood asthma, how is that 35% of monozygotic twin pairs are discordant for asthma?

76. (a) What is a direct-to-consumer genetic test? (b) What are some advantages of direct-to-consumer tests? (c) What are some disadvantages or problems associated with these tests? (d) Do you think states and/or the federal government should regulate direct-to-consumer genetic testing? Defend your position with facts and arguments.

77. Explain what it means for a genetic counselor to use "nondirected counseling."

78. Juliet and Bob have two children, Jack and Norma. No one in this family is red–green colorblind, although Juliet sometimes has difficulty passing the color-vision part of the driving test. Norma has a colorblind son with Tom, who is not colorblind. Draw a pedigree that includes all of this information. Below each person in the pedigree, write his or her genotype for the red–green colorblindness gene, using B for the normal allele and b for the colorblindness allele.

Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 06_7e 79. Tony, who is not diseased, has a sister with cystic fibrosis (CF). Neither of his parents has CF. Tony is expecting a child with Tina. Tina's family history is unknown. Draw a pedigree that includes all of this information. Below each person in the pedigree, write his or her genotype or possible genotypes, using A for the normal CF allele and a for the disease-causing recessive allele.

80. Is it possible that the characteristic in the pedigree below could be X-linked recessive? If so, what is the genotype of each individual? If not, explain why not, giving specific genotypes. Use alleles A and a.

81. A maternal blood screening test carried out in the second month of a pregnancy indicates a level of αfetoprotein that is significantly higher than normal. a. What is α-fetoprotein? b. What does a higher than normal level of α-fetoprotein in the mother's blood indicate? c. Critique the maternal blood screening test—specifically, what are its limitations? d. Given the limitations of the tests, why are they so commonly ordered? e. If you were the doctor who ordered this test, how would you proceed?

82. Explain the principle behind using concordance values for monozygotic (MZ) and dizygotic (DZ) twins to determine the influence of genetic factors on individual differences for a trait.

Copyright Macmillan Learning. Powered by Cognero.

Page 30

Name:

Class:

Date:

Chap 06_7e Answer Key 1. e 2. c 3. c 4. e 5. b 6. d 7. e 8. c 9. a 10. c 11. b 12. a 13. d 14. a 15. b 16. c 17. b 18. b 19. d 20. b 21. e 22. a 23. e 24. b 25. a 26. e Copyright Macmillan Learning. Powered by Cognero.

Page 31

Name:

Class:

Date:

Chap 06_7e 27. b 28. b 29. a 30. c 31. b 32. c 33. e 34. d 35. a 36. c 37. a 38. e 39. a 40. c 41. b 42. d 43. c 44. c 45. c 46. e 47. b 48. b 49. b 50. a 51. b 52. c 53. d 54. d Copyright Macmillan Learning. Powered by Cognero.

Page 32

Name:

Class:

Date:

Chap 06_7e 55. c 56. e 57. d 58. a, d 59. a, b, d, e 60. b, e 61. a, c, d, e 62. 2/3 × 1/50 × 1/4 = 2/600, or 1/300 (Tony: heterozygous) × (Tina: heterozygous) × (homozygous recessive from two heterozygotes) 63. Genetic counselors are trained to help patients understand the results of genetic tests and their implications. They provide information related to genetic disorders, their causes, and potential treatments so that patients can make informed decisions. Their services are often needed because genetic tests are complicated and their results are often not straightforward. For example, many genetic conditions can be caused by many different alleles, but typically genetic tests only can detect the most common causative alleles. Also, there often is not a simple relationship between a genotype and its associated phenotype. With incomplete penetrance, for example, the presence of a particular genotype is associated only with a probability of developing the associated medical condition. Many people have difficulty in understanding these probabilities and making appropriate decisions.

64. (1) A person with an inherited genetic condition might start early treatment to lessen the severity of the condition.

(2) The symptoms of some genetic conditions can be prevented by early treatment.

(3) If a person is at risk for inheriting a genetic condition, testing can alleviate the anxiety associated with uncertainty.

(4) Parents who are considering children can determine if they are heterozygous for a homozygous recessive condition.

(5) A woman who knows the genotype of the embryo or fetus she is carrying may use the information to decide whether to continue or terminate a pregnancy.

(6) A woman who knows the genotype of the embryo or fetus she is carrying may use the information to prepare for the birth of a child with an inherited condition.

(7) For embryos produced by in vitro fertilization, genetic testing is used to determine which embryos to implant.

65. X-linked dominant is possible, if I-1 is XBY, I-2 is Xb Xb , II-1 Xb Y, and II-2 is XB Xb .

66. 2/3. Tony's parents are Aa and Aa. Their children's genotypes, with probabilities, are 1/4 AA, 1/4 Aa, 1/4 aA, 1/4 aa. Tony does not have CF, so he cannot be genotype aa. Of the remaining genotypes, 2/3 are heterozygous.

67. Answers may vary. Large-scale databases with human genetic disorder information, such as the Online Mendelian Inheritance in Man web site, provide far more detailed information for patients, doctors, and scientists. Correlation of that information with genomics and proteomics data will significantly expedite molecular and biochemical understanding of genetic syndromes or diseases, and also the development of treatment.

Copyright Macmillan Learning. Powered by Cognero.

Page 33

Name:

Class:

Date:

Chap 06_7e 68. Answers may vary. Genetic mosaicism and a trait being too common to see sufficient unaffected individuals in the pedigree should be brought up. Other complicating factors include new mutations arising that affect the trait being studied, incomplete penetrance, variable expressivity, gene interactions, maternal effects, and environmental effects on phenotypes.

69. Autosomal dominant is possible, if I-1 is dd, I-2 is Dd, and II-1 and II-2 are dd.

70. Autosomal dominant is possible, if I-1 is Dd, I-2 is dd, II-1 is dd, and II-2 is Dd. 71. a. Newborn screening allows for early identification of medical conditions that can be treated more effectively when diagnosed early. For example, PKU can lead to mental retardation if an infant is fed a normal diet. However, if fed a strictly regulated diet from soon after birth, the brain will develop normally. Mandatory screening has reduced the incidence and severity of suffering from these conditions, along with social and economic costs associated with them, in the population as a whole. b. Some people are concerned about how genetic information from screening tests might be used and how the blood collected might be used. These concerns are understandable, given our history of abusing genetic information and misusing biological samples in the past. Also, blood samples taken for newborn screening have been used in research without consent of the individuals or their parents. Also, some parents might be concerned that children identified as having a genetic disorder might be stigmatized and their futures limited in some way. c. Obviously, answers will vary. Some students will argue in favor of mandatory screening and cite the benefits listed in (a). The best answers supporting screening will also discount the concerns cited in (b) by arguing that laws are now in place or could be enacted to limit the potential for such abuses. Some students will argue in support of screening on an economic basis. By identifying a disorder early when it can be treated successfully, the treatment will save money compared with the much higher cost of dealing with the untreated patient with the disorder. The PKU story illustrates this argument well. Some students will argue against mandatory screening and cite the potential for abuses cited in (b). They might also argue against government intrusion in the lives of families.

72. X-linked dominant is possible, if I-1 is Xb Y, I-2 is XBXb , II-1 is Xb Y, and II-2 is Xb Xb .

73. X-linked recessive is possible, if I-1 is Xa Y, I-2 is XAXa , II-1 XAY, and II-2 is Xa Xa .

74.

Copyright Macmillan Learning. Powered by Cognero.

Page 34

Name:

Class:

Date:

Chap 06_7e 75. a. Sixty-five percent concordance indicates that, in 65% of monozygotic twin pairs in which one twin has asthma, the other twin will also have asthma. b. The significantly higher concordance in monozygotic twins compared to dizygotic twins indicates that genes play a major role in contributing to childhood asthma. The difference can be attributed to the fact that monozygotic twins share 100% of their genes, whereas dizygotic twins only share 50% on average. c. The discordance between monozygotic twins suggests that the environment also plays an important part in determining asthma. Given that the monozygotic twins are genetically identical, their different phenotypes must be due to differences in their developmental environments.

76. (a) A direct-to-consumer genetic test is one that is offered directly to consumers without the involvement of healthcare professionals. (b) Patients often have better access to these tests and costs might be lower because patients do not have to pay for the services of a health-care professional. Also the tests can be carried out anonymously, alleviating concerns about violations of genetic privacy. (c) In many cases, the tests are offered without appropriate information and genetic counseling to allow the patients to understand the results and use the information effectively. Other problems are that the tests are not always accurate and confidentiality is not always guaranteed. (d) Answers will vary. Students advocating for regulation should emphasize the disadvantages and problems listed in (c) and underscore the need for consumer protections. Students advocating for keeping the market unregulated should emphasize freer access, reduced costs, and the responsibility of individuals for choosing the services and using the results appropriately. They might argue that informed consumers will choose companies that provide proper genetic counseling and guarantee confidentiality.

77. Nondirected counseling describes a style of counseling in which the genetic counselor provides information, calculates probabilities, explains risk factors, and explains reproductive options but does not offer his or her opinion or bring his or her personal values into the discussion.

78.

Copyright Macmillan Learning. Powered by Cognero.

Page 35

Name:

Class:

Date:

Chap 06_7e

79.

80. X-linked recessive is not possible because if I-2 is Xa Xa , then II-1 is Xa Y and should have the phenotype.

81. a. α-fetoprotein is a protein that is normally created by the fetus during development and is normally present in fetal blood, amniotic fluid, and the mother's blood during pregnancy. b. A higher than normal level indicates that there might be a problem with development of the fetus. Specifically, high levels are associated with neural-tube defects and certain other common disruptions to fetal development. c. Higher than normal levels of α-fetoprotein are associated with increased risk of genetic or developmental problems, but they do not prove that a problem is present. d. Maternal blood screening tests are not as invasive or as expensive as more definitive tests (e.g., amniocentesis). e. You should order additional tests to confirm or rule out the presence of a developmental problem. These tests might include additional blood tests, ultrasound, amniocentesis, and CVS. 82. MZ twins theoretically share 100% genetic information, and DZ twins share only 50%. So if genotype differences cause most of the variation in the population, concordance values for the characteristic will be higher in MZ twins and lower in DZ twins.

Copyright Macmillan Learning. Powered by Cognero.

Page 36

Name:

Class:

Date:

Chap 07_7e Indicate the answer choice that best completes the statement or answers the question.

You just bought two black guinea pigs, one male and one female, of the same genotype, that are known to be heterozygous (Bb). You also know that black fur (BB) is dominant over white fur (bb) and that a lethal recessive allele is located only 1 m.u. away from the recessive b allele, and your animals are both heterozygous for this gene also.

1. What is the probability of finding a white individual among the progeny if you cross these two animals? a. 0.25 b. 0.0025 c. 0.000025 d. 0.005 e. 0.495 2. Linkage disequilibrium is defined by which of the following? a. positions in the genome where people vary by a single nucleotide base b. the probability that two genes are linked c. the nonrandom association between alleles in a haplotype d. the occurrence of two alleles in the repulsion configuration e. crossing over that occurs between two genes that are located close to each other 3. Two linked genes, (A) and (B), are separated by 18 m.u. A man with genotype Aa Bb marries a woman who is aa bb. The man's father was AA BB. What is the probability that their first two children will both be ab/ab? a. 0.168 b. 0.0081 c. 0.032 d. 0.062 e. 0.13 4. A situation in which the coefficient of coincidence is greater than 1.0 would indicate that: a. the interference is high, and one crossover suppresses the occurrence of a second one. b. no double crossovers were found in the progeny of a testcross, even though some were expected based on probability. c. double crossovers were found in the progeny of a testcross, but there were fewer of them than would be expected based on probability. d. there were more double crossovers in the progeny than would be expected based on probability. e. the genes involved were actually assorting independently.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 07_7e You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue shells and long antenna or green shells and short antenna. Crossing these lines yields F1 progeny with blue shells and long antenna. Crossing F1 progeny with beetles that have green shells and short antenna yields the following progeny: Blue shell, long antenna Green shell, short antenna Blue shell, short antenna Green shell, long antenna Total

82 78 37 43 240

5. Assuming that the genes are linked, what is the map distance between them in m.u.? a. 33.3 m.u. b. 25.0 m.u. c. 49.5 m.u. d. 8.0 m.u. e. The genes are assorting independently.

You just bought two black guinea pigs, one male and one female, of the same genotype, that are known to be heterozygous (Bb). You also know that black fur (BB) is dominant over white fur (bb) and that a lethal recessive allele is located only 1 m.u. away from the recessive b allele, and your animals are both heterozygous for this gene also.

6. You decide to start raising your own guinea pigs, but after mating these animals several times, you discover they produce only black offspring among the first 12 progeny. How would you best explain this result? a. The B locus is on the X chromosome, so it can never produce a white phenotype. b. The B allele is actually codominant with the b allele, so a white phenotype cannot be produced. c. The recessive l allele is in tight coupling linkage with the b allele, so almost all of the bb offspring will also be ll and thus die before they can be observed. d. The dominant L allele is in tight repulsion linkage with the B allele, so it will be impossible to produce the Bb genotype that would express the white phenotype. e. Normally, it would be expected that 25% of the offspring would be white, but in this case, random deviations from the expectation resulted in no white offspring.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 07_7e 7. In flower beetles, pygmy (py) is recessive to normal size (py+), and red color (r) is recessive to brown (r+). A beetle heterozygous for these characteristics was testcrossed to a beetle homozygous for pygmy and red. The following are progeny phenotypes from this testcross. py+ r+ py+ r py r+ py r Total

180 22 19 191 412

Carry out a series of chi-square tests to determine if there is equal segregation of alleles at the py locus. What is the correct chi-square value, and how many degree(s) of freedom should be used in its interpretation? a. 0.16 with one degree of freedom b. 0.16 with three degrees of freedom c. 0.48 with one degree of freedom d. 0.48 with two degrees of freedom e. 4.56 with one degree of freedo 8. A genetic map shows which of the following? a. the distance in numbers of nucleotides between two genes b. the number of genes on each of the chromosomes of a species c. the linear order of genes on a chromosome d. the location of chromosomes in the nucleus when they line up at metaphase during mitosis e. the location of double crossovers that occur between two genes

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 07_7e You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue shells and long antenna or green shells and short antenna. Crossing these lines yields F1 progeny with blue shells and long antenna. Crossing F1 progeny with beetles that have green shells and short antenna yields the following progeny: Blue shell, long antenna Green shell, short antenna Blue shell, short antenna Green shell, long antenna Total

82 78 37 43 240

9. A chi-square test is done to test for independent assortment. What is the resulting chi-square value, and how many degree(s) of freedom should be used in its interpretation? a. 27.1 and one degree of freedom b. 14.9 and three degrees of freedom c. 14.9 and two degrees of freedom d. 27.1 and three degrees of freedom e. 0.42 and two degrees of freedom 10. Consider the following three-point (trihybrid) testcross:

Calculate the number of double crossovers you would expect to observe if 1000 progeny result from this testcross assuming a coefficient of coincidence of 0.25. a. about 14 b. about 26 c. about 10 d. about 4 e. none because of crossover interference

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 07_7e 11. You are studying two linked genes in lizards. You have two females, and you know that both are the same genotype, heterozygous for both genes (A/a and B/b). You testcross each female to a male that is fully homozygous recessive for both genes (a/a and b/b) and get the following progeny with the following phenotypes: Female 1 AB – 37 ab – 33 Ab – 4 aB – 6

Female 2 AB – 5 ab – 4 Ab – 35 aB – 36

How can you explain the drastic difference between these two crosses? a. The two genes are assorting independently in female 1 and are linked in female 2. b. The two genes are linked in female 1 and are assorting independently in female 2. c. The two alleles are in the coupling configuration in female 1 but in the repulsion configuration in female 2. d. The two alleles are in the repulsion configuration in female 1 but in the coupling configuration in female 2. e. The two genes are likely located on a sex chromosome in female 1 and are likely located on an autosome in female 2.

12. If a three-strand double crossover occurs between two genes during meiosis, what is the expected outcome with respect to these two genes? a. Two chromatids are parental, and two chromatids are recombinant. b. All four chromatid are parental. c. All four chromatids are recombinant. d. Three chromatids are recombinant, and one is parental. e. Three chromatids are parental, and one is recombinant.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 07_7e 13. In flower beetles, pygmy (py) is recessive to normal size (py+), and red color (r) is recessive to brown (r+). A beetle heterozygous for these characteristics was testcrossed to a beetle homozygous for pygmy and red. The following are progeny phenotypes from this testcross. py+ r+ py+ r py r+ py r Total

180 22 19 191 412

Carry out a series of chi-square tests to determine if the two loci are assorting independently. What is the correct chi-square value, and how many degree(s) of freedom should be used in its interpretation? a. 112 with one degree of freedom b. 265 with three degrees of freedom c. 367 with four degree of freedom d. 16.5 with three degrees of freedom e. 367 with three degrees of freedom 14. If the recombination frequency between genes (A) and (B) is 5.3%, what is the distance between the genes in map units on the linkage map? a. 53 m.u. b. 5.3 m.u. c. 0.53 m.u. d. 10.6 m.u. e. 25 m.u.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 07_7e Dr. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits. He mates these two strains with each other. Dr. Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males. In the resulting testcross progeny, he observes 500 flies that are of the following makeup: 41 207 210 42

with baby blue eyes and pink wings with baby blue eyes only with pink wings only with wild-type phenotype

15. What is the relationship with respect to location between the two genes? a. They are far apart on the same chromosome and assorting independently. b. They are linked, and the map distance between them is 41.5 m.u. c. They are on different chromosomes and assorting independently. d. They are linked and 16.6 m.u. apart. e. They are linked and 50.0 m.u. apart.

16. You are working with five disease-resistance genes (A, B, C, D, and E) that are on chromosome 4 of Arabidopsis. You cross a line that is fully homozygous recessive for these five genes to each of five lines (recessive lethal lines) that are heterozygous for a different deletion. You observe the following results (wt = wild type; m = 50% of progeny displayed mutant phenotype): Recessive lethal lines A Disease B Resistance C Phenotype D E

1 wt m m wt m

2 m m wt wt wt

3 m wt wt m wt

4 m m m wt wt

5 wt wt wt wt wt

What is the CORRECT gene order for your five mutants? a. C A B D E b. A D B C E c. A D E C B d. D A B C E e. E B A C D

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 07_7e 17. In maize (corn), assume that the genes A and B are linked and 30 map units apart. If a plant of Ab/aB is selfed, what proportion of the progeny would be expected to be of ab/ab genotype? a. 2.25% b. 15% c. 9% d. 30% e. 4.5% 18. What does lod stand for? a. linkage over DNA b. linkage of dihybrids c. long overall distances (with respect to map distances) d. linker of DNA e. logarithm of odds

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 07_7e 19. You and a colleague are working on a rare Peruvian llama that appears to be susceptible to diabetes, a disease related to insulin function. Your colleague has established a somatic-cell hybrid panel, and you would like to figure out to which llama chromosome the gene that encodes the llama insulin receptor maps. You also have an assay that allows you to detect which of the somatic-cell lines can produce the insulin receptor. You assay the colleague's somatic-cell hybrid panel and get the following results. A (+) or (–) in the following table indicates whether the insulin receptor is present or absent in each hybrid panel. The gene is on which of the 74 llama chromosomes?

Line # 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Llama chromosomes Assay 7, 10, 15, 29, 41, 55, 68 – 4, 8, 16, 31, 44, 56, 70 – 9, 12, 18, 52, 54, 61, 73 + 1, 8, 13, 22, 33, 45, 59 – 2, 11, 14, 18, 23, 49, 62 + 5, 12, 15, 24, 35, 47, 63 – 3, 17, 21, 32, 43, 58, 72 – 7, 21, 25, 36, 53, 64 – 1, 3, 9, 28, 37, 48, 65 – 13, 22, 27, 38, 50, 66, 19 – 4, 6, 12, 18, 20, 40, 67 + 2, 16, 26, 34, 51, 69 – 10, 14, 17, 39, 46, 57, 74 – 5, 11, 19, 30, 42, 60, 71 –

a. 55 b. 7 c. 41 d. 68 e. 18

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 07_7e 20. A study is performed on families in Sweden that are segregating a genetic disorder. Lod-score analysis indicates that the gene involved in the disorder shows a strong likelihood of linkage with a particular DNA marker locus. However, a second study performed in Italy with other families segregating the same genetic disorder results in lod-score values that strongly indicate the lack of linkage between the gene and the same DNA marker locus. Assuming that both studies were performed appropriately, what is the most likely explanation for the different outcomes? a. No recombinants were found in the families studied in Sweden. b. The allele that caused the disorder was in coupling linkage with one of the DNA marker alleles in the Swedish families but was in repulsion linkage in the Italian families. c. This disorder is caused by mutations in either of two different genes; one of these genes is linked to the DNA marker locus, and the other gene is not. d. In the Italian families, the gene involved with the disorder is near a lethal allele at another locus and most of the parental or nonrecombinant genotypes contain the lethal allele; this reduces the number of nonrecombinants observed. e. Linkage should have been observed in the Italian families, but there were only two alleles at the DNA marker locus that prevented recombinant offspring from appearing.

21. A cell possessing two nuclei derived from different cells through cell fusion is called: a. a heterokaryon. b. a haplotype. c. recombinant. d. nonrecombinant. e. None of the answers is correct.

22. Three-factor testcrosses are only informative in gene mapping when which of the following occurs? a. One parent is homozygous recessive for the three genes, and the other parent is homozygous dominant. b. All three genes are located on separate chromosomes, and one parent is homozygous dominant for at least two of these genes. c. Both parents are homozygous for the three genes. d. One parent is heterozygous for the three genes, and the other parent is homozygous recessive. e. One of the genes must be located on a sex chromosome and be heterozygous, and the other two genes must be located on an autosome and be homozygous.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 07_7e 23. The results of linkage analysis for DNA marker A and the p53 gene are shown below. What is the best estimate for the approximate genetic distance between marker A and the p53 gene in humans?

Recombination values (m.u.) lod score

1 2.13

5 2.54

10 3.14

20 4.10

30 4.96

40 3.22

a. 1 m.u. b. 5 m.u. c. 10 m.u. d. 20 m.u. e. 30 m.u.

24. Why are the progeny of a testcross generally used to map loci? Why not the F2 progeny from an F1 × F1 cross? a. Only recombinant offspring would be found in the progeny of an F1 × F1 cross. b. The progeny of an F1 × F1 cross would be found in a 9:3:3:1 ratio when two genes are involved, whereas the progeny of a testcross would result in a 1:1:1:1 ratio. c. It is easier to classify recombinant and parental offspring of a testcross than with the progeny of an F1 × F1 cross. d. In a testcross more of the progeny would be expected to display the dominant phenotype than in the progeny of an F1 × F1 cross. e. A testcross is more useful for mapping genes that are located near each other, but when genes are quite far apart on the same chromosome, an F1 × F1 cross actually is more useful.

25. Linked genes: a. assort randomly. b. cannot cross over and recombine. c. are allelic. d. cosegregate. e. will segregate independently.

26. The map distance between the genes C and D is 24 cM. A cross is set up as follows: CD/cd × cd/cd. 200 progeny are produced from the cross. What is the expected number of progeny with the genotype Cd/cd? a. 12 b. 24 c. 36 d. 48 e. 100 Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 07_7e 27. GGHH and ccdd individuals are crossed to each other and the F1 backcrossed to the gghh parent. The 4000 progeny included 1806 GgHh, 1794 gghh, 196 Gghh, and 204 ggHh individuals. GGhh and ggHH individuals are next crossed to each other and the F1 testcrossed to a gghh individual. A total of 1600 offspring appeared in the progeny. How many do you expect to be of GgHh genotype? a. 40 b. 160 c. 320 d. 80 e. 1600 28. Interference occurs when: a. two genes are assorting independently. b. two genes are far apart on a genetic map. c. one crossover inhibits another. d. the number of recombinant progeny classes in the testcross of a heterozygote exceeds the number of parental progeny. e. a crossover causes the termination of the meiosis event in which the crossover is occurring.

29. In mapping studies that use molecular markers, linked genes are: a. allelic. b. dominant. c. on different chromosomes. d. on the same chromosome. e. recessive lethal.

30. A low coefficient of coincidence indicates that: a. far fewer double-crossover recombinant progeny were recovered from a testcross than would be expected from the map distances of the genes involved. b. crossing over has been enhanced for genes that are located near the centromere of chromosomes because there is less interference of one crossover on the occurrence of a second crossover event. c. single-crossover recombinant classes in the progeny have been increased because the genes involved produce lethal phenotypes when in parental gene combinations. d. there is a large map distance between one of the outside genes in the heterozygous parent and the middle gene, while there is a short map distance between the middle gene and the other outside gene. e. the physical distance between two genes is very short compared with the genetic map distance between these two genes.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 07_7e 31. Compared with a physical map, a genetic map: a. is always more accurate. b. is always less accurate. c. is equally accurate. d. measures different things. e. cannot be made for humans.

32. Two linked genes, (A) and (B), are separated by 18 m.u. A man with genotype Aa Bb marries a woman who is aa bb. The man's father was AA BB. What is the probability that their first child will be Aa bb? a. 0.18 b. 0.41 c. 0.09 d. 0.25 e. 0.50 33. A panel of cell lines was created from mouse–human somatic-cell fusions. The following table indicates which human chromosomes are found in five cell lines (A, B, C, D, and E). A (+) or (–) in the following table indicates whether the enzyme glutathione S-transferase is present or absent in each cell line. Assume that a student is given the information provided in the table, along with information about the presence or absence of the enzyme in each cell line. On what human chromosome is the gene for glutathione S-transferase located?

a. X b. 2 c. 8 d. 6 e. The gene must be located on a human chromosome not present in any of the cell lines above.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 07_7e 34. Is it possible for two different genes located on the same chromosome to assort independently? a. No, if two genes are on the same chromosome, they will be linked and the recombination frequency will be less than 50%. b. Yes, if the two genes are close enough to each other, there will be a limited number of crossover events between them. c. No, there will be very high crossover interference such that the recombination frequency will be reduced significantly. d. Yes, if the genes are far enough apart on the same chromosome, a crossover will occur between them in just about every meiotic event. e. Yes, but only if the two genes are both homozygous.

In Drosophila melanogaster, cut wings (ct) are recessive to normal wings (ct+), sable body (s) is recessive to gray body (s+), and vermilion eyes (v) are recessive to red eyes (v +). All three recessive mutations are Xlinked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The following are the progeny resulting from the testcross.

V v+ v+ v+ v+ v v v

ct ct ct+ ct+ ct ct ct+ ct+

s s s s+ s+ s+ s s+

Total

510 1 14 500 73 20 81 1 1200

35. What is the interference value shown by this cross? a. 0.42 b. 0.25 c. 0.58 d. –0.42 e. 0.13

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 07_7e 36. You are examining the following human pedigree and want to determine if the rare dominant disease allele (D) is linked to a specific DNA sequence location you are using as a molecular marker. Parental and progeny genotypes and phenotypes are indicated. Note that the father is a dihybrid at both loci, but the mother is homozygous at both loci. There is complete penetrance of the trait and a linkage phase of D-R1/d-R2 in the father. Assuming that the marker and the gene are linked, what is the best estimate of the map distance between the two loci?

a. 12 m.u. b. 50 m.u. c. 16 m.u. d. 5 m.u. e. 25 m.u.

37. The map distances for genes that are close to each other are more accurate than map distances for genes that are quite far apart because: a. with genes that are far apart, double crossovers and other multiple-crossover events often lead to lethal recombinants that reduce the number of recombinant progeny. b. with genes that are far apart, double crossovers and other multiple-crossover events often lead to nonrecombinant or parental offspring and thus reduce the true map distance. c. crossover interference will cause more double crossovers and other multiple-crossover events to occur than would be expected and thus result in a higher number of recombinant progeny than expected to occur with genes that are far apart. d. double crossovers and other multiple-crossover events occur more often when genes are close to each other and can be readily detected, so these map distances are more accurate than those for genes that are far apart. e. when genes are far apart, single-crossover recombinant classes are more difficult to detect than when genes are close together.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 07_7e 38. You are studying three genes X, Y, and Z that are linked (in that order) in the imperial scorpion, Pandinus imperator. The distance between X and Y is 10 m.u., and the distance between Y and Z is 8 m.u. You conduct a testcross by crossing a heterozygous female with a homozygous recessive male and obtain 1500 testcross progeny. When the progeny are analyzed, you find five double-crossover offspring. What is the interference value shown by this cross? a. 0.008 b. 0. 42 c. 0.12 d. 0.58 e. 0.22 39. An individual has the following genotype. Gene loci (A) and (B) are 15 m.u. apart. What are the correct frequencies of some of the gametes that can be made by this individual?

a. Ab = 7.5%; AB = 42.5% b. ab = 25%; aB = 50% c. AB = 7.5%; aB = 42.5% d. aB = 15%; Ab = 70% e. aB = 70%; Ab = 15% 40. What major contribution did Barbara McClintock and Harriet Creighton make to the study of recombination? a. Genetic recombination of alleles is associated with physical exchange between chromosomes. b. Genes are located on chromosomes, and the map distance between them could often be measured by the number of nucleotides in the DNA. c. Determining map distances in humans could be done by using pedigrees and calculating lod scores. d. Association studies allow genes that have no obvious phenotype to be accurately mapped. e. Crossing over does not occur in male Drosophila, so there is no genetic recombination.

41. Assume that an individual of AB/ab genotype is involved in a testcross and four classes of testcross progeny are found in equal frequencies. Which of the following statements is TRUE? a. The genes A and B are on the same chromosome and closely linked. b. The genes A and B are on the same chromosome and very far apart. c. The genes A and B are probably between 10 and 20 map units apart on the same chromosome. d. The genes A and B are likely located on different chromosomes. e. The genes A and B could be located on different chromosomes or on the same chromosome and very far apart.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 07_7e 42. A physical map often measures _____, whereas a genetic map measures _____. a. distances between chromosomes; distances between genes b. map units; base pairs c. centiMorgans; base pairs d. base pairs; centiMorgans e. map units; centiMorgans 43. In corn, small pollen (sp) is recessive to normal pollen (sp+) and banded necrotic tissue, called zebra necrotic (zn), is recessive to normal tissue (zn+). The genes that produce these phenotypes are closely linked on chromosome 10. If no crossing over occurs between these two loci, give the types of progeny expected from the following cross:

a. sp+ zn+/sp zn; sp zn/sp zn b. sp+ zn/sp zn; sp zn+/sp zn c. sp+ zn+/sp+ zn+; sp+ zn+/sp zn; sp zn/sp zn d. sp+ zn/sp+ zn; sp+ zn/sp zn+; sp zn+/sp+ zn; sp zn+/sp zn+ e. sp+ zn+/sp zn; sp+ zn/sp zn 44. In addition to determining genotypes, two- and three-factor testcrosses can be used to: a. map gene loci. b. screen recessive mutants. c. measure heritability. d. determine parental origin. e. determine the physical location of genes.

45. Crossing over occurs during: a. late anaphase. b. prophase. c. metaphase. d. early anaphase. e. telophase.

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 07_7e 46. Assume that you are able to examine a total of 100 separate meiotic events in an animal species. You note that within 10 of these meiotic events, there was a crossover event occurring between genes A and B. In the remaining 90 events there was no crossover event between these two genes. What would be the expected map distance between genes A and B? a. 10 m.u. b. 5 m.u. c. 20 m.u. d. 45 m.u. e. 50 m.u.

47. Assume that A and B are two linked genes on an autosome in Drosophila. A testcross is made where AB/ab flies are crossed to ab/ab flies and the progeny are counted and shown below. However, it is known that the Aa/bb genotype is lethal before the flies hatch and does not appear among the testcross progeny counted.

What is the most precise map distance between the two genes that can be calculated from these data?

Aa Bb = 235 aa bb = 225 aa Bb = 20 a. 4.2 m.u. b. 4.0 m.u. c. 16.4 m.u. d. 8.0 m.u. e. 50 m.u.

48. Recombination frequencies can be calculated by: a. counting the number of recombinant and parental offspring when an individual who is heterozygous for two genes is involved in a testcross. b. performing a chi-square test on the F2 progeny when an individual who is homozygous for two genes is initially crossed to an individual who is homozygous recessive for these two genes. c. counting the number of offspring who are expressing the dominant phenotype when a heterozygous individual for two genes is involved in a testcross. d. performing a chi-square test of the progeny of a cross between parents who are both heterozygous for the same two genes. e. counting the number of offspring that are found in the cross of an individual who is heterozygous for two genes with another parent who is homozygous dominant for one of these genes and homozygous recessive for the other gene.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 07_7e 49. In using somatic-cell hybridization experiments, a human gene was found to be located on chromosome 6.

However, when lod-score analysis was performed to detect linkage between this gene and a DNA marker locus also known to be on chromosome 6, no linkage could be found between the marker locus and the gene.

What is the MOST likely explanation for this result? a. Somatic-cell hybridization experiments are not very accurate, and the gene may be on chromosome 5 or chromosome 7 instead of chromosome 6. b. Too few recombinants could be found to indicate linkage in the lod-score analysis. c. Lod-score analysis was not applicable to the case described. d. The gene and the DNA marker locus are so far apart on chromosome 6 that they assort independently. e. There were probably too few double-crossover events occurring between the two loci, so the lod score could not be determined accurately.

50. A testcross includes: a. one parent who is homozygous recessive for one gene pair and a second parent who is homozygous recessive for a second gene pair b. one parent who is homozygous dominant for one or more genes and a second parent who is homozygous recessive for these same genes c. two parents who are both heterozygous for two or more genes d. one parent who shows the dominant phenotype for one or more genes and a second parent who is homozygous recessive for these genes e. one parent who shows the recessive phenotype for one or more genes and a second parent who is homozygous dominant for these genes

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 07_7e 51. A series of two-point crosses among fruit flies is carried out between genes for brown eyes (bw), arc wings (a), vestigial wings (vg), ebony body color (e), and curved wings (cv). The following number of nonrecombinant and recombinant progeny were obtained from each cross: Genes in cross a, bw a, cv a, e a, vg bw, cv bw, e bw, vg cv, e cv, vg vg, e

Progeny (NR) 2224 2609 3200 5172 4614 4150 2796 3116 2102 4559

Progeny (R) 117 823 3200 2379 1706 4150 1434 3116 305 4559

Using these data from two-point crosses, what it the best genetic map (in m.u.) that can be developed? a. cv 5 bw 13 a 34 vg with e assorting independently b. bw 5 cv 24 vg 32 a with e assorting independently c. a 5 bw 13 vg 24 e with vg assorting independently d. cv 13 bw 5 a 27 vg with e assorting independently e. bw 5 a 24 cv 13 vg with e assorting independently 52. What is a major difference in using lod-score analysis compared to using association studies in determining gene locations in humans? a. Lod-score analysis relies on family or pedigree data, while association studies use population data. b. Lod-score analysis requires that the loci being mapped must be on different chromosome arms, while association studies can map genes on different chromosomes. c. Association studies compare genotypes between parents and their children, while lod-score analysis compares genotypes between siblings of the same family. d. Lod-score analysis requires isolated human populations, while association studies require very large family pedigrees. e. Lod-score analysis requires a large number of genes with multiple alleles, while association studies can use genes that have only two alleles.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 07_7e 53. Consider the following three-factor (trihybrid) testcross:

Calculate the number of individuals of a+a bb c+c genotype if 1000 progeny result from this testcross. a. about 102 b. about 46 c. about 130 d. about 65 e. about 250 54. Two genes, A and B, are located 30 map units apart. The dihybrid shown below is mated to a tester aa bb.

What proportion of the offspring is expected to be dominant for both traits?

a. 0% b. 15% c. 30% d. 35% e. 70%

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 07_7e Dr. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits. He mates these two strains with each other. Dr. Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males. In the resulting testcross progeny, he observes 500 flies that are of the following makeup: 41 207 210 42

with baby blue eyes and pink wings with baby blue eyes only with pink wings only with wild-type phenotype

55. Assuming the wild-type alleles for these two genes are b+ and pw+, what is the correct testcross of the F1 flies? a. b+ pw+/b pw × b pw/b pw b. b+ pw+/b pw × b pw+/b+ pw c. b+ pw/b pw+ × b pw/b pw d. b+ pw/b pw+ × b+ pw+/b pw e. b+ pw+/b pw × b+ pw/b pw+ 56. Recombination occurs through: a. crossing over and chromosome interference. b. chromosome interference and independent assortment. c. somatic-cell hybridization and chromosome interference. d. complete linkage and chromosome interference. e. crossing over and independent assortment.

57. A double heterozygote for two linked genes in the mouse is testcrossed by crossing it to a homozygous recessive parent. In the offspring, the two parental classes appear in a frequency of 45% each, and the two recombinant classes appear in a frequency of 5% each. What is the distance in map units between the two genes? a. 45 m.u. b. 22.5 m.u. c. 10 m.u. d. 5 m.u. e. 2.5 m.u.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 07_7e 58. A testcross is carried out with the parent AB/ab. The genes A and B are completely linked. What are the expected genotypes of the progeny? a. AB/AB, ab/ab b. AABB, AaBb, aabb c. AB/ab, ab/ab d. AB/ab, ab/ab, Ab/ab, aB/ab e. AB/AB, ab/ab 59. A testcross is performed on an individual to examine three linked genes. The most frequent phenotypes of the progeny were Abc and aBC, and the least frequent phenotypes were abc and ABC. What was the genotype of the heterozygous individual that is testcrossed with the correct order of the three genes? a. Abc aBC b. BAC/bac c. bcA/BCa d. aBc/AbC e. bAc/BaC 60. Lod scores measure: a. the relatedness of two individuals. b. the number of crossover events that occur along an entire chromosome. c. how often double crossovers occur. d. the length of a linkage group. e. the likelihood of linkage between genes.

61. In soreflies (a hypothetical insect), the dominant allele, L, is responsible for resistance to a common insecticide called Loritol. Another dominant allele, M, is responsible for the ability of soreflies to sing like birds. A truebreeding mute, Loritol-resistant sorefly was mated to a true-breeding singing, Loritol-sensitive sorefly, and the singing, Loritol-resistant female progeny were testcrossed with true-breeding wild-type (i.e., mute, Loritolsensitive) males. Of the 400 total progeny produced, 117 were mute and Loritol-resistant, 114 could sing and were Loritol-sensitive, 83 could sing and were Loritol-resistant, and 86 were mute and Loritol-sensitive. A chi-square test is done to determine if there is equal segregation of alleles at the L locus. What will be the chisquare value obtained and how many degrees of freedom would be used to interpret this value? a. 0.09 and one degree of freedom b. 0.56 and two degrees of freedom c. 0 and one degree of freedom d. 9.72 and four degrees of freedom e. A chi-square test is not the appropriate statistical test to answer this question.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 07_7e In Drosophila melanogaster, cut wings (ct) are recessive to normal wings (ct+), sable body (s) is recessive to gray body (s+), and vermilion eyes (v) are recessive to red eyes (v +). All three recessive mutations are Xlinked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The following are the progeny resulting from the testcross.

V v+ v+ v+ v+ v v v

ct ct ct+ ct+ ct ct ct+ ct+

s s s s+ s+ s+ s s+

Total

510 1 14 500 73 20 81 1 1200

62. What is the CORRECT genetic map with respect to gene order and distances (in m.u.) for these three genes? a. s – 13 – ct – 3 – v b. s – 3 – v – 13 – ct c. v – 13 – ct – 3 – s d. s – 26 – v – 3 – ct e. ct – 13 – s – 3 – v 63. In soreflies (a hypothetical insect), the dominant allele, L, is responsible for resistance to a common insecticide called Loritol. Another dominant allele, M, is responsible for the ability of soreflies to sing like birds. A truebreeding mute, Loritol-resistant sorefly was mated to a true-breeding singing, Loritol-sensitive sorefly, and the singing, Loritol-resistant female progeny were test-crossed with true-breeding wild-type (i.e., mute, Loritolsensitive) males. Of the 400 total progeny produced, 117 were mute and Loritol-resistant, 114 could sing and were Loritol-sensitive, 83 could sing and were Loritol-resistant, and 86 were mute and Loritol-sensitive. What will be the results of a chi-square test for independent assortment? a. 9.70 with three degrees of freedom b. 4.63 with three degrees of freedom c. 6.48 with four degrees of freedom d. 2.54 with one degree of freedom e. Because there are four classes of offspring, the genes must be assorting independently.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 07_7e 64. Linked genes always exhibit: a. phenotypes that are similar. b. recombination frequencies of less than 50%. c. homozygosity when involved in a testcross. d. a greater number of recombinant offspring than parental offspring when involved in a testcross. e. a lack of recombinant offspring when a heterozygous parent is involved in a testcross.

Indicate one or more answer choices that best complete the statement or answer the question.

65. Genetic distances within a given linkage group: (Select all that apply.) a. cannot exceed 100 m.u. b. are dependent on crossover frequencies between paired, nonsister chromatids. c. can be measured in centiMorgans or map units. d. cannot be determined. e. can only be determined by physical mapping techniques.

66. Which of the following statements about SNPs are CORRECT? (Select all that apply.) a. SNPs are positions where individuals differ in DNA sequence at a single nucleotide. b. SNPs are found only on autosomes. c. SNPs are always located within genes. d. SNPs may be associated with specific traits. e. Genome-wide association studies (GWAS) of SNPs with human disease are typically used to identify specific genes involved in disease development.

67. Which of the following statements about recombination rates is CORRECT? (Select all that apply.) a. Recombination rates are very similar between species. b. In humans, genotypic sex influences recombination rates. c. Within one species, recombination rates can vary between chromosomes. d. Within one chromosome, recombination rates can vary. e. In humans, the highest rates of recombination occur near centromeres.

68. A geneticist finds that a human gene and a particular DNA marker locus are located on chromosome 8 on the basis of somatic-cell hybridization studies. However, when lod-score analysis is performed with these two loci using family pedigrees, no evidence for linkage between the two loci can be found. Assuming that both types of studies were performed correctly and the results are valid, how would you explain the different outcomes?

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 07_7e 69. Discuss the differences and at least one similarity between recombination and independent assortment.

70. In a two-point linkage analysis, genes a and b have been found to be 26 m.u. apart on the same chromosome.

A third gene, c, has just been discovered and found to be located between a and b. A three-point linkage analysis with a, b, and c indicates that a and b are actually 33 m.u. apart rather than 26 m.u. Why does the three-point analysis give a different map distance for a and b than does the two-point linkage analysis, and which is more accurate?

In the introduction to this chapter, you read about a study conducted by Zapata and colleagues mapping genes that affect fear and aggression in dogs.

71. Describe the technique that they used and why that approach was successful.

72. Assume that you discover a new human gene that you believe is located on the Y chromosome although not in the region (pseudoautosomal) of the Y that is homologous with part of the X chromosome. How would you map this gene with respect to the other genes on the Y chromosome?

73. Geneticists often assume that map distances less than 7 to 8 map units (m.u.) are quite accurate. Map distances that exceed this threshold significantly are assumed to be less accurate, and the level of accuracy declines as map distances increase. Briefly explain this observation.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 07_7e 74. What is the chi-square test used for, and what does it tell you?

In the introduction to this chapter, you read about a study conducted by Zapata and colleagues mapping genes that affect fear and aggression in dogs.

75. How did Zapata and colleagues corroborate their initial findings? Do you think corroboration is important for validation of scientific findings? How does the scientific community at large view corroboration?

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 07_7e Answer Key 1. d 2. c 3. a 4. d 5. a 6. c 7. a 8. c 9. d 10. d 11. c 12. a 13. b 14. b 15. d 16. d 17. a 18. e 19. e 20. c 21. a 22. d 23. e 24. c 25. d 26. b Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 07_7e 27. d 28. c 29. d 30. a 31. d 32. c 33. d 34. d 35. c 36. e 37. b 38. d 39. c 40. a 41. e 42. d 43. c 44. a 45. b 46. b 47. d 48. a 49. d 50. d 51. e 52. a 53. b 54. d Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 07_7e 55. c 56. e 57. c 58. c 59. e 60. e 61. c 62. b 63. a 64. b 65. b, c 66. a, d 67. b, c, d 68. Somatic-cell hybridization studies allow for the physical location of genetic loci to be assigned, which normally means revealing on which chromosome they are located. The two loci in this case have been both assigned to chromosome 8, but it is not known how close to each other they are. The lack of linkage revealed by the lod-score analysis indicates that the two loci are quite far apart on chromosome 8 and are not linked. It is not uncommon for two loci to be on the same chromosome but not be linked.

69. Genetic recombination (i.e., recombination due to crossing over) involves precise cleavage, physical exchange, and splicing together of chromatin between paired homologous chromosomes. By contrast, independent assortment (i.e., recombination in the case of physically unlinked genes) involves random alignment of homologous chromosome pairs along the metaphase plate, which results in a random assortment of haploid sets of individual homologs in the gametes. Both cases involve random rearrangement of alleles and contribute to the immense genetic variation present in the gametes of sexually reproducing species.

70. Multiple crossovers, particularly double crossovers, are likely to occur between a and b. In the two-point linkage analysis, some of these double crossovers are not detected since they result in parental or nonrecombinant genotypes in the offspring. This leads to an underestimation of the true map distance between a and b since they will not be counted as recombinants even though crossing over has taken place. With three-point linkage analysis, many of these double crossovers that went undetected in the two-point analysis are now detected as recombinants because of the middle gene, c. This results in a longer and more accurate map distance between a and b. Because of this, three-point linkage analysis is more accurate and is an important reason why three-point linkage analysis is preferred when it can be performed.

Copyright Macmillan Learning. Powered by Cognero.

Page 30

Name:

Class:

Date:

Chap 07_7e 71. They used a genome-wide associated study (GWAS) looking for statistically significant correlation between a behavioral trait and SNPs in 11 breeds of dogs. Using SNP correlations, chromosomal regions were mapped.

From chromosomal regions, several genes were identified. Some of the genes show a dog brain expression profile supporting their roles in fear and aggression and supporting the validity of the study.

Answers about why the approach was successful may vary. Points that can be brought up are that dogs of the same breed are genetically similar while dogs of different breeds are genetically distinct, allowing specific associations to be investigated. A sufficient number of SNPs exist and are scattered throughout the dog genome, allowing associations to be found. SNP analysis of individual dogs is fast and cheap (as it is for people).

(Note: A similar question is in the Chapter 3 Test Bank, which may reinforce the concepts of the usefulness of GWAS studies for students.)

72. A gene located on the Y chromosome should be strictly transmitted from fathers to sons. Mapping such genes is difficult because there is normally only one copy of the Y chromosome and, therefore, one copy of a gene on the Y chromosome. This means that no normal recombination is possible, and analysis of recombination is the main way that genetic mapping studies are done. One possibility is to find rare XYY men where two copies of genes on the Y chromosome will be present, and recombination may be observed in the sons of such men who are fertile.

73. The further apart two genes are, the more likely it is that multiple crossovers will occur. Double crossovers (or an even number of crossovers) between two genes may lead to parental genotypes in the offspring that will not be counted as recombinants, even though crossing over has taken place. Since genetic maps are created by counting observable recombinants in the offspring, double crossovers will lead to an underestimation of the true map distance.

74. The chi-square test is a statistical tool for analyzing data generated from genetic experiments to determine if observed results are consistent with a hypothesis proposed to explain them. The calculated chi-square statistic may be used to determine whether or not to accept or reject the proposed hypothesis within predetermined confidence limits.

Copyright Macmillan Learning. Powered by Cognero.

Page 31

Name:

Class:

Date:

Chap 07_7e 75. Zapata and colleagues did several things to corroborate their initial findings. They repeated their GWAS study on another cohort of dogs and found the same chromosomal loci associated with fear and aggression as in the original study. They also created a model which, importantly, made predictions. They tested the prediction that individual dogs will show fear and aggression based on the dog's genetic makeup and achieved a 58% success rate compared to a 7% random success rate.

Corroboration of results is extremely important for validation of scientific findings. The original group reporting findings should be able to replicate their own results and, perhaps more importantly, independent research groups should as well. Corroboration is achieved by conducting experiments (which may be the same or different from the original set) and comparing findings to previously reported findings. The scientific community relies heavily on corroboration. A lack of corroboration or refutation of findings has led scientific journals and/or authors to withdraw published findings in several cases.

One useful reference to launch a discussion of corroboration and validation of scientific findings is: Berg, Jeremy. (2019). Replication challenges. Science 365: 957.

DOI: 10.1126/science.aaz2701

Copyright Macmillan Learning. Powered by Cognero.

Page 32

Name:

Class:

Date:

Chap 08_7e Indicate the answer choice that best completes the statement or answers the question.

1. The centromere is located approximately in the middle of which type of chromosome? a. acrocentric b. metacentric c. paracentric d. submetacentric e. telocentric 2. Wild-type Arabidopsis has five chromosomes (2n = 10). Trisomic plants are designated as "Tr" followed by the trisomic chromosome number—that is, Tr1 is trisomic for chromosome 1. Assuming that trisomy is fully viable and that all possible pairing configurations (including nonpairing) are possible at meiosis, what proportion of the progeny from the cross wt × Tr1;Tr2 will have a wild-type chromosomal complement? a. 1/2 b. 1/3 c. 1/4 d. 1/9 e. 1/81 3. The centromere is at or very near the end in which type of chromosome? a. acrocentric b. metacentric c. paracentric d. submetacentric e. telocentric 4. Arabidopsis thaliana has 2n = 10 chromosomes, and a close relative, Capsella rubella, has 2n = 16. You have created a hybrid between them and suspect that it is an allotetraploid. If you are right, what possible chromosome numbers could the hybrid have? a. 20 b. 23 c. 26 d. 29 e. 32 f. 20 or 32 g. 23 or 26 or 29

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 08_7e 5. Which type of polyploidy is synonymous with amphidiploid? a. allodiploid b. allotriploid c. autotriploid d. allotetraploid e. autotetraploid 6. A human embryo is produced by the fusion of two gametes that have a normal complement of sex chromosomes. This individual develops into an adult and is diagnosed with Turner syndrome. How can this be? a. An X chromosome was lost soon after fertilization, leading to a mosaic individual with a mixture of XX and XO cells. b. Nondisjunction produced a gamete that lacked any sex chromosome. This gamete was fertilized, which resulted in the individual with Turner syndrome. c. This individual is a gynandromorph with a mixture of both male and female characteristics. d. The embryo was exposed to colchicine soon after fertilization, which produced a polyploidal individual.

Dosage compensation caused an imbalance between autosomal and X chromosome gene expression. e. A nonreciprocal translocation between the X chromosome and an autosome resulted in the loss of one of the X chromosomes.

7. Which type of chromosome mutation decreases the amount of genetic material for one entire chromosome? a. translocation b. aneuploidy c. polyploidy d. inversion e. transversion

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 08_7e 8. Two chromosomes have the following segments, where · represents the centromere: KLM·NOPQR STUV·WXYZ What type of chromosome mutation would result in the following chromosomes?

KLM·NOXYZ STUV·WPQR a. deletion b. tandem duplication c. displaced duplication d. reverse duplication e. pericentric inversion f. paracentric inversion g. nonreciprocal translocation h. reciprocal translocation 9. How is colchicine, the chemical used in preparing karyotypes, useful for studying chromosomal mutations? a. Colchicine prevents cells from entering anaphase, stalling them in metaphase with condensed chromosomes. b. Colchicine induces chromosome condensation during interphase, which allows the visualization of interphase chromosomes. c. Colchicine causes chromosomal breakage, leading to inversions and translocations that can be observed with a microscope. d. Colchicine ruptures the cell membrane, allowing the spreading of cells on microscope slides. e. Colchicine causes cells to exit the cell cycle, making the chromosome easier to visualize.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 08_7e 10. Two chromosomes have the following segments, where · represents the centromere: KLM·NOPQR STUV·WXYZ What type of chromosome mutation would result in the following chromosomes?

KLM·NORQP STUV·WXYZ a. deletion b. tandem duplication c. displaced duplication d. reverse duplication e. pericentric inversion f. paracentric inversion g. nonreciprocal translocation h. reciprocal translocation 11. What would be the consequence of a diploid gamete (resulting from meiotic nondisjunction) being fertilized by a haploid gamete from the same species? a. allodiploid b. allotriploid c. autotriploid d. allotetraploid e. autotetraploid

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 08_7e 12. Two chromosomes have the following segments, where · represents the centromere: KLM·NOPQR STUV·WXYZ What type of chromosome mutation would result in the following chromosomes?

KLM·NOPQRYZ STUV·WX a. deletion b. tandem duplication c. displaced duplication d. reverse duplication e. pericentric inversion f. paracentric inversion g. nonreciprocal translocation h. reciprocal translocation 13. Chimpanzees, gorillas, and orangutans all have 48 chromosomes, whereas humans have 46. Human chromosome 2 is a large, metacentric chromosome with G-banding patterns that closely match those found on two different acrocentric chromosomes of the apes. Which statement would BEST explain these findings? a. A translocation took place in a human ancestor, creating a large metacentric chromosome from the two long arms of the ancestral acrocentric chromosomes. The other small chromosome produced by this translocation was lost. b. Meiotic nondisjunction gave rise to a nullisomic gamete. The fusion of two nullisomic gametes produced the ancestor of the human species with 46 chromosomes instead of 48. c. A nonreciprocal translocation and subsequent fusion of the chromosomal fragments created a genome of 46 chromosomes without the loss of any genetic information. d. Infection by a primate virus created a new chromosome when the viral DNA became a permanent part of the genome. Humans were not affected by this virus, so they did not acquire the extra chromosome. e. Humans have the correct number of chromosomes. The extra chromosome pair in the other apes is a classic case of tetrasomy as a result of meiotic nondisjunction in the primate ancestor.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 08_7e 14. A man has a condition where all of his gametes undergo nondisjunction of the sex chromosomes in meiosis I, but meiosis II proceeds normally. He mates with a woman who produces all normal gametes. What is the probability that the fertilized egg will develop into a child with Turner syndrome (XO)? Assume that all gametes and zygotes are viable. a. 1/8 b. 1/4 c. 1/3 d. 1/2 e. 0 15. Which type of chromosome mutation increases the amount of genetic material for all chromosomes? a. translocation b. aneuploidy c. polyploidy d. inversion e. transversion 16. What is responsible for primary Down syndrome? a. disomy b. inversion c. nondisjunction d. polyploidy e. translocation

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 08_7e 17. Two chromosomes have the following segments, where · represents the centromere: KLM·NOPQR STUV·WXYZ What type of chromosome mutation would result in the following chromosomes?

KLM·NOPQR STUV·WZ a. deletion b. tandem duplication c. displaced duplication d. reverse duplication e. pericentric inversion f. paracentric inversion g. nonreciprocal translocation h. reciprocal translocation 18. An individual heterozygous for a reciprocal translocation possesses the following chromosomes: A B · C DE F G A B · C DVW X R S · T UE F G R S · T UVW X Will the products from alternate, adjacent-1, or adjacent-2 segregation be missing some genes? a. alternate segregation b. adjacent-1 segregation c. alternate and adjacent-1 segregation d. alternate and adjacent-2 segregation e. adjacent-1 and adjacent-2 segregation 19. Most strains of cultivated bananas were created by crossing plants within and between two diploid species: Musa acuminata (genome = AA) and Musa balbisiana (genome = BB). Some bananas have genome AAB, which is an example of which kind of polyploidy? a. allodiploid b. allotriploid c. autotriploid d. allotetraploid e. autotetraploid Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 08_7e 20. Two chromosomes have the following segments, where · represents the centromere: KLM·NOPQR STUV·WXYZ What type of chromosome mutation would result in the following chromosomes?

KLM·NOPRQQR STUV·WXYZ a. deletion b. tandem duplication c. displaced duplication d. reverse duplication e. pericentric inversion f. paracentric inversion g. nonreciprocal translocation h. reciprocal translocation 21. Which type of aneuploidy is represented in gametes labeled A in the figure?

a. nullisomy b. monosomy c. trisomy d. tetrasomy

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 08_7e 22. Which of the following statements about chromosomal inversions is NOT true? a. The individual organisms have neither lost nor gained any genetic material. b. An inversion can break a gene into two parts and separate each to different locations. c. The expression of the gene may be altered due to the positional effect. d. The inversion mutations will not likely have pronounced phenotypic effects since there is no loss of genetic material. e. Individuals heterozygous for inversions may end up producing abnormal gametes.

23. Which type of chromosome mutation DECREASES the amount of genetic material? a. deletion b. duplication c. inversion d. translocation e. transversion 24. Which type of chromosome mutation results in a chromosome segment that is turned 180 degrees? a. deletion b. duplication c. inversion d. translocation e. transversion 25. Which of the following chromosome rearrangements represents inversion?

a. A b. B c. C d. D Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 08_7e 26. Wild-type Arabidopsis has five chromosomes (2n = 10). Trisomic plants are designated as "Tr" followed by the trisomic chromosome number—that is, Tr1 is trisomic for chromosome 1. Assuming that trisomy is fully viable and that all possible pairing configurations (including nonpairing) are possible at meiosis, what proportion of the progeny from the cross Tr1 × Tr1 will have a wild-type chromosomal complement? a. 1/2 b. 1/3 c. 1/4 d. 1/9 e. 1/81 27. _____ can cause genes to move from one linkage group to another. a. Inversions b. Deletions c. Polyploidy d. Translocations e. Unequal crossing over 28. Which type of aneuploidy is represented in gametes labeled B in the figure?

a. nullisomy b. monosomy c. trisomy d. tetrasomy 29. Which of the following statements about allopolyploidy from two species is INCORRECT? a. Allopolyploidy arises from hybridization between the individuals of very different species. b. The species that undergo allopolyploidy have to be sufficiently related for the hybridization. c. The mitotic nondisjunction in sterile hybrids may result in allotetraploidy. d. The hybrid is functionally haploid and sterile. e. The sterile haploid hybrid can perpetuate itself through asexual reproduction.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 08_7e 30. A newly discovered species of dung beetle has 2n = 16 chromosomes. It mates with a closely related beetle species that has 2n = 12 chromosomes. How many chromosomes would there be in an allotriploid beetle produced from this cross? a. 13 or 17 b. 19 or 25 c. 18 or 24 d. 20 or 22 e. 36 or 48 31. Which of the following statements regarding the Drosophila Bar phenotype is NOT true?

a. Unequal crossing-over events may give rise to the duplication of the Bar region. b. The trait is inherited as completely dominant, where both homozygous and heterozygous individuals show similar phenotype. c. The more Bar segment copy the fly has, the more reduced the facets in the eyes get. d. The Bar phenotype reflects the genetic phenomenon called gene dosage effect. e. The balance of gene product is critical to cell function and normal development.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 08_7e 32. A woman has a condition where all of her gametes undergo nondisjunction of chromosome 21 in meiosis I, but meiosis II proceeds normally. She mates with a man who produces all normal gametes. What is the probability that the fertilized egg will develop into a child with Down syndrome? Assume that all gametes and zygotes are viable. a. 1/8 b. 1/4 c. 1/3 d. 1/2 e. 0 33. A plant species has 2n = 18 chromosomes. How many chromosomes would you expect to find in a tetrasomic individual of this species? a. 17 b. 19 c. 20 d. 22 e. 36 34. A man has a condition where all of his gametes undergo meiosis I normally, but there is nondisjunction of the sex chromosomes in meiosis II. He mates with a woman who produces all normal gametes. What is the probability that the fertilized egg will develop into a child with Klinefelter syndrome (XXY)? Assume that all gametes and zygotes are viable. a. 1/8 b. 1/4 c. 1/3 d. 1/2 e. 0 35. Which of the following is NOT a rearrangement that causes chromosomal mutations? a. deletion b. duplication c. inversion d. translocation e. All these rearrangements cause chromosomal mutations.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 08_7e 36. Wild-type Arabidopsis has five chromosomes (2n = 10). Trisomic plants are designated as "Tr" followed by the trisomic chromosome number—that is, Tr1 is trisomic for chromosome 1. Assuming that trisomy is fully viable and that all possible pairing configurations (including nonpairing) are possible at meiosis, what proportion of the progeny from the cross Tr1;Tr2 × Tr1;Tr2 will have a wild-type chromosomal complement? a. 1/2 b. 1/3 c. 1/4 d. 1/9 e. 1/81 37. Two chromosomes have the following segments, where · represents the centromere: KLM·NOPQR STUV·WXYZ What type of chromosome mutation would result in the following chromosomes?

KLM·NOPQR STXYUV·WXYZ a. deletion b. tandem duplication c. displaced duplication d. reverse duplication e. pericentric inversion f. paracentric inversion g. nonreciprocal translocation h. reciprocal translocation 38. Approximately _____ of the gametes produced by an individual heterozygous for a translocation will be nonviable. a. all b. half c. none d. a quarter e. an eighth

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 08_7e 39. Two chromosomes have the following segments, where · represents the centromere: KLM·NOPQR STUV·WXYZ What type of chromosome mutation would result in the following chromosomes?

KLM·NOPQR STXW·VUYZ a. deletion b. tandem duplication c. displaced duplication d. reverse duplication e. pericentric inversion f. paracentric inversion g. nonreciprocal translocation h. reciprocal translocation 40. Which form of aneuploidy describes an organism that is missing a single chromosome? a. nullisomy b. monosomy c. disomy d. trisomy e. tetrasomy 41. Which of the following is a form of aneuploidy in which four members of the same homologous pair are present? a. nullisomy b. monosomy c. disomy d. trisomy e. tetrasomy 42. Which of the following is a form of aneuploidy in which two members of the same homologous pair are absent? a. nullisomy b. monosomy c. disomy d. trisomy e. tetrasomy Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 08_7e 43. The wheat genome is larger than the human genome. a. True b. False 44. Which type of chromosome mutation is responsible for familial Down syndrome? a. disomy b. inversion c. nondisjunction d. polyploidy e. translocation 45. Which form of aneuploidy causes primary Down syndrome? a. nullisomy b. monosomy c. disomy d. trisomy e. tetrasomy 46. Wild-type Arabidopsis has five chromosomes (2n = 10). Trisomic plants are designated as "Tr" followed by the trisomic chromosome number—that is, Tr1 is trisomic for chromosome 1. Assuming that trisomy is fully viable and that all possible pairing configurations (including nonpairing) are possible at meiosis, what proportion of the progeny from the cross wt × Tr1 will have a wild-type chromosomal complement? a. 1/2 b. 1/3 c. 1/4 d. 1/9 e. 1/81 47. A man has a condition where all of his gametes undergo nondisjunction of the sex chromosomes in meiosis I, but meiosis II proceeds normally. He mates with a woman who produces all normal gametes. What is the probability that the fertilized egg will develop into a child with Klinefelter syndrome (XXY)? Assume that all gametes and zygotes are viable. a. 1/8 b. 1/4 c. 1/3 d. 1/2 e. 0

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 08_7e 48. Which of the following is represented in the figure?

a. pericentric inversion b. reciprocal translocation c. nonreciprocal translocation d. duplication e. equal crossing over 49. A newly discovered species of dung beetle has 2n = 16 chromosomes. It mates with a closely related beetle species that has 2n = 12 chromosomes. How many chromosomes would there be in an amphidiploid beetle produced from this cross? a. 12 b. 14 c. 16 d. 28 e. 12 or 16 50. Many strains of cultivated bananas were created by crossing plants within and between two diploid species: Musa acuminata (genome = AA) and Musa balbisiana (genome = BB). The Cavendish banana, also a cultivated banana and the variety most often sold in grocery stores, has a genome that is AAA. How would you describe this genome? a. allodiploid b. allotriploid c. autotriploid d. allotetraploid e. autotetraploid

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 08_7e 51. The inheritance of both chromosomes from the same parent is a condition called: a. displaced duplication. b. uniparental disomy. c. tandem duplication. d. unbalanced polymorphism. e. nondisjunction.

52. What typically happens to the small fragment generated by a Robertsonian translocation? a. The small fragment duplicates continuously. b. The small fragment remains stable. c. The small fragment often gets lost. d. The small fragment pairs up with its homologous chromosome. e. The small fragment tends to recombine back to a larger chromosome.

53. Which type of chromosome mutation INCREASES the amount of genetic material? a. deletion b. duplication c. inversion d. translocation e. transversion 54. Two chromosomes have the following segments, where · represents the centromere: KLM·NOPQR STUV·WXYZ What type of chromosome mutation would result in the following chromosomes?

KLM·NOPOPQR STUV·WXYZ a. deletion b. tandem duplication c. displaced duplication d. reverse duplication e. pericentric inversion f. paracentric inversion g. nonreciprocal translocation h. reciprocal translocation

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 08_7e 55. Which type of chromosome has a centromere displaced toward one end, creating a long arm and a short arm? a. acrocentric b. metacentric c. paracentric d. submetacentric e. telocentric 56. Which of the following chromosome rearrangements represents duplication?

a. A b. B c. C d. D 57. Which of the following is NOT a form of aneuploidy in diploid organisms? a. nullisomy b. monosomy c. disomy d. trisomy e. tetrasomy 58. The complete set of chromosomes possessed by an organism is called a: a. polyploid. b. chromatin complement. c. karyotype. d. genotype. e. diploid.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 08_7e 59. The centromere is near one end, producing a long arm and a knob, in which type of chromosome? a. acrocentric b. metacentric c. paracentric d. submetacentric e. telocentric 60. Which of the following statements about wheat is CORRECT? a. The wheat genome is haploid. b. The wheat genome is diploid. c. The wheat genome is triploid. d. The wheat genome is hexaploid and descends from an allopolyploid. e. The wheat genome of a plant growing today is considered allopolyploid.

61. What type of organism results from the hybridization of a haploid gamete from one species with a diploid gamete from a different species? a. allodiploid b. allotriploid c. autotriploid d. allotetraploid e. autotetraploid

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 08_7e 62. In order to perform karyotype analysis, chromosomes are obtained from actively dividing cells and treated with chemicals to keep them at the _____ stage of mitosis.

a. prophase b. interphase c. anaphase d. telophase e. metaphase 63. The wheat genome has more genes than the human genome. a. True b. False Indicate one or more answer choices that best complete the statement or answer the question.

64. Which of the following statements about karyotypes is CORRECT? (Select all that apply.) a. Karyotypes are arranged in order of ascending chromosome size. b. Karyotypes are prepared from chromosomes in interphase. c. To prepare a karyotype, mitotic progression in cells must be halted before chromosome isolation. d. Karyotyping involves staining after chromosome isolation. e. Chromosome size and banding patterns identify chromosomes within a karyotype.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 08_7e 65. What are three ways that aneuploidy can arise?

66. Species I is diploid (2n = 6) with chromosomes AABBCC; a related species II is diploid (2n = 6) with chromosomes MMNNOO. Indicate the chromosomes that would be found in individuals with the following chromosome mutations. a. trisomic for chromosome A b. tetrasomic for chromosome N c. an autotriploid of species I d. an allotetraploid e. an autotetraploid of species II f. monosomic for chromosome B g. a double trisomic for chromosomes A and C h. a nullisomic for chromosome O

67. In rare cases some children that express recessive genetic diseases, such as cystic fibrosis, have only one parent that is heterozygous for the disease allele. If the parents are the true biological parents of the affected child, offer a genetic explanation for this observation.

68. An individual that is heterozygous for a paracentric inversion has the following chromosomes: A B •C D E F G H I A B •C F E D G H I a. Sketch the pairing of these two chromosomes during prophase I of meiosis, showing all four strands. b. Draw the chromosomes that would ultimately result from a single crossover between the E and F segments. c. What will happen when the chromosomes separate in anaphase I of meiosis?

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 08_7e 69. Given an individual with a single recessive allele, explain how a deletion could result in the expression of the recessive phenotype.

70. You are studying the inheritance of two autosomal recessive traits, curly wings (cy) and ebony body color (e), in Drosophila. The cy locus is on chromosome 2 and the e locus is on chromosome 3. In the course of this study, you identify a strain of flies that carries a translocation between chromosome 2 and chromosome 3.

Neither cy nor e is located in the translocation region. You generate flies with the following genotype: N2(cy)/T2(cy+); N3(e)/T3(e+) N2 indicates a normal chromosome 2; T2 indicates a translocation chromosome with the centromere of chromosome 2 and a portion of chromosome 3; cy is the mutant allele associated with the curly wing phenotype, cy+ is the wild-type allele. These flies have normal wings and normal body color. a. Draw the pairing of these chromosomes during meiosis I. b. Indicate the gametes that would be generated by alternate and adjacent-1 segregation patterns during meiosis I. c. You mate two of these flies to one another. Indicate the expected genotypes and phenotypes (and their proportions) in the offspring.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 08_7e 71. Isochromosomes have the structure ABC·CBA (where · represents the centromere and A, B, and C represent wild-type alleles of three different genes). In some cases, such isochromosomes are derived from two copies of one half of a metacentric chromosome, which has a centromere near its center. For example, a wild-type metacentric chromosome ABC·DEF might form two distinct isochromosomes, ABC·CBA and FED·DEF. The centromeres of such distinct isochromosomes can be homologous so that they cause the segregation of the two distinct isochromosomes from each other during meiosis. a. A mutant of genotype ABC·CBA / FED·DEF is viable. If you cross this mutant with a wild-type individual, what would you expect to be the genotype of the offspring? Would these offspring be viable or not? Explain. b. What would you expect if you cross this mutant with another individual of genotype ABC·CBA / FED·DEF? Explain. c. You are given a pericentric inversion mutant in which this same chromosome is of genotype ABD·CEF. In an individual heterozygous for this inversion and a wild-type chromosome there is a crossover between C and the centromere. Draw the crossover event and indicate the gametes that would be generated. d. Assume that the inversion interval is very small so that both duplications and deletions for this interval (from C to D) do not cause inviability. Indicate the genotypes expected in offspring of a cross between the inversion heterozygote and an individual of genotype ABC·CBA / FED·DEF.

72. You are studying the inheritance of the recessive red-spotted (r) trait in pea plants. You identify a phenotypically wild-type plant that, when crossed to a red-spotted plant, produces 5/6 wild-type offspring and 1/6 red-spotted offspring. When you look at the chromosomes of the wild-type plant, you note that it contains an extra copy of chromosome 2, which is where the red-spotted locus is located. a. What is the genotype of the wild-type plant? What gametes will this plant generate? What are the genotypes of the wild-type and red-spotted offspring? b. You identify a plant that you think is genotypically Rrr. If you cross this plant to an rr individual, what proportion of wild-type and red-spotted offspring would you expect?

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 08_7e In the model organism C. elegans, maintaining recessive lethal mutations in a strain to allow further study can be a challenge.

73. Thus, in order to maintain recessive lethal mutations in C. elegans,balancer chromosomes are often used.

Balancer chromosomes have chromosomal rearrangements such as duplications, deletions, inversions, or translocations. Homozygotes for the balancer chromosome can be identified either by a visible phenotype or are not viable due to recessive lethality. Imagine a let/balancer chromosome hermaphrodite strain is constructed. If the balancer chromosome has a rearrangement that covers the chromosomal region let mutation, explain how the let mutation can be maintained in a heterozygous state indefinitely over many generations.

74. You are studying two different mutant fish lines (A and B). You suspect that both have mutations that cause nondisjunction for a particular chromosome, which when aneuploid results in an altered egg phenotype. You notice that all of the eggs from the A mutants have the altered phenotype. In contrast, only half the eggs laid by B mutants have the altered phenotype. If you are right and the mutants are suffering nondisjunction during meiosis, what could explain the frequency difference in egg phenotype?

75. A female rat that is heterozygous for an autosomal reciprocal translocation has 36 eggs that were generated from the following nine meioses: four by alternate segregation, four by adjacent-1 segregation, and one by adjacent-2 segregation. She is mated to a chromosomally wild-type male. What is the probability that her offspring will inherit a chromosome bearing the translocation?

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 08_7e 76. Species A and B are closely related. Species A has 2n = 12 chromosomes, and species B has 2n = 14 chromosomes. Chromosome numbers (2n) for the following related species are thought to have arisen through polyploidy. For each species, indicate what type of polyploid it is and how it may have formed. a. Species C 18 b. Species D 24 c. Species E 26 d. Species F 28 e. Species G 36 f. Species H 52 g. Species I 56

77. How can a chromosome deletion be detected?

78. Duchenne muscular dystrophy (DMD) is normally an X-linked recessive human disease affecting boys. Girls afflicted with DMD are rare. Cytogenetic studies of several girls with DMD have in each case revealed that these individuals carry X-autosome translocations. The autosomes vary, but the breakpoint on the X in every case is in band p21, which is the location of the DMD gene. Cytogenetic studies further revealed that in all cells studied in such DMD girls the normal X chromosome exists as a Barr body. How might these observations account for the existence of DMD-affected girls? Why is only the normal X, but not the translocated X, inactivated?

79. Describe two ways in which an inversion can alter gene expression.

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 08_7e 80. Explain the genetic basis for sterility in autopolyploid organisms.

81. Down syndrome has, in some cases, been found to run in families. Explain the genetic basis for this inherited form of the condition.

In the model organism C. elegans, maintaining recessive lethal mutations in a strain to allow further study can be a challenge.

82. Write out the genotypes of the F1 progreny produced from a self-fertilizing C. elegans hermaphrodite mother of the genotype +/let where '+' represents the wild-type allele and let represents a recessive lethal allele.

Using this information, explain why the +/let worms might be lost over time if only a few progeny are picked from each successive generation for maintenance. Keep in mind, each wild-type C. elegans hermaphrodite mother produces a few to several hundred progeny. (See A7, The Nematode Worm Caenorhabditis elegans.)

83. What practical applications may knowledge of the wheat genome sequence bring to society?

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 08_7e In the model organism C. elegans, maintaining recessive lethal mutations in a strain to allow further study can be a challenge.

84. Another issue in maintaining recessive lethal mutations is that recombination often occurs in the +/letstrain such that the let mutation is lost, even if the + chromosome is marked with a linked mutation. For example, imagine a strain with the genotype + let/x + where x is a recessive mutation that produces a visible phenotype. In theory, heterozygotes could be maintained by picking wild-type progeny that must be heterozygotes. Explain why.

85. Two nonhomologous chromosomes have the following segments: A •BCDEFG R •STUVWX Draw chromosomes that would result from the following chromosome rearrangements. a. Reciprocal translocation of CD and TU b. Reciprocal translocation of CD and W c. Robertsonian translocation

In the model organism C. elegans, maintaining recessive lethal mutations in a strain to allow further study can be a challenge.

86. However, recombination may occur between the let and x genes if they are not too closely linked, and that is often the case. Such a recombination event would produce two recombinant chromosomes: ++ and x let.

Explain why the let mutation would be lost by a progeny worm carrying either of the recombinant chromosomes.

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 08_7e 87. List the four basic types of chromosome rearrangements.

Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 08_7e Answer Key 1. b 2. d 3. e 4. g 5. d 6. a 7. b 8. h 9. a 10. f 11. c 12. g 13. a 14. d 15. c 16. c 17. a 18. e 19. b 20. d 21. c 22. d 23. a 24. c 25. b 26. d Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 08_7e 27. d 28. b 29. a 30. d 31. b 32. d 33. c 34. e 35. e 36. e 37. c 38. b 39. e 40. b 41. e 42. a 43. a 44. e 45. d 46. b 47. d 48. b 49. d 50. c 51. b 52. c 53. b 54. b Copyright Macmillan Learning. Powered by Cognero.

Page 30

Name:

Class:

Date:

Chap 08_7e 55. d 56. a 57. c 58. c 59. a 60. d 61. b 62. e 63. a 64. c, d, e 65. (1) Aneuploidy can arise from fertilization between balanced gametes from one parent and unbalanced gametes from the other parent. The unbalanced gametes were formed by chromosome nondisjunction during meiosis.

(2) Aneuploidy can arise from balanced gametes from one parent combining with unbalanced gametes present in individuals that are translocation carriers (e.g., familial Down syndrome). The unbalanced gametes were formed through the anomalous segregation of chromosomes during meiosis.

(3) Aneuploidy can arise from nondisjunction of chromosomes during a mitotic division in autosomal cells. This can create patches of aneuploid cells next to patches of cells with normal chromosome complements, forming mosaics.

66. a. AAABBCC b. MMNNNNOO c. AAABBBCCC d. AABBCCMMNNOO (or AAABBBCCC, MNO or ABC, MMMNNNOOO) e. MMMMNNNNOOOO f. AABCC g. AAABBCCC h. MMNN 67. The most likely (in fact, for cystic fibrosis, confirmed) explanation for these observations is an uncommon condition called uniparental disomy in which both or one of the chromosomes are inherited from the same parent. For example, in a child produced from a mating between parents that include a heterozygous carrier and a homozygous wild-type individual, two copies of the chromosome with the recessive disease allele (both from the heterozygous parent) are passed to the child. The corresponding chromosome and allele from the other parent is not transmitted to the child. Uniparental disomy is thought to originate from trisomic embryos that lose one of their chromosomes early during development and retain the two remaining chromosomes from the same parent.

Copyright Macmillan Learning. Powered by Cognero.

Page 31

Name:

Class:

Date:

Chap 08_7e

68. c. The acentromeric segment will not be directed to the nucleus and will be degraded; two chromosomes will contain full complements of genetic information; one chromosome (A B • C D E F C) will contain a duplication of locus C and be deficient for loci G, H, and I. 69. If a recessive allele for a locus is present on one chromosome, and the homologous chromosome contains a deletion of that locus, then the recessive allele will be expressed, because no wild-type product is produced to mask the recessive phenotype. The normally recessive alleles expressed in individuals heterozygous for deletions are said to display pseudodominance.

70. a.

b.

Alternate Adjacent-1 N2(cy) N3(e) N2(cy) T3(e+) T2(cy+) T3(e+) T2(cy+) N3(e) c. Flies of the genotype above produce the gametes indicated in (b). If two of these flies mate, a Punnett square of the gametes can be used to determine the genotypes and phenotypes of offspring.

N2(cy) N3(e) N2(cy) N3(e) N2(cy)/ N2(cy); Copyright Macmillan Learning. Powered by Cognero.

T2(cy+) T3(e+) N2(cy)/

N2(cy) T3(e+)

T2(cy+) N3(e)

Page 32

Name:

Class:

Date:

Chap 08_7e N3(e)/ N3(e) Curly, ebony T2(cy+) T3(e+)

N2(cy) T3(e+)

T2(cy+) N3(e)

N2(cy)/ T2(cy+); N3(e)/T3(e+) WT DEAD

DEAD

T2(cy+); N3(e)/T3(e+) WT T2(cy+)/ T2(cy+); T3(e+)/T3(e+) WT DEAD

DEAD

DEAD

DEAD

DEAD

DEAD

DEAD

N2(cy)/ T2(cy+); N3(e)/T3(e+) WT

N2(cy)/ T2(cy+); N3(e)/T3(e+) WT DEAD

1/6 of the offspring will have curly wings, ebony body color, and normal chromosome karyotype.

5/6 of the offspring will have noncurly wings and normal body color (1/6 will be homozygous for the translocation chromosomes; 4/6 will be heterozygous for the translocation chromosomes).

Copyright Macmillan Learning. Powered by Cognero.

Page 33

Name:

Class:

Date:

Chap 08_7e 71. a. The gametes from the mutant would be 1/2 ABC·CBA and 1/2 FED·DEF. All gametes from a wild-type strain would be ABC·DEF. The resultant progeny would be (i) 1/2 ABC·DEF/ABC·CBA (ii) 1/2 ABC·DEF/FED·DEF Individuals of genotype (i) are duplicated for the ABC arm of the chromosome and deficient for the DEF arm.

Likewise, individuals of genotype (ii) are duplicated for DEF and deficient for ABC. Duplications and deletions of such large regions of the chromosomes would likely result in lethality, so the offspring would not be expected to be viable. b. If the mutant were crossed with an individual carrying the same isochromosomes, three genotypes of progeny would be generated: (i) 1/4 ABC·CBA/ABC·CBA (ii) 1/2 ABC·CBA/FED·DEF (iii) 1/4 FED·DEF/FED·DEF Individuals of genotypes (i) and (iii) would likely be inviable. Individuals of genotype (ii) are identical to the parents. c. The following diagrammed crossover will generate the following gametes in the inversion heterozygote:

ABD·CEF (parental) ABD·CBA (recombinant) FED·CEF (recombinant) ABC·DEF (parental) d. In a cross with the ABC·CBA/FED·DEF individual, the parental gametes would all result in lethality due to significant duplications and deletions: (1) ABC·CBA/ABD·CEF (2) ABC·CBA/ABC·DEF (3) FED·DEF/ABD·CEF (4) FED·DEF/ABC·DEF However, the recombinant gametes from the inversion heterozygote can result in four classes of progeny: (i) ABC·CBA/ABD·CBA lethal (ii) ABC·CBA/FED·CEF Dp(C), Df(D) (iii) FED·DEF/FED·CEF lethal (iv) FED·DEF/ABD·CBA Dp(D), Df(C) Since duplication and deletion of the inverted region do not cause inviability, classes (ii) and (iv) will be viable.

Class (ii) individuals have an extra copy of C and are missing a copy of D. Class (iv) individuals are missing a copy of C and have an extra copy of D.

Copyright Macmillan Learning. Powered by Cognero.

Page 34

Name:

Class:

Date:

Chap 08_7e 72. a. The wild-type plant is RRr. To be wild-type in phenotype, it must possess at least one R allele. However, if it were Rr, the expected offspring from the testcross to the rr individual would be 1/2 wild-type and 1/2 red-spotted.

The extra chromosome suggests that the wild-type individual could be either RRr or Rrr. The gametes produced by the first plant would be RR, Rr, R, and r, and by the second plant, Rr, rr, R, and r. However, the distribution of the gametes in either case will not be even. If gametes with the extra chromosome 2 are just as viable as those with only one copy, then we would expect approximately half of the gametes to carry a single copy and half of the gametes to carry two copies (very rarely there could be gametes with either three copies of a single chromosome or no copies of a chromosome, but these outcomes are significantly less likely and so will not be treated in the probability calculations that follow). We can calculate the probability of each gamete produced by an RRr individual as follows: p(r) = 1/3 [probability of the r allele] × 1/2 [probability of only getting one chromosome during segregation] = 1/6 p(R) = 2/3 [probability of the R allele] × 1/2 = 1/3 p(RR) = 2/3 [probability of the first chromosome carrying the R allele] × 1/2 [probability of the second chromosome carrying the R allele] × 1/2 [probability of getting two chromosomes during segregation] = 1/6 p(Rr) = (2/3 [probability of the first chromosome carrying the R allele] × 1/2 [probability of the second chromosome carrying the r allele] × 1/2) + (1/3 [probability of the first chromosome carrying the r allele] × 2/2 [probability of the second chromosome carrying the R allele] × 1/2) = 1/3 Therefore, the offspring from a testcross to rr should be 1/6 rr (red-spotted), 1/3 Rr (wild-type), 1/6 RRr (wildtype), and 1/3 Rrr (wild-type). b. The fraction of gamete types produced by the Rrr individual should be as follows: p(R) = 1/3 × 1/2 = 1/6 p(r) = 2/3 × 1/2 = 1/3 p(Rr) = (1/3 × 2/2 x 1/2) + (2/3 × 1/2 × 1/2) = 1/3 p(rr) = 2/3 × 1/2 × 1/2 = 1/6 Thus, the offspring would be 1/6 Rr (wild-type), 1/3 rr (red-spotted), 1/3 Rrr (wild-type), and 1/6 rr (redspotted) or, phenotypically, 1/2 wild-type and 1/2 red-spotted. Therefore, in this cross it is difficult to distinguish this aneuploid individual from a normal heterozygous individual, since both produce a 1:1 ratio of wild-type : redspotted in a testcross.

73. The let/balancer chromosome hermaphrodite mother will produce progeny that are let/let homozygotes and balancer chromosome/balancer chromosome homozygotes. The first class will not be viable due recessive lethality. The second class will either be easily identifiable due to a visible phenotype or not viable due to recessive lethality. The hermaphrodite let/balancer chromosome mother will also produce let/balancer chromosome progeny. These will have a wild-type phenotype which is distinct from the two classes of homozygotes and thus easily identifiable. Also, because there is chromosomal rearrangement on the homolog of the let chromosome, recombination will be suppressed and the let mutation will be maintained.

74. In mutant A nondisjunction occurs in meiosis I, resulting in half the gametes being n + 1 and half being n – 1. In mutant B nondisjunction occurs in meiosis II, resulting in a fourth of the eggs being n + 1, a fourth being n – 1, and half being n.

Copyright Macmillan Learning. Powered by Cognero.

Page 35

Name:

Class:

Date:

Chap 08_7e 75. Both adjacent-1 and adjacent-2 segregation will produce nonviable offspring because the resulting zygote will have a duplication or deletion of large portions of the reciprocal chromosomes. Therefore, only the offspring derived from gametes of the alternate segregation need be considered. Half of the meiotic products from alternate segregation have translocation chromosomes, so the probability is 50% that the viable offspring will bear these chromosomes.

76. a. Species C: This is a triploid that could have arisen through nondisjunction of the entire chromosome set in a gamete from species A that subsequently fertilized with a normal haploid (n = 6) gamete. b. Species D: This is a tetraploid of species A that could have arisen through chromosome doubling of both gametes prior to their fusion or of the diploid cell(s) just after gamete fusion. c. Species E: This is an allotetraploid (also called an amphidiploid) formed by fusion of gametes from species A and species B followed by chromosome doubling or by two chromosome doublings followed by gamete fusion. d. Species F: This is an autotetraploid of species B that could have arisen through chromosome doubling of both gametes prior to their fusion or of the diploid cell(s) just after gamete fusion. e. Species G: This is an autohexaploid of species A that could have arisen through fertilization of a tetraploid-like species D by a gamete having doubled chromosomes (possibly also caused by fusion of two gametes containing tripled chromosomes). f. Species H: This is an allooctaploid that could have arisen through doubling of chromosomes in tetraploid species D and tetraploid species F plants followed by gamete fusion or gamete fusion followed by chromosome doubling. g. Species I: This is an autooctaploid that could have arisen through doubling of chromosomes in two different plants of tetraploid species F and subsequent fusion of their gametes or normal gamete fusion in species F followed by chromosome doubling.

77. (1) Large deletions can be easily seen by means of microscopy because the chromosome will be shortened. In heterozygotes, this will be obvious because sections of the longer chromosome (corresponding to the region missing in the truncated chromosome) will form unpaired loops of chromatin during meiotic pairing.

(2) If the deletions are relatively small and contain known genetic markers, no recombination will be observed for the markers located within the deleted section.

(3) In deletion heterozygotes, phenotypic expression of recessive alleles, present on the undeleted chromosome, can be expressed. This phenomenon is called pseudodominance.

(4) Deletions can also cause imbalances in the formation of quantifiable gene products (i.e., gene dosage difference). For example, quantification of specific gene products (e.g., levels of protein or mRNA, biochemical assays for production, or activities of specific enzymes) in individuals with different genotypes (e.g., heterozygote versus homozygote) could indicate the number of functional alleles present. Note in this case, however, that the presence of a nonfunctional allele would cause the same phenotype as a deleted allele because in neither case would product be formed.

Copyright Macmillan Learning. Powered by Cognero.

Page 36

Name:

Class:

Date:

Chap 08_7e 78. The translocation breakpoint likely falls in or near the wild-type DMD gene, such that these DMD girls are heterozygous for a wild-type allele and a disrupted DMD gene. Inactivation of the normal X chromosome in all cells of these females eliminates wild-type DMD gene activity. The females thus exhibit the DMD disease phenotype. Preferential inactivation of the untranslocated X chromosome may occur because autosomal sequences adjacent to X chromosomal sequences on the translocated chromosome prevent inactivation. In this case, the wildtype X chromosome would be inactivated in all cells by default. Alternatively, X inactivation may occur randomly in the translocation heterozygotes. However, cells that inactivate the translocated chromosome would also be expected to inactivate autosomal genes; these cells may die due to inappropriate reduced gene dosage of autosomal genes. In either case, consistent inactivation of the wild-type X chromosome would eliminate wild-type DMD expression in these girls, resulting in the development of DMD. 79. (1) Position effect: Inversions may reposition alleles in different genomic contexts, which may significantly alter their expression. For example, alleles may be repositioned to a heterochromatic region and be inhibited or to a highly active region and be induced.

(2) Gene disruption: Because inversions involve cutting pieces of chromosomes and splicing pieces of chromosomes back together, gene sequences may be disrupted. Breakpoints can occur in gene regulatory regions, coding domains, introns, and so forth, any or all of which may ultimately alter phenotypic expression.

80. In autopolyploids, all the chromosome sets come from the same species; thus, all the chromosomes are homologous and attempt to align during meiotic prophase I. The result is that some of the chromosomes align, others do not align, and sometimes all the homologs align. During anaphase, the chromosomes segregate unevenly, producing gametes containing various unpredictable numbers of each homolog in them; some gamete cells may even be missing specific homologs. Subsequent fertilization frequently results in formation of aneuploid zygotes, which are genetically unbalanced and usually unviable.

81. The most common form of Down syndrome involves trisomy 21, which results from nondisjunction during meiosis in one parent. The rare cases of heritable familial Down syndrome are caused by the translocation of most of chromosome 21 to the end of chromosome 14, forming a 14/21 translocation. The compound chromosome (14/21 translocation) can be transmitted through an individual's gametes and has been confirmed to run in families.

82. The three genotypes descending from a +/let mother are +/+, +/let, and let/let. The let/let worms are not viable since there are homozygous for the recessive lethal mutation. Phenotypically, the +/+ and +/let worms will look identical to each other, that is, are wild-type. So, if a geneticist picks a few worms to a new plate, both +/+ and +/let worms will be picked. +/+ worms will produce 100% +/+ progeny. Of viable progeny, +/let worms will segregate 1/3 +/+ and 2/3 +/letprogeny. However, the brood size will be smaller compared to its +/+ siblings because 1/4 of progeny are lost due to recessive lethality. Thus, the chances of picking predominantly or exclusively +/+ worms becomes higher at each successive generation, and eventually worms with the genotype +/let will be lost.

83. Feedback may vary, and there are many possible answers. Genetic information may allow scientists to improve wheat yield, disease resistance, stress tolerance, nutritional value, plant size (reduce), lifecycle length (decrease), nutrient requirements (decrease or find cheaper supplements), and so forth.

84. All heterozygotes should be wild-type. One of the two homozygous classes (+ let/+ let) would be inviable, and the other (x +/x+) would show the “X” visible phenotype and not be picked.

Copyright Macmillan Learning. Powered by Cognero.

Page 37

Name:

Class:

Date:

Chap 08_7e 85. a.

A •BTUEFG R •SCDVWX b.

A •BWEFG R •STUVCDX c. X W V U T S • B C D E F G (A and R segments will be lost) 86. At fertilization, if a ++ recombinant chromosome pairs with a + let chromosome, the same problem as a) will be encountered. If the ++ recombinant chromosome pairs with x + chromosome, then the let mutation is already lost.

If a x let recombinant chromosome pairs with a + let chromosome, the worm will not be viable. If a x let recombinant chromosome an x + chromosome, the worm will show the x phenotype and, thus, not be picked.

87. Deletions, duplications, inversions, and translocations

Copyright Macmillan Learning. Powered by Cognero.

Page 38

Name:

Class:

Date:

Chap 09_7e Indicate the answer choice that best completes the statement or answers the question.

1. The transfer of DNA from a donor cell to a recipient cell through a cytoplasmic connection is called: a. transformation. b. transduction. c. the lysogenic cycle. d. the lytic cycle. e. conjugation.

2. Which type of transduction is used to map distances between phage genes? a. generalized transduction b. specialized transduction c. targeted transduction d. random transduction e. discontinuous transduction 3. leu– bacteria are mixed in a flask with leu+ bacteria, and soon all bacteria are leu+. However, if the leu– cells are on one side of a U-tube, which is divided into two compartments by a filter with fine pores, and the leu+ cells are on the other, the leu– cells do not become prototrophic. Which process is likely to produce this observed result? a. conjugation b. transduction c. transformation d. reciprocal translocation e. transfection 4. Which of the following statements is FALSE? a. When Hfr cells are crossed with F– cells, most of the recipient cells remain F– . b. The F plasmid or factor is capable of integrating into the bacterial chromosome at several different sites. c. F+ cells are normally capable of undergoing conjugation with F– and Hfr recipient cells. d. F+ cells can convert F– cells to F+ upon conjugation. e. Hfr cells are capable of transferring part of the bacterial chromosome to recipient cells upon conjugation.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 09_7e 5. A bacterium of genotype a+b+c+d+ is the donor in a cotransformation mapping. The recipient is a– b– c– d– .

Data from the transformed cells (genotype and percent of each cotransformant) are shown below. What is the order of the genes?

Genotype a+ and b+ a+ and c+ a+ and d+ b+ and c+ b+ and d+ c+ and d+

Cotransformants 2% 0% 5% 5% 0% 0%

a. a c b d b. a d c b c. c b a d d. c a d b e. b c d a 6. Which of the following statements about bacterial genomes is NOT true? a. All bacteria contain a single circular double-stranded DNA as their genome. b. Some bacteria may have linear chromosomes instead of a circular one. c. In addition to a chromosome, many bacteria possess small extrachromosomal DNA called plasmids. d. Each plasmid contains an origin of replication that allows independent replication for its maintenance. e. The F factor, which is important for bacterial conjugation, is found as a circular episome of E. coli.

7. The human immunodeficiency virus (HIV) is what type of virus? a. a DNA virus that can infect a type of human blood cell b. a double-stranded RNA virus that can recombine with the genomes of similar viruses from other species c. a single-stranded RNA virus that can undergo reverse transcription d. a single-stranded DNA virus that uses reverse transcriptase to make a provirus e. a double-stranded DNA virus that can infect both humans and chimpanzees simultaneously

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 09_7e 8. Bacterial strains that can produce all the necessary compounds and therefore grow on minimal media are called: a. autotrophs. b. heterotrophs. c. prototrophs. d. omnitrophs. e. auxotrophs.

9. Plaques produced by the temperate phage lambda on E. coli are turbid, since some bacteria remain viable within the plaque area. Plaques produced by the virulent phage T4 are usually clear. Which statement is the BEST explanation for this difference? a. Lambda can propagate via the lytic or lysogenic cycle. b. T4 carries a smaller DNA genome than lambda. c. Recombination occurs more frequently between lambda genomes than T4 genomes. d. Lambda often carries a portion of the sex factor. e. T4 will not infect E. coli cells that are growing.

10. What does the enzyme reverse transcriptase do? a. Using the amino acid sequence of a protein as a template, it makes an RNA molecule. b. Using RNA as a template, it makes a DNA molecule. c. Using RNA as a template, it makes an RNA molecule. d. Using DNA as a template, it makes an RNA molecule. e. Using DNA as a template, it makes a DNA molecule.

11. Which of the following will have the LEAST influence on the efficiency of transformation in E. coli bacteria? a. calcium chloride treatment b. heat shock c. electrical field d. chilling on ice e. using high concentrations of DNA 12. Which of the following statements about antibiotic resistance in bacteria is NOT true? a. Antibiotic resistance cannot be conferred by conjugation as conjugation only affects the fertility of bacteria. b. The antibiotic resistance gene can be transmitted to bacteria via transformation or transduction. c. Environments where antibiotics are frequently used, such as hospitals, are under higher risk of giving rise to bacteria with antibiotic resistance. d. Antibiotic resistance often originates from the microbes that produce antibiotics for their own survival. e. The plasmid containing the antibiotic resistance gene can pass the gene to genetically unrelated bacteria.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 09_7e 13. Some retroviruses contain genes that can stimulate cell division and cause tumors in host cells. Such genes are called: a. oncogenes. b. lytic genes. c. prophage genes. d. capsid genes. e. envelope genes.

14. Which explanation is one of the reasons for the ability of the influenza virus to evolve rapidly? a. Its single-stranded DNA genome is sensitive to environmental factors that create mutations. b. The enzyme that replicates its RNA genome is prone to making mistakes that result in mutations. c. Its ability to integrate into the host genome allows it to remain in host cells for long periods of time. d. It can convert its RNA genome into a DNA copy by using the enzyme reverse transcriptase. e. It has the ability to replicate its genome both within the host cell and outside of it.

15. Cotransformation between two genes is more likely if they are: a. close to one another. b. far apart from one another. c. both next to the F factor. d. both oriented in the same direction. e. not located on the same chromosome.

16. To MOST readily demonstrate transformation of bacteria in the laboratory, one could extract DNA from: a. an auxotroph and add it to prototrophic cells. b. arg– cells and add it to arg+ cells. c. streps (sensitive) and add it to strepr (resistant) cells. d. arg+ cells and add it to arg– cells. e. both arg– and arg+ cells, mix them, and select for recombinants.

17. Which of the following horizontal gene transfer mechanisms specifically uses time as a basic unit of mapping? a. transformation b. crossing over c. transduction d. conjugation e. recombination

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 09_7e 18. A wild-type strain (h+r+) of phage T2 was mixed with a double mutant (hr) for host range (h) and rapid lysis (r) and used with a high multiplicity of infection to infect E. coli B. The progeny from the cross were titered (counted) on a mixture of E. coli B and E. coli B/2 strains. The following plaques were scored: hr,212; h+r,114; hr+,106; h+r+,188 What is the approximate recombination frequency for these two loci? a. 35% b. 24% c. 14% d. 11% e. 65% 19. What is the result of conjugation between F´ and F– cells? a. one F+ cells b. two F´ cells c. two F+ cells d. one Hfr cell and one F– cell e. two Hfr cells 20. You perform interrupted-mating experiments on three Hfr strains (A, B, and C). Genes are transferred (from last to first) in the following order from each strain: strain A, thi-his-gal-lac-pro; strain B, azi-leu-thr-thi-his; strain C, lac-gal-his-thi-thr. How are the F factors in these strains oriented? a. A and B are oriented in the same direction. b. B and C are oriented in the same direction. c. A and C are oriented in the same direction. d. All of them are oriented in the same direction. e. It cannot be determined from the information given.

21. Which of the following statements about genetic exchange in bacteria is NOT true? a. In conjugation there has to be a physical connection between the donor cell and the recipient cell. b. Plasmids do not have to integrate into the host cell chromosome in order to be replicated. c. Interrupted conjugation results in the production of Hfr strains. d. The order of gene transfer is not the same for different Hfr strains. e. Antibiotic resistance can be transferred from one bacterial cell to another by conjugation.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 09_7e 22. Which of the following statements about bacteria is INCORRECT? a. Bacteria in oceans provide 50% of oxygen in the air. b. Bacterial first arose approximately 3.5 billion years ago. c. Bacteria are found only in temperate environments. d. Bacteria are the most diverse biological group of organisms. e. Agricultural soil typically has fewer bacterial species than forest soils.

23. The life cycle of virulent phages that always kill their host cell and never become inactive prophages would be the _____ cycle. a. lethal b. lytic c. temperate d. strict e. lysogenic 24. Two different strains of a mutant phage infect a single bacterium. One phage strain is d– and the other is e– .

Some of the progeny phages are genotype d+e+, and some are d– e– . What genetic phenomenon does this demonstrate? a. complementation b. specialized transduction c. generalized transduction d. recombination e. conjugation 25. Which of the following statements about retroviruses is FALSE? a. All retroviruses contain oncogenes, which can induce the formation of tumors. b. All retroviruses contain gag genes whose product forms the viral protein coat. c. All retroviruses require pol genes, which are critical for retrotranscription. d. All retroviral genomes have gag, pol, and env genes. e. Not all RNA viruses are retroviruses.

26. How are Hfr strains of bacteria different from F+ strains? a. Cells of F+ strains are able to transfer chromosomal genes, whereas cells of Hfr strains cannot. b. Cells of Hfr strains cannot initiate conjugation with F– cells. c. The F factor is integrated into the bacterial chromosome in all or most cells of an Hfr strain but in only a few cells in an F+ strain. d. Cells of Hfr strains carry F´ plasmids, whereas F+ cells do not. e. Cells of Hfr strains can initiate conjugation with F+ cells or other Hfr cells.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 09_7e 27. A new strain of H1N1 influenza virus that can infect humans consists of RNA parts from which three strains of influenza virus? a. swine, human, and chimpanzee b. bird, human, and monkey c. human, bird, and chimpanzee d. human, bird, and swine e. human, monkey, and chimpanzee 28. Two mutations that affect plaque morphology in a particular phage (a– and b– ) have been isolated. Phages carrying both mutations (a– b– ) are mixed with wild-type phage (a+b+) and added to a culture of bacterial cells. Following infection and lysis, samples of the phage lysate are collected and cultured on bacterial cells.

The following numbers of plaques are observed: Plaque Phenotype a+b+ a+ba-b+ a-b-

Number 2043 320 357 2134

What is the frequency of recombination between the a and b genes? a. 7% b. 14% c. 42% d. 28% e. 36% 29. A bacterial cell transfers chromosomal genes to F– cells, but it rarely causes them to become F+. The bacterial cell is: a. Hfr. b. lysogenic. c. auxtrophic. d. lytic. e. F+.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 09_7e 30. Which of the following statements concerning the role of reverse transcriptase in the life cycle of certain RNA viruses is FALSE? a. It is produced from the pol gene. b. It makes DNA from an RNA template. c. It is essential for producing a provirus. d. It is part of the life cycle of retroviruses. e. It is very accurate and leads to a low mutation rate.

31. You cross an Hfr strain of ala+ his– val+ genotype with an F– strain of ala– his+ val– genotype and wish to recover F– ala+ his+val– recombinants. Which of the following media should you use that will allow the recombinants to grow and form colonies but not allow growth of the two parental strains? a. Minimal medium with histidine and valine but no alanine b. Minimal medium with valine but no alanine or histidine c. Minimal medium with histidine and alanine but no valine d. Minimal medium with no histidine, alanine, or valine e. Minimal medium with alanine but no histidine or valine 32. The figure below shows a partial chromosome map of an E. coli Hfr strain. Each mark equals 10 minutes. If transfer of genes begins at "*" and goes in the direction of the arrow, which of the predicted results from this map is MOST likely to be observed?

a. gal will be the first and ton will be the last gene to be transferred. b. lac and azi will rarely be transferred together. c. Ten minutes after transfer of ton, azi will be transferred. d. It would take 30 minutes to transfer all of the genes shown. e. All the chromosomal genes will be transferred by the end.

33. Bacterial mutants that require supplemental nutrients in their growth media are called: a. autotrophs. b. heterotrophs. c. prototrophs. d. omnitrophs. e. auxotrophs.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 09_7e 34. Which of the following features would NOT be considered an advantage for using organisms such as bacteria and viruses for genetic studies? a. rapid reproduction and high progeny number b. haploid genome for expressing mutations c. complete absence of recombination, which maintains the integrity of the genome d. low cost to maintain and little storage space required e. genomes being small and readily subjected to genetic manipulation 35. What of the following is NOT an advantage of using genomic approaches to examine bacteria? a. Bacteria that cannot be cultured in the lab can still be studied. b. Genomic data can provide information about bacterial diversity. c. Genomic data can provide information about bacterial evolution. d. Genomic data can provide information about genome organization. e. Genomic data have replaced the requirement for bacterial lab culture in disease diagnosis.

36. Two different strains of a mutant phage infected a single bacterium. One phage strain is d– e+ and the other is d+e– . The phages go through the lytic cycle and produce progeny. What will be the recombinant genotypes in the progeny? a. d– e+ and d+e– b. d+e+ and d– e– c. d– e+ and d+e+ d. d+e+ and d+e– e. d– e+ and d– e– 37. Which of the following statements about nutritional requirements and growth of bacteria is NOT true? a. Culture media developed for bacteria must contain a carbon source and essential elements for the survival of the bacteria. b. Auxotrophic mutants can grow on medium that lacks a carbon source because they can synthesize their own nutrients. c. Each bacterium has specific nutritional needs and conditions for successful cultivation. d. Prototrophic bacterial strains can grow on minimal media. e. The growth rate of bacteria on specific media can be assessed by the number and size of bacterial colonies.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 09_7e 38. Bacterial cells containing an F plasmid that has acquired bacterial chromosomal genes are called: a. F+. b. F´. c. F– . d. Hfr. 39. The process of transferring DNA from one bacterium to another through a bacteriophage is called: a. conjugation. b. induction. c. transformation. d. transduction. e. infection.

40. When the F factor integrates into the E. coli chromosome, the result is an _____ strain. a. Hfr b. F– c. F+ d. F´ e. F+/– 41. Integrated, inactive phage DNA is called a: a. progeny. b. prophage. c. transformant. d. transductant. e. conjugate.

42. HIV belongs to a group of viruses called _____ viruses. a. dsDNA b. ssDNA c. ssRNA-RT d. dsDNA-RT e. ssRNA

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 09_7e Indicate one or more answer choices that best complete the statement or answer the question.

43. Why is the bacterium Mycobacterium leprae of interest to geneticists? (Select all that apply.) a. It is the cause of leprosy, a current public health problem. b. It has a large number of pseudogenes which is unusual. c. It has no close relatives in the genus Mycobacterium. d. It has an unusual shape. e. All of the above.

44. Candidate phyla radiation (CPR) is a newly discovered group of bacteria. Which of the following statements about this group are CORRECT? (Select all that apply.) a. The group contain up to 90 different phya. b. The group has been studied primarily through specialized lab culture methods. c. Members of this group may be symbionts. d. The genome size of this group is considered large. e. The cell size of group members is smaller than most bacteria previously studied.

45. In order to better understand arginine biosynthesis in bacteria, a microbial geneticist might first isolate mutant bacterial strains. a. What characteristics must these mutant bacteria have? b. Outline a strategy for isolating such mutants. c. List three possible methods for mapping the genetic location of the mutations in these strains.

46. (a) Explain the mechanism that leads to rapid evolution of the virus that causes influenza. (b) Distinguish between antigenic drift and antigenic shift, and explain the significance of each to influenza evolution and the occurrence of influenza in humans.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 09_7e 47. Both retroviruses and lysogenic bacteriophages employ a mechanism that allows them to be replicated and passed from cell to cell without producing viruses. What is the common mechanism that these two very different viruses use?

48. (a) Explain how chromosomal genes are transferred from donors to recipients when cells of an F+ strain are mixed with F– cells. (b) Explain why transfer of chromosomal genes occurs at a higher frequency when cells of an Hfr strain are mixed with F– cells.

49. What causes an F– cell to be converted to F+?

50. What causes an F– cell to be converted to Hfr in the presence of F+ cells?

51. (a) What is an F´ plasmid, and how is it formed? (b) Explain how an F´ can be used to construct a bacterial strain that is partially diploid. (c) Explain how partial diploid strains can be used to assess interactions between different alleles (e.g., lac + and lac– ).

52. Explain the significance of horizontal gene transfer to bacterial evolution and to our ability to discern relationships between different groups of bacteria.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 09_7e 53. What is the difference between specialized and generalized transduction?

54. Two phage phenotypes are controlled by the genes a and b. In a mapping experiment, a culture of bacteria is infected simultaneously with an a– b+ strain and an a+b– strain. When plaques are analyzed, five out of 1000 have the a+b+ or a– b– phenotype. Based on the information, how far apart are genes a and b?

55. List and describe three different ways that DNA from one bacterium can be transferred into bacterial cells.

56. HIV has a high mutation rate. What causes this, and how might this be advantageous to the virus?

57. You are studying a new phage that infects H. pylori. You have isolated two mutant strains of the phage, each producing a different plaque phenotype due to a specific mutation: rough (r) and big (b). You co-infect H. pylori with both strains by adding a mixture of phages to a culture of cells. You collect the cell lysate containing progeny phages, plate diluted phages on a lawn of H. pylori cells, and observe 970 rough plaques, 890 big plaques, 0 rough and big plaques, and 0 normal, wild-type plaques. a. What is the recombination frequency between the r locus and the b locus? b. How can you explain the results?

58. How does a virulent phage differ from a temperate phage?

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 09_7e 59. Outline the steps involved in mapping a bacterial chromosome by cotransformation.

60. A retrovirus has an RNA genome but integrates into the DNA chromosome of a host cell. Explain how it does this.

61. What are plasmids, and what purposes do they serve?

62. A virulent bacteriophage is used to infect a prototrophic bacterial culture. Phages are collected from the culture and are used to infect a new bacterial strain that has several auxotrophies. After infection, rare prototrophs are found: met+ leu+ met+ pro+ met+ his+ pro+ leu+ pro+ his+ leu+ his+

0 1 0 2 0 1

How are the auxotrophy genes organized on the bacterial chromosome?

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 09_7e Answer Key 1. e 2. a 3. a 4. c 5. c 6. a 7. c 8. c 9. a 10. b 11. d 12. a 13. a 14. b 15. a 16. d 17. d 18. a 19. b 20. a 21. c 22. c 23. b 24. d 25. a 26. c Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 09_7e 27. d 28. b 29. a 30. e 31. b 32. b 33. e 34. c 35. e 36. b 37. b 38. b 39. d 40. a 41. b 42. c 43. a, b 44. a, c, e 45. a. Each mutant strain must be auxotrophic for the amino acid arginine. This characteristic of mutant strains can be used to determine the number of genes, the locations of genes, and the functions of genes that encode proteins used for arginine biosynthesis. b. Bacteria would first be plated on complete media (i.e., media containing arginine and other essential nutrients).

Next, bacteria would be replica-plated both to Petri plates containing media that lacked arginine and to duplicate plates containing media that contained arginine. Analyzing the growth of the strains on these two types of media would reveal auxotrophs: those that grow on media containing arginine but do not grow on media lacking arginine. c. Three methods are (1) interrupted conjugation, (2) cotransformation, and (3) generalized transduction.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 09_7e 46. (a) The enzyme that copies the RNA genome of the influenza virus is prone to replication errors, leading to a high rate of mutation and rapid production of new genetic variants upon which natural selection can act. (b) Antigenic drift is the continual change in influenza proteins that results from mutations due to replication errors.

Antigenic drift involves fairly small changes in the virus from year to year. Although the changes are not dramatic, they are significant in that most humans are not immune to the new variants, and a new flu vaccine must be created each year. Antigenic shift is the production of a fundamentally new type of influenza virus through reassortment that occurs when two or more different strains of virus infect a single cell. Some of the progeny will carry fundamentally new combinations of genetic traits. Typically, very few people will be immune to the new strain, giving rise to the potential for widespread infection or pandemic. Significantly, reassortment is unpredictable, so that vaccines typically are not available until after the new strain has spread throughout the world.

47. Both integrate their viral genome into a host genome, so that when the cell DNA is replicated, the viral genetic information is replicated as well.

48. (a) Within a few cells of an F+ strain, the F factor will integrate into one of several positions of the bacterial chromosome and become Hfr. When the integrated F factor initiates conjugation, genes of the bacterial chromosome will be transferred behind the leading part of the F factor. Once inside the recipient, the transferred chromosomal genes can be recombined into the recipient's chromosome. Transfer of chromosomal genes occurs at only a low frequency because relatively few cells in an F+ strain are Hfr. (b) Almost all of the cells of an Hfr strain contain an F factor that is integrated into the bacterial chromosome. All of these cells have the ability to transfer chromosomal genes to F– cells, so the frequency of transfer is high.

49. The F factor plasmid in an F+ cell is replicated, and one copy is transferred to the F– cell through conjugation.

50. The F– cell must first receive an F factor plasmid by conjugation with an F+ cell. Once inside the recipient cell, the F plasmid can integrate into the bacterial chromosome, converting the cell to Hfr. 51. (a) An F´ is an F factor that carries one or two genes from the bacterial chromosome. An F´ forms when an F factor excises imprecisely from the chromosome, carrying a small part of the chromosome with it. (b) When a cell carrying an F´ contacts an F– cell, the F´ is copied and transferred to the F– cell. The recipient cell will then possess two copies of the gene or genes that are carried on the F´. (c) It is more difficult to assess interactions between alleles in bacteria because bacteria normally are haploid and the alleles normally do not occur together in a cell or strain. However, F´ strains can be used to construct partial diploid strains that allow for these interactions to be tested. For example, a strain can be established that is heterozygous for the lac gene, allowing one to determine the dominance relationship between lac+ and lac– .

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 09_7e 52. Horizontal gene transfer is the transfer of genes from one type of bacteria to another through conjugation, transformation, or (less likely) transduction. In some cases, genes can be transferred horizontally between bacteria that are not closely related. Horizontal gene transfer allows bacteria of different types to exchange genetic variation and to acquire new genetic traits that can be tested by natural selection. Some evidence suggests that horizontal gene transfer is extensive in nature. For example, about 17% of the E. coli genome is thought to have come from other types of bacteria. As a result, some bacteriologists question whether concepts of species even apply in bacteria, given that a particular bacterium can contain parts of a variety of genomes from different groups.

Horizontal gene transfer complicates molecular analysis of relationships between different types of bacteria because such analyses assume that the genes being compared have been acquired by each group through vertical gene transfer (inheritance). To the extent that assumption is violated by horizontal gene transfers, molecular analysis of relationships is made much more difficult. In reality, such analyses are valid only for that part of the genome that has been transferred vertically by inheritance. In other words, the history of the group is not reflected perfectly in the history of its genome.

53. Generalized transduction transfers fragments from anywhere in the chromosome; specialized transduction transfers only DNA that is adjacent to the prophage insertion site.

54. They are 0.5 map units apart.

55. (1) Transformation: Competent cells take up DNA from the environment.

(2) Transduction: DNA is carried into the cell in a virus.

(3) Conjugation: DNA from one cell is transferred in a pilus to another cell.

56. The reverse transcriptase of HIV has an unusually high error rate, so that many mutations occur as the HIV genome is converted from RNA to DNA. A high mutation rate, though potentially creating nonfunctional viral proteins, also leads to rapid evolution of the virus, allowing it continually to adapt to new threatening conditions even within one host.

57. a. No recombinant plaques were observed, so the recombination frequency is 0. b. The genes are very close to each other or could be overlapping so that no recombination occurs between them.

58. Virulent phages reproduce using only the lytic cycle, whereas temperate phages can use the lytic or lysogenic cycle.

59. (1) A donor strain is chosen that has a number of prototrophies or antibiotic resistances.

(2) The recipient strain's genotype is different from that of the donor.

(3) DNA from the donor strain is fragmented and is used to transform the recipient strain.

(4) Transformants are selected—for instance, for a prototrophy or antibiotic resistance.

(5) Transformants are tested for other genes they acquired in the transformation.

(6) Order and distance of transferred genes are determined by comparing cotransformation frequencies.

60. The retrovirus genome encodes a reverse transcriptase enzyme. RTase makes a DNA copy of the viral RNA genome. The DNA copy integrates into the host cell's DNA. 61. Plasmids are small, circular, extrachromosomal DNA molecules found naturally in bacteria. They carry extra genes and can transfer these genes from one bacterial cell to another. They are also used extensively in genetic engineering.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 09_7e 62. (1) met and pro are close to each other.

(2) pro and leu are close to each other.

(3) leu and his are close to each other.

(4) The gene order is met pro leu his.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 10_7e Indicate the answer choice that best completes the statement or answers the question.

1. In the following DNA molecule, how many hydrogen bonds are present?

AATAGCGGATGCCCGAATACGAG TTATCGCCTACGGGCTTATGCTC a. 24 b. 48 c. 58 d. 0 e. 3 2. You are a research assistant in a lab that studies nucleic acids. Your advisor gave you four tubes for analysis.

Each of these tubes differs in its contents by the source of its nucleic acids: mouse cytoplasm (single-stranded RNA), yeast nuclei (double-stranded DNA), rotavirus (double-stranded RNA), and parvovirus (singlestranded DNA). The approximate nucleotide base composition of each sample is given in the table below.

Tube 1 2 3 4

A 32 34 30 33

C 17 16 21 16

U 0 15 0 34

T 32 0 26 0

G 19 35 23 17

Which tube MOST likely contains rotavirus? a. tube 1 b. tube 2 c. tube 3 d. tube 4 3. How did Albert Kossel contribute to our understanding of DNA? a. He used X-ray diffraction to examine the structure of DNA. b. He determined that DNA contains four different nitrogenous bases. c. He found that the "transforming principle" is destroyed by enzymes that hydrolyze DNA. d. He found that the phosphorus-containing components are the genetic material of phages. e. He discovered the "transforming principle" that could genetically alter bacteria.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 10_7e 4. The bonds that connect nucleotides in a single strand are called _____ bonds. a. phosphodiester b. peptide c. ionic d. hydrogen e. glycosidic 5. What is the difference in hydrogen bonding between A/T pairs versus G/C pairs? a. A/T pairs have one more hydrogen bond than G/C pairs. b. G/C pairs have one more hydrogen bond than A/T pairs. c. A/T pairs have two more hydrogen bonds than G/C pairs. d. G/C pairs have two more hydrogen bonds than A/T pairs. e. G/C pairs have three more hydrogen bonds than A/T pairs.

6. In the following DNA molecule, how many purines are present?

AATAGCGGATGCCCGAATACGAG TTATCGCCTACGGGCTTATGCTC a. 23 b. 25 c. 48 d. 11 7. What type of secondary structure is formed by the pairing of three strands of DNA? a. A-DNA b. B-DNA c. C-DNA d. H-DNA e. Z-DNA

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 10_7e 8. Which circle shows a noncovalent bond?

a. circle a b. circle b c. circle c d. circle d 9. A DNA molecule of 50 base pairs contains 15 cytosine bases (C). How many thymine bases will it have? a. 10 b. 15 c. 30 d. 35 e. 60

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 10_7e 10. The following table shows Chargaff's data that demonstrate base composition of DNA from different biological sources.

Source E. coli Yeast Sea urchin Rat Human

A 26.0 31.3 32.8 28.6 30.3

T 23.9 32.9 32.1 28.4 30.3

G 24.9 18.7 17.7 21.4 19.5

C 25.2 17.1 18.4 21.5 19.9

Which of the following is NOT a general conclusion that is supported by these data? a. DNA consists of a series of four-nucleotide units, each containing all four bases—ATGC—in a fixed sequence. b. The amount of adenine is always equal to the amount of thymine. c. The amount of guanine is always equal to the amount of cytosine. d. Although the ratio of A/T and G/C is the constant, the relative amount of any particular base varies between species. e. (A + G) / (T + C) = 1 11. You are a research assistant in a lab that studies nucleic acids. Your advisor gave you four tubes for analysis.

Each of these tubes differs in its contents by the source of its nucleic acids: mouse cytoplasm (single-stranded RNA), yeast nuclei (double-stranded DNA), rotavirus (double-stranded RNA), and parvovirus (singlestranded DNA). The approximate nucleotide base composition of each sample is given in the table below.

Tube 1 2 3 4

A 32 34 30 33

C 17 16 21 16

U 0 15 0 34

T 32 0 26 0

G 19 35 23 17

Which tube MOST likely contains yeast nuclei? a. tube 1 b. tube 2 c. tube 3 d. tube 4

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 10_7e 12. Heat can disrupt hydrogen bonding between DNA strands. Which of the following DNA strands would denature at the HIGHEST temperature? a. 10% AT and 90% GC b. 30% AT and 70% GC c. 50% AT and 50% GC d. 70% AT and 30% GC e. 90% AT and 10% GC 13. Which of the following is NOT characteristic of A-form DNA compared to B- or Z-form DNA? a. has right-handed helixes b. exists when less water is present c. is long and narrow d. has 50% purines, 50% pyrimidines 14. Which diagram shows a nucleotide that would be used to make RNA?

a. diagram A b. diagram B c. none of these diagrams d. diagram D e. diagram E

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 10_7e 15. Which term CORRECTLY describes the molecule below?

a. thymine base b. purine base c. pyrimidine base d. nucleotide e. amino acid 16. How did Rosalind Franklin contribute to our understanding of DNA? a. She used X-ray diffraction to show that the structure of DNA is helical. b. She determined that DNA contains four different nitrogenous bases. c. She found that the "transforming principle" is destroyed by enzymes that hydrolyze DNA. d. She found that the phosphorus-containing components are the genetic material of phages. e. She used models to show that DNA is a double helix.

17. How did Chargaff's rules contribute to Watson and Crick's elucidation of the structure of DNA? a. The rules suggested an equal concentration of sugars and phosphates. b. The rules suggested the amounts of all four bases were equal. c. The rules suggested the base-pairing combinations of adenine with thymine and guanine with cytosine. d. The rules suggested that each base corresponds to an amino acid.

18. In eukaryotic DNA, regions called "CpG islands" are often associated with unusual gene expression patterns, particularly decreased expression. What could be different about the DNA structure in these regions to account for decreased gene expression? a. methylated cytosine b. methylated guanine c. methylated phosphate d. a triple-stranded region

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 10_7e 19. You are a research assistant in a lab that studies nucleic acids. Your advisor gave you four tubes for analysis.

Each of these tubes differs in its contents by the source of its nucleic acids: mouse cytoplasm (single-stranded RNA), yeast nuclei (double-stranded DNA), rotavirus (double-stranded RNA), and parvovirus (singlestranded DNA). The approximate nucleotide base composition of each sample is given in the table below.

Tube 1 2 3 4

A 32 34 30 33

C 17 16 21 16

U 0 15 0 34

T 32 0 26 0

G 19 35 23 17

Which tube MOST likely contains mouse cytoplasm? a. tube 1 b. tube 2 c. tube 3 d. tube 4 20. Which secondary structure of DNA would MOST likely be seen in a dehydrated tissue sample? a. A b. B c. Z d. H 21. Which of these sequences could form a hairpin? a. 5′-GGGGTTTTCCCC-3′ b. 5′-AAAAAAAAAAAA-3′ c. 5′-AAAAGGCCCCCC-3′ d. 5′-TTTTTTCCCCCC-3′ e. 5′-GGGTTTGGGTTT-3′

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 10_7e 22. You are a research assistant in a lab that studies nucleic acids. Your advisor gave you four tubes for analysis.

Each of these tubes differs in its contents by the source of its nucleic acids: mouse cytoplasm (single-stranded RNA), yeast nuclei (double-stranded DNA), rotavirus (double-stranded RNA), and parvovirus (singlestranded DNA). The approximate nucleotide base composition of each sample is given in the table below.

Tube 1 2 3 4

A 32 34 30 33

C 17 16 21 16

U 0 15 0 34

T 32 0 26 0

G 19 35 23 17

Which samples would be destroyed by a DNase? a. tubes 1 and 3 b. tubes 2 and 4 c. tubes 1 and 2 d. tubes 3 and 4 e. all the tubes 23. Indicate which of the following statements is TRUE. a. There are three phosphates between each sugar in a molecule of DNA. b. A-, B-, and Z-form DNA are all right-handed helixes. c. There are three hydrogen bonds between AT pairs. d. Ribose sugars have a hydroxyl on the 2′ carbon. e. All organisms contain DNA that is roughly 25% A, 25% T, 25% G, and 25% C.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 10_7e 24. Which diagram shows a nucleotide with a purine base?

a. diagram A b. diagram B c. diagram C d. diagram D e. diagram E 25. How did Fred Griffith contribute to our understanding of DNA? a. He used X-ray diffraction to examine the structure of DNA. b. He determined that DNA contains four different nitrogenous bases. c. He found that the "transforming principle" is destroyed by enzymes that hydrolyze DNA. d. He found that the phosphorus-containing components are the genetic material of phages. e. He discovered the "transforming principle" that could genetically alter bacteria.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 10_7e 26. You are a research assistant in a lab that studies nucleic acids. Your advisor gave you four tubes for analysis.

Each of these tubes differs in its contents by the source of its nucleic acids: mouse cytoplasm (single-stranded RNA), yeast nuclei (double-stranded DNA), rotavirus (double-stranded RNA), and parvovirus (singlestranded DNA). The approximate nucleotide base composition of each sample is given in the table below.

Tube 1 2 3 4

A 32 34 30 33

C 17 16 21 16

U 0 15 0 34

T 32 0 26 0

G 19 35 23 17

Which tube MOST likely contains parvovirus? a. tube 1 b. tube 2 c. tube 3 d. tube 4 27. A molecule that consists of a nitrogenous base bonded to the 1'-carbon of a ribose or deoxyribose is a(n): a. nucleoside. b. hairpin. c. isotope. d. polynucleotide. e. nucleotide.

28. Which DNA modification is the addition of a –CH3 group? a. triple strand b. methylation c. looping d. cytosine substitution e. hydroxylation 29. Which of the following is NOT an example of secondary structure in nucleic acids? a. hairpin b. stem c. H-DNA d. B-DNA e. C-DNA

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 10_7e 30. While investigating a gene that might be responsible for pathogen resistance in the plant Arabidopsis, you discover that many of the nucleotides in the gene sequence are methylated. Which nucleotide is MOST likely to be methylated? a. A b. T c. C d. G 31. Which circle shows a bond that would also be found in an RNA transcribed from one strand of this DNA?

a. circle a b. circle b c. circle c

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 10_7e 32. Which circle shows a phosphodiester bond?

a. circle a b. circle b c. circle c d. circle d 33. Why was the idea that genes are made of nucleic acids NOT widely accepted until after 1950? a. Proteins are more abundant than nucleic acids, so it seemed more logical that proteins would carry genetic information. b. Until the structure of DNA was understood, how DNA could store and transmit genetic information was unclear. c. Amino acids existed in the prebiotic environment, so they would have most likely been the first to carry genetic information for life. d. DNA was not chemically stable for long enough to be a good method of storing genetic information. e. Nothing was known about the chemistry of DNA until after 1950.

34. While investigating a gene that might be responsible for pathogen resistance in the plant Arabidopsis, you discover that many of the nucleotides in the gene sequence are methylated. What might this methylation do to the expression of this gene? a. nothing b. increase c. decrease

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 10_7e 35. Hershey and Chase determined whether DNA or protein was the genetic material in bacteriophages. What isotope did they use to label the viral DNA? a. 14C b. 15N c. 18O d. 32P e. 35S 36. Which figure shows one of the amino acids that was key to distinguishing DNA from protein in the Hershey and Chase experiment?

a. diagram A b. diagram B c. diagram C d. diagram D e. diagram E 37. How many hydrogen bonds will be involved in base pairing in a DNA molecule of 50 base pairs that contains 15 cytosine bases? a. 45 b. 100 c. 115 d. 135 e. 150 Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 10_7e 38. Upon removal of bacteriophage coats from infected bacterial cells, where was the label for the DNA? a. in the cell pellet b. in the supernatant c. in bacteriophage coats d. in both the cell pellet and supernatant 39. Which secondary structure of DNA is typically left-handed? a. A b. B c. Z d. H 40. In the following DNA molecule, how many 3´ hydoxyls are present?

AATAGCGGATGCCCGAATACGAG TTATCGCCTACGGGCTTATGCTC a. 4 b. 2 c. 0 d. 24 41. If a DNA molecule contains 27% cytosine bases (C), then what percentage of thymine bases will it have? a. 10% b. 27% c. 46% d. 23% e. 52% 42. Two double-stranded fragments of DNA are exactly the same length. At 89°C, fragment A has completely denatured, which means that the two strands have separated. At that temperature, fragment B is still doublestranded. How might these fragments differ to result in different denaturation temperatures? a. Fragment A has a higher C+C content than fragment B. b. Fragment B has a higher C+G content than fragment A. c. Fragment A has a secondary structure of B form, while fragment B is an A form. d. Fragment B is methylated more than fragment A.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 10_7e 43. A high-resolution X-ray diffraction technique was used to obtain detailed secondary structure of a DNA molecule. The characteristics of the DNA showed it was a left-handed helix with a general shape that was longer and narrower than the classic Watson–Crick model of DNA. Which form of DNA was resolved? a. A-DNA b. B-DNA c. Z-DNA d. H-DNA e. single-stranded DNA 44. Hershey and Chase determined whether DNA or protein was the genetic material in bacteriophages. What isotope did they use to label the viral protein? a. 14C b. 15N c. 18O d. 32P e. 35S 45. You are a researcher studying the genetic basis of colon cancer. You have been working with a colon cancer cell line to determine the expression levels of different genes that might contribute to cancer formation. You obtain the DNA methylation status of five genes of interest (the data are shown in the table below). The plus (+) sign indicates the level of DNA methylation; more plus signs correlate with increased methylation levels.

Gene 1 2 3 4 5

Methylation levels ++ +++++ +++ ++ +

Based on the information shown above, which gene would you predict to have the lowest rate of transcription? a. gene 1 b. gene 2 c. gene 3 d. gene 4 e. gene 5

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 10_7e 46. With respect to their 3' and 5' ends, the two polynucleotide chains of a double-stranded DNA molecule are: a. antiparallel. b. parallel. c. methylated. d. complementary. e. nitrogenous.

47. In the following DNA molecule, how many ribose sugars are present?

AATAGCGGATGCCCGAATACGAG TTATCGCCTACGGGCTTATGCTC a. 48 b. 24 c. 4 d. 2 e. 0 48. How did Avery, MacLeod, and McCarty contribute to our understanding of DNA? a. They used X-ray diffraction to examine the structure of DNA. b. They determined that DNA contains four different nitrogenous bases. c. They found that the "transforming principle" is destroyed by enzymes that hydrolyze DNA. d. They found that the phosphorus-containing components are the genetic material of phages. e. They discovered the "transforming principle" that could genetically alter bacteria.

49. If the sequence of one strand of DNA is 5′-GCTAGCGTCG-3′, what is the sequence of the complementary strand? a. 3′-GCTAGCGTCG-5′ b. 5′-GCTGCGATCG-3′ c. 3′-CGATCGCAGC-5′ d. 5′-CGATCGCAGC-3′ e. 5′-CGAUCGCAGC-3′

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 10_7e 50. You are a researcher studying the genetic basis of colon cancer. You have been working with a colon cancer cell line to determine the expression levels of different genes that might contribute to cancer formation. You obtain the DNA methylation status of five genes of interest (the data are shown in the table below). The plus (+) sign indicates the level of DNA methylation; more plus signs correlate with increased methylation levels.

Gene 1 2 3 4 5

Methylation levels ++ +++++ +++ ++ +

Based on the information shown above, which gene would you predict to have the highest rate of transcription? a. gene 1 b. gene 2 c. gene 3 d. gene 4 e. gene 5 51. The concept that genetic information passes from DNA to RNA to protein is called the: a. central dogma. b. nitrogenous base. c. transforming principle. d. polynucleotide strand. e. reverse transcription.

52. Which hypothesis contributed to the mistaken idea that protein is the genetic material because, with its 20 different amino acids, protein structure could be more variable than that of DNA? a. tetranucleotide hypothesis b. central dogma hypothesis c. RNA world hypothesis d. one gene–one enzyme hypothesis e. adaptor hypothesis

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 10_7e 53. If a DNA molecule is 30% cytosine (C), what is the percentage of guanine (G)? a. 30% b. 60% c. 35% d. 70% e. 15% 54. Which of the following is NOT a key characteristic that genetic material must possess? a. Genetic material must contain complex information. b. Genetic material must replicate faithfully. c. Genetic material must encode the phenotype. d. Genetic material must have the capacity to vary. e. Genetic material must contain nitrogen but not sulfur.

55. While doing research on deep-sea vents, you discover a very simple new life form. After some initial analysis, you find that this life form contains small fragments of DNA, small complementary RNA fragments, and proteins. Fortuitously, you collected two strains, one that is purple and one that is yellow. You wish to discover which of those three molecules could be the genetic material. The classic experiment of which of the following scientists would be the MOST appropriate to mimic? a. Hershey, Chase b. Avery, MacLeod, and McCarty c. Franklin d. Griffith e. Fraenkel-Conrat and Singer 56. How did Alfred Hershey and Martha Chase contribute to our understanding of DNA? a. They used X-ray diffraction to examine the structure of DNA. b. They determined that DNA contains four different nitrogenous bases. c. They found that the "transforming principle" is destroyed by enzymes that hydrolyze DNA. d. They found that the phosphorus-containing components are the genetic material of phages. e. They discovered the "transforming principle" that could genetically alter bacteria.

57. Heat can disrupt hydrogen bonding between DNA strands. Which of the following DNA strands would denature at the LOWEST temperature? a. 10% AT and 90% GC b. 30% AT and 70% GC c. 50% AT and 50% GC d. 70% AT and 30% GC e. 90% AT and 10% GC Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 10_7e 58. In the Hershey–Chase experiment, proteins and DNA were differentially labeled in order to follow where they were present. The specific labels were: a. 3H for protein and 15N for DNA. b. 35S for protein and 15N for DNA. c. 35S for protein and 32P for DNA. d. 32P for protein and 35S for DNA. e. 15N for protein and 32P for DNA. 59. Indicate which of the following statements is FALSE. a. Covalent bonds connect nucleotides in a strand; noncovalent interactions hold strands into a doublestranded structure. b. Uracil is similar to thymine except that uracil lacks a methyl group on the carbon at position 5 on the carbon–nitrogen ring. c. Frederick Griffith demonstrated that a transforming chemical from dead bacteria could change the genetic information of living bacteria. d. Avery, MacLeod, and McCarty showed that DNA is the genetic information of cells and that RNA is the genetic information of viruses. e. The pyrimidine bases in nucleic acids are cytosine, thymine, and uracil.

60. Which diagram shows a nucleotide as it would appear in DNA?

a. diagram A b. diagram B c. diagram C d. diagram D e. diagram E Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 10_7e 61. Which of the following forces acts between stacked base pairs to stabilize the double-helix structure of DNA? a. hydrogen bonding b. van der Waals forces c. dipole–dipole interactions d. ionic bonds e. covalent bonds 62. While doing research on deep-sea vents, you discover a very simple new life form. After some initial analysis, you find that this life form contains small fragments of DNA, small complementary RNA fragments, and proteins. Fortuitously, you collected two strains, one that is purple and one that is yellow. You wish to discover which of those three molecules could be the genetic material. You heat-kill some of the purple life form and subject three different homogenized samples to different enzymes: DNase, RNase, or protease. Which sample will NOT transform yellow into purple? a. DNase b. RNase c. protease d. All will cause transformation. e. None will cause transformation.

63. Which of the following would NOT necessarily be true for a DNA molecule? a. A = T b. C = G c. A + G = C + T d. A + C = G + T e. A + T = G + C Indicate one or more answer choices that best complete the statement or answer the question.

64. While doing research on deep-sea vents, you discover a very simple new life form. After some initial analysis, you find that this life form contains small fragments of DNA, small complementary RNA fragments, and proteins. Fortuitously, you collected two strains, one that is purple and one that is yellow. You decide to attempt a transformation: seeing if you can convert the purple form into the yellow form. From which sample(s) would you find some yellow form? (Select all that apply.) a. live yellow form b. heat-killed yellow form c. purple form d. heat-killed purple form

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 10_7e 65. Why were bacteriophages used in the Hershey–Chase experiment? (Select all that apply.) a. They had a protein coat and an internal DNA molecule. b. They had a DNA coat and an internal protein molecule. c. Their proteins and DNA were mixed together. d. They injected protein inside bacterial cells. e. They injected their genetic material into bacterial cells.

66. Which of the following chemical or structural characteristics of RNA are different from those of DNA? (Select all that apply.) a. The RNA sugar is ribose instead of deoxyribose. b. RNA is usually a single-stranded molecule instead of a hydrogen-bonded double strand like DNA. c. The bases in RNA include uracil instead of thymine. d. RNA molecules are generally shorter in length than DNA macromolecules. e. The 2′ carbon of ribose has an H, unlike the OH in that position of deoxyribose.

67. In 2018 sequence analysis was carried out on a 90,000-year-old DNA sample from a 13-year-old female named Denny. The analysis indicates she was the daughter of parents of two distinct species: a Neanderthal mother and a Denisovan father. Which of the following statements must be TRUE? (Select all that apply.) a. The chemical nature of genetic material is the same in Neanderthals and Denisovans. b. Somatic DNA samples were extracted from Denny. c. Germline DNA samples were extracted from Denny. d. The base sequence of DNA from Neanderthals and Denisovans is sufficiently distinct to be identified by sequencing. e. Chromosomes from Neanderthals and Denisovans were able to pair as sets of homologs during cell division.

68. Describe two characteristics of Z-DNA that distinguish it from B-DNA.

69. How did the work of Hershey and Chase contribute to the conclusion that DNA is the genetic material? What technique helped them to distinguish between viral DNA and protein?

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 10_7e 70. How many of each of the following does this DNA molecule have?

AATAGCGGATGCCCGAATACGAG TTATCGCCTACGGGCTTATGCTC a. 3′ hydroxyls b. hydrogen bonds c. purines d. ribose sugars

71. While investigating a gene that might be responsible for pathogen resistance in the plant Arabidopsis, you discover that many of the nucleotides in the gene sequence are methylated. (1) Which nucleotide (A, T, C, or G) is most likely to be methylated? (2) Draw the structure of this methylated nucleotide. (3) What might this methylation do to the function of this gene?

72. The discovery of the "transforming principle" in experiments by Fred Griffith was followed up by a key set of experiments performed by Avery, McCarty, and MacLeod. Describe the link between the two sets of experiments and how Avery, McCarty, and MacLeod built on Griffith's work.

73. What would be the sequence of an RNA produced by using the DNA sequence shown as a template?

3′-TACCGTGCGTGACATTAAGCC-5′ Write the sequence from 5′ to 3′, left to right.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 10_7e 74. List and briefly describe the three different secondary structures of DNA and discuss the physiological significance of each.

75. How did Chargaff's rules contribute to Watson and Crick's elucidation of the structure of DNA?

76. Two double-stranded fragments of DNA are exactly the same length. At 89°C, fragment A has completely denatured, which means that the two strands have separated. At that temperature, fragment B is still double stranded. How might these fragments differ to result in different denaturation temperatures?

77. Draw a dinucleotide of two DNA molecules. Label the 5′ end, the 3′ end, and the phosphodiester bond. Draw the ring(s) of a purine base on one nucleotide and a pyrimidine base on the other. Then label them.

78. In eukaryotic DNA, regions called "CpG islands" are often associated with unusual gene expression patterns, particularly decreased expression. What could be different about the DNA structure in these regions to account for decreased gene expression?

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 10_7e 79. What would be the sequence of a single-stranded DNA produced by using the DNA sequence shown as a template?

3′-TACCGTGCGTGACATTAAGCC-5′ Write the sequence from 5′ to 3′, left to right.

80. Summarize the evidence that RNA serves as the genetic material in tobacco mosaic virus (TMV).

81. Describe the secondary structure that DNA might form in an ancient, dehydrated tissue sample.

82. List at least four errors in this drawing of two deoxyribonucleotides in a phosphodiester bond.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 10_7e 83. While doing research on deep-sea vents, you discover a very simple new life form. After some initial analysis, you find that this life form contains small fragments of DNA, small complementary RNA fragments, and proteins. Fortuitously, you collected two strains, one that is purple and one that is yellow. What experiments would you perform to determine which of these three cellular constituents serves as the genetic material in your new organism?

84. In the diagram of a DNA molecule below, label the 5′ and 3′ ends of each strand, a hydrogen bond, a phosphodiester bond, a purine, a pyrimidine, and a deoxyribose sugar.

85. Draw the complementary strand to the following single-stranded DNA and label the 5′ and 3′ ends: 5′ATAGCATGGGCCATACGATTACTGA-3′.

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 10_7e 86. List four characteristics required of genetic material. For each characteristic, indicate how the structure of DNA helps us to understand the characteristic.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 10_7e Answer Key 1. c 2. d 3. b 4. a 5. b 6. a 7. d 8. b 9. d 10. a 11. a 12. a 13. c 14. c 15. b 16. a 17. c 18. a 19. b 20. a 21. a 22. a 23. d 24. d 25. e 26. c Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 10_7e 27. a 28. b 29. e 30. c 31. a 32. a 33. b 34. c 35. d 36. b 37. c 38. a 39. a 40. b 41. d 42. b 43. a 44. e 45. b 46. a 47. e 48. c 49. c 50. e 51. a 52. a 53. a 54. e Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 10_7e 55. b 56. d 57. e 58. c 59. d 60. d 61. b 62. a 63. e 64. a, b 65. a, e 66. a, b, c, d 67. a, b, d, e 68. (1) Z-DNA takes the form of a left-handed helix.

(2) Z-DNA's shape is in general longer and narrower than is B-DNA's shape.

69. It distinguished between protein and DNA as genetic material in phages. It is the DNA that is used to produce new phages within host cells. Radioactive isotopes were used to distinguish between DNA and protein. They used radioactive 32P to label the viral DNA, and they used radioactive 35S to label the viral proteins.

70. a. 2 b. 58 c. 23 d. 0

Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 10_7e 71. (1) Higher eukaryotes, including plants, are typically methylated at cytosine.

(2) _

(3) Methylation is often associated with low levels of gene expression.

72. Griffith demonstrated that the genetic material of heat-killed virulent IIIS bacteria could transform live cells of the originally nonvirulent IIR bacteria into virulent cells, showing there must be a transforming substance ("transforming principle"). Griffith did not know the chemical identity of the transforming principle and postulated that it might be a polysaccharide in the bacterial coat.

Avery, McCarty, and McLeod followed up on Griffith's work by providing the identity of the transforming principle. They purified an extract which was capable of the transformation observed by Griffith. There were a few candidate chemicals in that extract that might be the actual transforming principle. To identify they correct candidate, they used a process of elimination. They add proteases to the extract, which destroyed proteins, and found it retained transforming activity. They added RNase to the extract, which destroyed RNA, and found it retained transforming activity. Thus, the transforming principle did not appear to be protein or RNA. They added DNase to the extract and found it lost transforming activity. Thus, the group identified DNA as the transforming principle.

73. 5′-AUGGCACGCACUGUAAUUCGG-3′ 74. The three different secondary structures of DNA are B form, which is thought to be the most common form under physiological conditions, along with A and Z forms. All three forms consist of two antiparallel DNA polynucleotide strands connected in the middle by hydrogen bonds. The B form and A form are both twisted into a right-hand helix, but the A form is shorter and wider than B-DNA, and its bases are tilted away from the main axis of the helix. The A form probably doesn't exist under physiological conditions but exists when less water is present.

Unlike A- and B-DNA, Z-DNA forms a left-handed helix. Z-DNA is long and thin compared to B-DNA. Z-DNA exists when the salt concentration is high, and it probably exists for certain sequences (stretches of alternating C and G nucleotides) under physiological conditions. Z-DNA might be associated with actively transcribed DNA, suggesting that it might play some role in gene expression.

75. The rules suggested the base-pairing combinations of adenine with thymine and guanine with cytosine.

76. Fragment B has a higher G + C content (i.e., fragment A has a higher A + T content). Three hydrogen bonds form between G–C pairs, so more energy is required to separate a G–C pair. Only two hydrogen bonds form per A–T pair.

Copyright Macmillan Learning. Powered by Cognero.

Page 30

Name:

Class:

Date:

Chap 10_7e

77.

78. CG sequences tend to be methylated on the cytosine. Methylated sequences tend to be transcribed less actively than are unmethylated sequences.

79. 5′-ATGGCACGCACTGTAATTCGG-3′ 80. TMV consists of a single RNA molecule surrounded by a cylinder of protein molecules. Fraenkel-Conrat and Singer created hybrid TMV by mixing RNA and protein from different strains. When the hybrid TMV particles infected tobacco leaves, the progeny viruses were the same as the strain from which the RNA had been isolated, demonstrating that the genetic information is carried on the RNA molecule and not the proteins. Later it was shown that the RNA by itself can infect tobacco plants and direct the production of new TMV, confirming the role of RNA in carrying genetic information.

81. In a dehydrated sample, DNA likely would take on the A form secondary structure. This secondary structure is a right-handed helix that is shorter and wider than is the physiological B-DNA form with bases tilted from the main central axis.

82. (1) Deoxyribonucleotides should not have 2′-OH groups.

(2) The phosphate on the top nucleotide does not have enough O atoms.

(3) The phosphodiester bond does not have any O atoms.

(4) The phosphate groups do not have a negative charge.

(5) 5′ carbons are not shown.

(6) No double bonds are shown in the phosphate groups.

(7) No O atoms are shown on the bottom phosphate.

83. You could do an experiment similar to that performed by Avery, MacLeod, and McCarty. After heat-killing a culture of the purple strain, you could treat aliquots with compounds that specifically degrade DNA, RNA, or proteins and then use those treated aliquots to try to transform the yellow strain into one that is purple, and vice versa.

Copyright Macmillan Learning. Powered by Cognero.

Page 31

Name:

Class:

Date:

Chap 10_7e

84.

85. 3′-TATCGTACCCGGTATGCTAATGACT-5′ 86. (1) Genetic material must have the capacity to contain complex information to encode all the functions of an organism. Because DNA molecules in cells normally are very long and because the four base pairs can occur in any sequence, DNA molecules are able to store a large amount of information in the form of the sequence of nucleotide base pairs.

(2) Genetic material must have a mechanism to be copied accurately for reproduction. The double helix model of DNA suggested to Watson and Crick that DNA could be replicated by semiconservative replication. In this mechanism, the two strands of a parental molecule are separated, and each parental strand acts as a template to direct the synthesis of a new strand through complementary base pairing.

(3) Genetic material must be able to encode phenotypes. Changes in the sequence of base pairs in DNA (mutations) allow for the existence of different versions of genes (alleles) that can encode alternative phenotypes.

(4) Genetic material must be able to vary in base sequence.

Copyright Macmillan Learning. Powered by Cognero.

Page 32

Name:

Class:

Date:

Chap 11_7e Indicate the answer choice that best completes the statement or answers the question.

1. You are studying a small eukaryotic gene of about 2000 bp in length. Estimate how many copies of histone H1 you would find along this region of the chromosome. a. 10 b. 20 c. 40 d. 80 e. 100 2. How many different types of histones are found in the nucleosome that packages mitochondrial DNA? a. zero b. one c. two d. three e. four 3. You are studying a small eukaryotic gene of about 2000 bp in length. Estimate how many copies of histone H4 you would find along this region of the chromosome. a. 10 b. 20 c. 40 d. 80 e. 100 4. How many base pairs per turn of the helix would MOST likely correspond to a positively supercoiled DNA molecule? a. 0 b. 5 c. 10 d. 15 e. 100 5. Which statement about mitochondrial genomes is NOT true? a. In most animals, the mitochondrial genome consists of a single circular DNA molecule. b. Plant mitochondrial genomes often include multiple circular DNA molecules. c. Each mitochondrion typically contains many copies of the mitochondrial genome. d. All copies of the mitochondrial genome within a cell are identical.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 11_7e 6. The _____ membrane of the chloroplast bears the enzymes and pigments required for photophosphorylation. a. outer b. middle c. thylakoid d. plasma e. double 7. Which of the following descriptions is NOT true of heterochromatin? a. It remains in a highly condensed state throughout the cell cycle. b. It makes up most chromosomal material and is where most transcription occurs. c. It exists at the centromeres and telomeres. d. It occurs along one entire X chromosome in female mammals when this X becomes inactivated. e. It is characterized by the absence of crossing over and replication late in the S phase.

8. Pea plants produce both pollen and eggs. A pea plant inherits a mutation for cytoplasmic male sterility. How will this affect the plant and/or its progeny? a. The plant will be able to reproduce only by self-fertilization. b. The plant will be able to reproduce only by cross-fertilization. c. The plant will be unable to produce progeny. d. The plant will produce progeny, but the progeny will not be able to reproduce.

9. A normal chromosome in a higher eukaryotic species would NOT be expected to contain: a. one centromere. b. one copy of telomere. c. two copies of histone 2A per nucleosome. d. satellite DNA. e. tandem repeat sequences.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 11_7e 10. Which pair in the pedigree shares the same mitochondrial DNA?

a. I-2 and III-1 b. II-2 and III-2 c. III-1 and III-2 d. II-2 and III-1 e. I-3 and II-6 11. A ribosomal RNA gene is an example of which type of DNA sequence in eukaryotes? a. moderately repetitive DNA b. highly repetitive DNA c. short interspersed elements d. long interspersed elements e. unique-sequence DNA 12. How many complete rotations would MOST likely correspond to a relaxed DNA molecule that is 100 bp in length? a. 0 b. 5 c. 10 d. 15 e. 100 13. A 2011 genetic study carried out in Romania suggests that childhood adversity results in shorter telomere lengths in the cells of children aged 6–10 years old. Which of the following statements is CORRECT? a. The study is likely only to be relevant to children raised in Eastern Europe. b. The study indicates at least 6 years of adversity is required to observe a shortening of telomeres. c. The study suggests environmental factors can influence chromosome structure. d. The study suggests environmental factors can alter genotypes of children as they develop. e. All of the above statements are correct.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 11_7e 14. A tRNA gene is an example of which type of DNA sequence in eukaryotes? a. moderately repetitive DNA b. highly repetitive DNA c. short interspersed elements d. long interspersed elements e. unique-sequence DNA 15. Assuming there are no heteroplasmic individuals but that people from different families have different DNA, how many different mitochondrial DNAs are there in the pedigree below?

a. 2 b. 4 c. 5 d. 6 e. 7

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 11_7e 16. DNA can be isolated from cells, and the different types of DNA from the cell can be separated by density gradient centrifugation. The following figures show the density gradient centrifugation profiles for DNA isolated from four different cell types: plant, animal, wild-type yeast with a plasmid, and petite yeast.

Which figure has the DNA profile for the petite yeast? a. 1 b. 2 c. 3 d. 4 17. How many base pairs per turn of the helix would MOST likely correspond to a relaxed DNA molecule? a. 0 b. 5 c. 10 d. 15 e. 100

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 11_7e 18. A gene-encoding sequence is an example of which type of DNA sequence in eukaryotes? a. moderately repetitive DNA b. highly repetitive DNA c. short interspersed elements d. long interspersed elements e. unique-sequence DNA 19. The human Y chromosome is about 50 million base pairs long. About how many nucleosomes would you expect to find associated with this chromosome? a. 2500 b. 50,000 c. 250,000 d. 1,000,000 e. 50,000,000 20. Which of the following does NOT fit the description of euchromatin? a. less condensed state b. transcriptionally inactive c. chromosomal arms d. common crossing-over sites

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 11_7e 21. DNA can be isolated from cells, and the different types of DNA from the cell can be separated by density gradient centrifugation. The following figures show the density gradient centrifugation profiles for DNA isolated from four different cell types: plant, animal, wild-type yeast with a plasmid, and petite yeast.

The large peak to the left in all the figures is: a. mitochondrial DNA. b. genomic DNA. c. chloroplast DNA. d. plasmids.

22. The agouti locus helps determine coat color in mice, and this phenotype can vary from light to dark between genetically identical individuals. You have discovered a drug that reduces the variation in the agouti phenotype. What is a likely explanation for this drug's mechanism of action? a. It inhibits DNA polymerases. b. It inhibits DNA methyl transferases. c. It activates shelterin proteins. d. It activates mitochondrial transcription. e. It causes DNA damage.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 11_7e 23. While working on Drosophila, you find a new mutant strain with an abnormal histone H3 gene. This novel histone mutant is predicted to cause the nucleosomes to bind an extra 15 bp of DNA compared to wild type.

If you digest isolated chromatin with a nuclease to release the core nucleosomes, what size DNA fragments would you expect from wild-type and mutant flies? a. Approximately 50 bp is known to bind to the normal core nucleosome, so the additional 15 bp binding to H3 would give a rise to ~65 bp. b. Approximately 100 bp is known to bind to the normal core nucleosome, so the additional 15 bp binding to H3 would give a rise to ~115 bp. c. Approximately 125 bp is known to bind to the normal core nucleosome, so the additional 15 bp binding to H3 would give a rise to ~140 bp. d. Approximately 145 bp is known to bind to the normal core nucleosome, so the additional 15 bp binding to H3 would give a rise to ~160 bp. e. The size of the DNA fragments cannot be determined.

24. What kind of gene would NOT be found in a chloroplast genome? a. a tRNA gene b. a gene for a subunit of the photosynthesis enzyme RuBisCO c. a gene for a ribosomal protein d. a gene for ribosomal RNA e. a gene for a histone protein 25. The _____ theory states that the ancestors of mitochondria and chloroplasts were free-living bacteria. a. phylogenetic b. endosymbiotic c. cell d. cytoplasmic inheritance e. old world 26. Paternal transmission of mitochondria is common in which group? a. humans b. mice c. most gymnosperms d. most flowering plants e. insects

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 11_7e 27. How does histone acetylation affect chromatin? a. It loosens the chromatin and allows increased transcription. b. It allows DNA to become resistant to damage. c. It helps the histones have a greater affinity for DNA. d. It inhibits DNA replication by making it more difficult to separate the DNA strands. e. It causes the chromatin to become more condensed in preparation for metaphase.

28. DNA can be isolated from cells, and the different types of DNA from the cell can be separated by density gradient centrifugation. The following figures show the density gradient centrifugation profiles for DNA isolated from four different cell types: plant, animal, wild-type yeast with a plasmid, and petite yeast.

The peak farthest to the right in Figure 1 is: a. plasmids. b. mitochondrial DNA. c. genomic DNA. d. chloroplast DNA. 29. Which of the following statements is NOT true? a. Both the mitochondria and the chloroplast generate ATP. b. A single eukaryotic cell may contain thousands of copies of the mitochondrial genome. c. According to the endosymbiotic theory, chloroplasts are thought to have evolved from cyanobacteria. d. The mutation rate of mitochondrial DNA is higher than the mutation rate of nuclear DNA. e. Oxidative phosphorylation capacity is constant throughout a person's lifetime.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 11_7e 30. Which of the following amino acids has a positive charge that helps to hold the DNA in contact with the histones? a. alanine b. arginine c. leucine d. valine e. serine 31. How many membranes separate the mitochondrial matrix from the cytoplasm? a. zero b. one c. two d. three e. four 32. The presence of more than one variation of DNA in the organelles of a single cell is called: a. homoplasmy. b. heteroplasmy. c. hemiplasmy. d. pseudoplasmy. e. paraplasmy.

33. Where would you expect to find the variant histone CenH3? a. telomere b. euchromatin c. centromere d. mitochondria e. chloroplast 34. Which statement is NOT true of negatively supercoiled DNA? a. It eases the separation of nucleotide strands during replication and transcription. b. It allows DNA to be packed into small spaces. c. It has less than 10 bp per turn of its helix. d. It is more negatively charged due to additional phosphates per turn of the helix. e. It is found in most cells.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 11_7e 35. Copies of a gene that arose by gene duplication are part of a gene: a. complex. b. family. c. tandemoplex. d. structure. e. chromosome.

36. Which of the following is an example of an epigenetic change in eukaryotes? a. a loss of an AT base pair from a gene b. the addition of methyl groups to cytosines in the promoter region of a gene c. the substitution of an AT base pair by a GC base pair in a gene as a result of a mistake during DNA replication d. a deletion that simultaneously removes two genes from the genome e. None of these examples represents epigenetic changes.

37. Alterations of chromatin of DNA structure that are stable and inheritable in offspring via DNA methylation or alteration of histone proteins are referred to as _____ changes. a. genetic b. mutational c. sensitivity d. epigenetic 38. How many complete rotations would MOST likely correspond to a positively supercoiled DNA molecule that is 100 bp in length? a. 0 b. 5 c. 10 d. 15 e. 100 39. Telomeres exist to help with the _____ of the ends of eukaryotic chromosomes. a. transcription b. replication c. metabolism d. destabilization e. translation

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 11_7e 40. How many base pairs per turn of the helix would MOST likely correspond to a negatively supercoiled DNA molecule? a. 0 b. 5 c. 10 d. 15 e. 100 41. Which of the following terms describes the chromatin of telomeres and centromeres? a. euchromatin b. constitutive heterochromatin c. facultative heterochromatin d. closed chromatin e. end chromatin

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 11_7e 42. DNA can be isolated from cells, and the different types of DNA from the cell can be separated by density gradient centrifugation. The following figures show the density gradient centrifugation profiles for DNA isolated from four different cell types: plant, animal, wild-type yeast with a plasmid, and petite yeast.

Which figure has the DNA profile that would most closely match DNA from a human cell? a. 1 b. 2 c. 3 d. 4 43. A telomere is an example of which type of DNA sequence in eukaryotes? a. moderately repetitive DNA b. highly repetitive DNA c. short interspersed elements d. long interspersed elements e. unique-sequence DNA

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 11_7e 44. How many complete rotations would MOST likely correspond to a negatively supercoiled DNA molecule that is 100 bp in length? a. 0 b. 5 c. 10 d. 15 e. 100 45. Which of the following statements is NOT true of bacterial DNA? a. Most bacterial genomes consist of a single circular DNA molecule. b. Packed bacterial DNA does not involve proteins. c. Bacterial DNA is confined to a region in the cell called the nucleoid. d. Many bacteria contain additional DNA in the form of small circular molecules called plasmids. e. About 3 to 4 million base pairs of DNA are found in a typical bacterial genome.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 11_7e 46. DNA can be isolated from cells, and the different types of DNA from the cell can be separated by density gradient centrifugation. The following figures show the density gradient centrifugation profiles for DNA isolated from four different cell types: plant, animal, wild-type yeast with a plasmid, and petite yeast.

Which figure has the DNA profile for the plant cell? a. 1 b. 2 c. 3 d. 4 47. An Alu sequence is an example of which type of DNA sequence in eukaryotes? a. moderately repetitive DNA b. highly repetitive DNA c. short interspersed elements d. long interspersed elements e. unique-sequence DNA

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 11_7e 48. A centromere is an example of which type of DNA sequence in eukaryotes? a. moderately repetitive DNA b. highly repetitive DNA c. short interspersed elements d. long interspersed elements e. unique-sequence DNA 49. Which of the following statements is TRUE? a. Most proteins in the human mitochondrion are encoded by nuclear genes. b. One piece of evidence supporting the endosymbiotic theory is the extreme similarity between mitochondrial DNAs from different organisms. c. Heteroplasmy refers to the presence of different alleles in a single organelle. d. Plants contain chloroplasts, not mitochondria. e. cpDNA evolves faster than nuclear DNA. 50. What is the BEST evidence that centromeres are critical for chromosome segregation at mitosis? a. Centromeres are observed on the chromosomes of mitotic cells. b. Spindle microtubules are observed attached to centromeres in mitotic cells. c. Chromosomes in nonmitotic cells lack centromeres. d. If a chromosome breaks during mitosis, the fragment that retains the centromere attaches to a spindle microtubule and moves to the pole whereas the fragment without the centromere fails to attach to a spindle microtubule and is lost. e. None of these are evidence.

Indicate one or more answer choices that best complete the statement or answer the question.

51. Which of the following has/have repetitive DNA sequences in heterochromatin state? (Select all that apply.) a. telomere b. centromere c. mitochondria d. chloroplast 52. Which of the following terms describe the feature(s) of mtDNA and cpDNA that differ from the eukaryotic nuclear DNA? (Select all that apply.) a. uniparental inheritance b. circular c. heteroplasmy d. homoplasmy e. high copy number Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 11_7e 53. If a bacterial chromosome were inserted into a eukaryotic cell, would it be stable and segregate like eukaryotic chromosomes do during mitosis and meiosis? (Select all that apply.) a. It would not be stable due to the lack of a eukaryotic-specific origin of replication; hence, it could not replicate properly in a eukaryotic cell. b. It would be generally stable because the chemical nature of DNA is the same regardless of the cell type. c. Due to the lack of centromeres on prokaryotic chromosomes, the chromosomes will not segregate normally during cell division. d. The prokaryotic chromosome can be induced to be stabilized by cleavage of circular form to mimic linear eukaryotic chromosome. e. The bacterial chromosome would be lost and eventually degraded.

54. Which of the following terms CORRECTLY describe(s) the inheritance pattern of mtDNA and cpDNA in eukaryotic cells? (Select all that apply.) a. maternal inheritance b. circular c. heteroplasmy d. homoplasmy e. high copy number

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 11_7e 55. The classic experiment that examined DNAse I sensitivity of chicken embryonic DNA from different tissues and at different developmental stages shows that _____. (Select all that apply.)

a. the chromatin structure changes in the course of development. b. the gene expression pattern changes during development. c. DNAse I sensitivity comes from sporadic mutations occurring during development. d. DNAse I sensitivity only occurs in chicken but in no other organisms.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 11_7e 56. The petite mutations in S. cerevisiae, which were discovered by Boris Ephrussi and his colleagues in the late 1940s, result in much smaller colony size, reflecting the defect in the cellular growth rate (Figure 11.15). Most petite mutations are known to occur on mitochondrial DNA. Which of the following statements offer a logical explanation of the petite phenotype? Select all that apply. (Photo credit: [From Xin Jie Chen and G. Desmond Clark-Walker, Genetics 144: 1445–1454, Fig 1, 1996. © Genetics Society of America. Courtesy of Xin Jie Chen, Department of Biochemistry and Molecular Biology, SUNY Upstate Medical University.])

a. The defect in the cellular growth comes from an inability to generate enough ATP. b. The growth defect is known to come from having excessive copies of mitochondria, resulting in toxicity from excess ATP. c. The mutations on the mtDNA can result in deficiency of the enzymes involved in aerobic respiration. d. They have no means to make any ATP because of mtDNA defects that affect the normal mitochondrial functions. e. The petite mutants only have to rely on anaerobic processes such as fermentation and glycolysis.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 11_7e 57. Based on the pedigree below, which of the following statements correctly interpreted the pedigree with respect to X-linked dominant, X-linked recessive, Y-linked, mitochondrial, autosomal recessive, and autosomal dominant inheritance? (Select all that apply.)

a. The phenotype presented is X-linked dominant because II-3 has the same phenotype as I-1. b. The phenotype presented is X-linked recessive because III-3 and III-4 do not have the same phenotype as II-5. c. The phenotype cannot be Y-linked because II-2 doesn't have the same phenotype as I-1. d. The phenotype could be autosomal recessive as III-3 and III-4 don't have the same phenotype as II-5. e. The phenotype could be autosomal dominant if II-5 and II-6 are heterozygotes, and III-3 and III-4 are homozygous recessive. f. The phenotype cannot be mitochondrial. II-3 and II-5 don't have the same phenotype as I-2. III-3 and III-4 don't have the same phenotype as II-5.

58. A plant has green leaves with multiple yellow spots. When used as an egg donor in a cross with a normal plant that has all green leaves, some of the progeny have green and yellow leaves and some have all green leaves.

When used as the pollen donor in a cross with a normal plant, all the progeny have all green leaves. Which of the following statements explain(s) the result of this cross? (Select all that apply.) a. The original plant with green leaves with multiple yellow spots is likely heteroplasmic for a mutation in the chloroplast genome. b. The yellow spots are cells that, by replicative segregation, have received only mutant chloroplast genomes. c. The plants with yellow leaves that originate from the plant with yellow spots as the egg donor received the mutant chloroplast maternally. d. Presumably, eggs that are heteroplasmic for mutant chloroplasts will not produce viable plants. e. When the plant is the pollen donor, the plant with nonmutant chloroplast DNA will contribute the chloroplasts, and all progeny will have all green leaves.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 11_7e 59. When chromatin from any eukaryote is digested with micrococcal nuclease (an endonuclease) and fractionated using electrophoresis, DNA fragments of approximately 200 base pairs in length are observed.

Which of the following statements explain(s) the observation? (Select all that apply.) a. The 200-base-pair-long DNA fragments represent the approximate length of DNA wrapped around the histone core. b. The 200-base-pair-long DNA fragment is a characteristic behavior of micrococcal nuclease on any given free DNA strand. c. The eukaryotic DNA has an enormous number of repetitive sequences, and the nuclease is cleaving certain repetitive sequences, generating these fragments. d. The result reveals the conserved composition of the nucleosome, which is the repeating unit that makes up chromatin in all eukaryotes. e. The cleavage occurs at the exposed linker region between adjacent nucleosomes that does not directly interact with the histone core.

60. Chromosomal puffs observed on polytene chromosomes indicate the region that is most likely _____. (Select all that apply.) a. transcriptionally inactive b. transcriptionally active c. DNAse I sensitive d. DNAse I insensitive 61. There are some genomes that have been reported to be positively coiled instead of negatively supercoiled, which is the status of most genomes that we have studied. The genomes that are positively supercoiled seem to belong to viruses and cells that exist at very high temperatures. Why might positive supercoiling be an advantage at high temperatures? (Select all that apply.) a. Negative supercoiling makes it more difficult for strands to separate while positive supercoiling would allow the strands to separate more readily. b. At high temperatures, the condition is more conducive for the strands to denature. c. The high temperature would increase the formation of the hydrogen bonds between bases. d. Positive supercoiling would allow the DNA to maintain its double-stranded structure at higher temperature. e. Positive supercoiling would allow the DNA to readily separate for transcription and replication.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 11_7e 62. Jack and Jill's son Jake has a severe case of myclonic epilepsy and ragged-red fiber (MERRF) syndrome. His case includes frequent and disabling myclonic seizures (involuntary twitching of the muscles) along with hearing loss, exercise intolerance, and poor night vision. Like most cases of MERRF, his case is associated with a mitochondrial mutation that he inherited from his mother Jill. His mother doesn't know that she harbors the MERRF mutation among her mtDNA molecules, but she has experienced occasional mild muscle twitching throughout her life and she does not see very well at night. What is/are the MOST likely explanation(s) for the difference in the severity of MERRF between Jake and his mother? (Select all that apply.) a. Heteroplasmy for mtDNA molecules in the cells of his mother is responsible. b. Some random mutations took place in Jack's mitochondria, which caused MERFF syndrome as his mother does not have full symptoms. c. It is likely that Jake has a higher proportion of mutant mtDNA molecules in his cells compared to his mother. d. The expression pattern of the mutant gene may be different in males from in females.

63. Draw a pedigree using the following information, filling in symbols for the mitochondrial disease: In generation I, the mother is affected with a mitochondrial disease but the father is not. They have two children. The older, a male, has a son with an unaffected woman. The younger, a female, has a daughter with an unaffected man.

64. Explain why DNA–DNA hybridization might be useful in helping to assess evolutionary relationships.

65. Explain why the symptoms for human mitochondrial diseases usually appear in adulthood and become worse with age.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 11_7e 66. A plant has green leaves with multiple yellow spots. When used as an egg donor in a cross with a normal plant that has all green leaves, some of the progeny have green and yellow leaves and some have all green leaves.

When used as the pollen donor in a cross with a normal plant, all the progeny have all green leaves. One of the progeny with green and yellow leaves has one branch with all green leaves and one branch with all yellow leaves. A flower on the all-green branch is fertilized with pollen from a flower on the all-yellow branch. Predict the results of this cross and the reciprocal cross.

67. While doing fieldwork, you discover two new closely related species of oysters. You measure the DNA content per cell of each species and find that the second species has significantly more DNA than the first species. DNA hybridization analysis of both species yields the following results. Based on the graphs shown, suggest an explanation for the difference in C-value between the species.

68. A plant has green leaves with multiple yellow spots. When used as an egg donor in a cross with a normal plant that has all green leaves, some of the progeny have green and yellow leaves and some have all green leaves.

When used as the pollen donor in a cross with a normal plant, all the progeny have all green leaves. Explain the results of the crosses with the plant.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 11_7e 69. "Mitochondrial Eve" is the name given to the idea that all humans alive today can trace their mitochondrial ancestry to a single African female alive around 150,000 years ago. What feature of mitochondrial inheritance makes this a reasonable conjecture?

70. Describe the relationship between nucleosomes and chromosome structure.

71. There are some genomes that have been reported to be positively coiled instead of negatively supercoiled, which is the status of most genomes that we have studied. The genomes that are positively supercoiled seem to belong to viruses and cells that exist at very high temperatures. Why might positive supercoiling be an advantage at high temperatures?

72. How does the organization of the eukaryotic chromosome differ from the organization of a bacterial chromosome? Include in your answer (a) classes of DNA sequences, (b) special features of the chromosome, (c) organization of the genes within the chromosome, and (d) proteins that interact with chromosomal DNA.

73. At the end of the nineteenth century, American bison were bred with domestic cattle in an attempt to rescue their declining populations. The resulting hybrids were bred with true bison. After many generations, animals that look like bison may still contain some ancestral cattle DNA. Conservation geneticists have considered isolating bison that show no cattle ancestry by the mitochondrial DNA test and breeding them to maintain a group of "pure" bison. The number of bison with no cattle genes is estimated at 15,000 animals (The New York Times, 23 April 2002, "Genetically, bison don't measure up to frontier ancestors"). List one reason this might not be a good idea.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 11_7e 74. What is "satellite" DNA? Explain why satellite DNA anneals rapidly after it is denatured.

75. Two haploid strains of petite yeast mutants are obtained independently. Each is crossed to a wild-type strain, and the resulting diploid is sporulated (goes through meiosis to produce haploid spores). Use the following results to explain the difference between the two strains and why the crosses give different results.

76. At the end of the nineteenth century, American bison were bred with domestic cattle in an attempt to rescue their declining populations. The resulting hybrids were bred with true bison. After many generations, animals that look like bison may still contain some ancestral cattle DNA. Ward et al. (Animal Conservation 2: 51– 57, 1999) tested current bison herds for a mitochondrial DNA marker specific to cattle. Of the North American bison they tested, 5.2% had the cattle-specific mitochondrial DNA marker. Does this under- or overestimate the number of bison–cattle hybrids? Explain your answer.

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 11_7e 77. Using the pedigree below, explain whether each of the following inheritance patterns is possible for the phenotype being followed citing specific individuals in your answer: X-linked dominant, X-linked recessive, Ylinked, mitochondrial, autosomal recessive, and autosomal dominant.

78. A new Drosophila phenotype is investigated with a series of crosses. P (parental) organisms are truebreeding. The following is the first cross:

Predict the F2 results if the allele that causes the mutant phenotype is X-linked recessive. Then predict the results if the allele causing the mutant phenotype is mitochondrial.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 11_7e 79. If a bacterial chromosome were inserted into a eukaryotic cell, would it be stable? Would it segregate like eukaryotic chromosomes do during mitosis and meiosis? Why or why not?

80. While working on Drosophila, you find a new mutant strain with an abnormal histone H3 gene. This novel histone mutant is predicted to cause the nucleosomes to bind an extra 15 bp of DNA compared to wild type.

If you digest isolated chromatin with a nuclease to release the core nucleosomes, what size DNA fragments would you expect from wild-type and mutant flies?

81. List and describe the three major classes of DNA sequences in the eukaryotic genome.

82. A new Drosophila phenotype is investigated with a series of crosses. P (parental) organisms are truebreeding. The following is the first cross:

A reciprocal cross to the one just shown is performed. Predict the results in the F1 and F2 generations if the mutant phenotype is X-linked recessive. Then predict the results if the allele causing the phenotype is mitochondrial.

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 11_7e 83. Describe the structure and packing of a bacterial chromosome.

84. Jack and Jill's son Jake has a severe case of myclonic epilepsy and ragged-red fiber (MERRF) syndrome. His case includes frequent and disabling myclonic seizures (involuntary twitching of the muscles) along with hearing loss, exercise intolerance, and poor night vision. Like most cases of MERRF, his case is associated with a mitochondrial mutation that he inherited from his mother Jill. His mother didn't know that she harbors the MERRF mutation among her mtDNA molecules, but she has experienced occasional mild muscle twitching throughout her life, and she does not see very well at night. a. What is the most likely explanation for the difference in the severity of MERRF between Jake and his mother? b. Jack and Jill would like to have another child. They want to know the probability that their next child will also suffer from MERRF syndrome. How would you counsel them?

85. When chromatin from any eukaryote is digested with micrococcal nuclease (an endonuclease) and fractionated using electrophoresis, DNA fragments of approximately 200 base pairs in length are observed.

Explain this result.

86. A plant has green leaves with multiple yellow spots. When used as an egg donor in a cross with a normal plant that has all green leaves, some of the progeny have green and yellow leaves and some have all green leaves.

When used as the pollen donor in a cross with a normal plant, all the progeny have all green leaves. Explain the phenotype of the plant's leaves.

Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 11_7e Answer Key 1. a 2. a 3. b 4. b 5. d 6. c 7. b 8. b 9. b 10. b 11. a 12. c 13. c 14. a 15. c 16. b 17. c 18. e 19. c 20. b 21. b 22. b 23. d 24. e 25. b 26. c Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 11_7e 27. a 28. d 29. e 30. b 31. c 32. b 33. c 34. c 35. b 36. b 37. d 38. d 39. b 40. d 41. b 42. c 43. b 44. b 45. b 46. a 47. c 48. b 49. a 50. d 51. a, b 52. a, b, c, e 53. a, c, e 54. a, d Copyright Macmillan Learning. Powered by Cognero.

Page 30

Name:

Class:

Date:

Chap 11_7e 55. a, b 56. a, c, e 57. c, e, f 58. a, b, c, e 59. a, d, e 60. b, c 61. d 62. a, c

63.

64. Two single-stranded DNA molecules will anneal if complementary; the strength of the association will be greater the more the two strands are complementary. We assume that the greater the similarity of DNA sequence, the more closely related the two species.

65. Mitochondrial DNA mutations affect oxidative phosphorylation and the production of ATP. This process declines with age, so someone with weakened mitochondrial function would show symptoms when his or her capacity for oxidative phosphorylation declined past a critical threshold. Symptoms would become worse as the ability to perform oxidative phosphorylation declined.

66. When the all-green branch supplies the egg, the progeny will all have normal chloroplasts and have all green leaves.

When the all-yellow branch supplies the egg, the progeny will all have mutant chloroplasts and will not survive.

67. Both species appear to have the same amount of highly repetitive and unique DNA, but the second species appears to have significantly more moderately repetitive DNA. One possible explanation for the presence of more moderately repetitive DNA in the second species is accumulation of replicative transposons in that species but not in the first.

Copyright Macmillan Learning. Powered by Cognero.

Page 31

Name:

Class:

Date:

Chap 11_7e 68. The chloroplasts are inherited maternally in this plant. When the plant is the egg donor, by replicative segregation, some eggs will contain only normal chloroplasts and the progeny will have all green leaves. Some eggs will be heteroplasmic and have the same phenotype as the maternal parent. Presumably, eggs that are homoplasmic for mutant chloroplasts will not produce viable plants. When the plant is the pollen donor, the plant with nonmutant chloroplast DNA will contribute the chloroplasts, and all progeny will have all green leaves.

69. Mitochondrial Eve is the most recent common matrilineal ancestor of all humans as measured through the mitochondria. mtDNA is maternally inherited, but more importantly, there is no recombination between parental genomes as is the case with nuclear genes. Therefore, it is possible to easily follow the inheritance of the mitochondria from generation to generation.

70. Chromosomes appear to be able to form in the absence of nucleosomes. Nucleosomes do, however, appear to play at least some role in chromosome structure within the cell. In the absence of nucleosomes, researchers observed chromosomes that are wider and more fragile than those associated with nucleosomes. The main role of nucleosomes appears to be in compacting and protecting the chromosome, but they may not be essential in creating the chromosome shape.

71. Negative supercoiling makes it easier for the two strands of DNA to separate (denature), whereas positive supercoiling makes it more difficult for strands to separate. At high temperatures the heat energy is already producing a tendency for the strands to separate by breaking the hydrogen bonds. While separation is needed for replication and transcription to occur, it is necessary for DNA to retain its double-stranded structure at other times, and positive supercoiling likely facilitates this.

Copyright Macmillan Learning. Powered by Cognero.

Page 32

Name:

Class:

Date:

Chap 11_7e 72. a. Bacterial chromosomes generally consist of mainly unique DNA sequences with very little moderately repetitive or highly repetitive DNA. In contrast, eukaryotic chromosomes typical contain unique, moderately repetitive DNA and highly repetitive DNA. The unique component contains many of the protein-encoding genes as well as other poorly defined unique sequences. The moderately repetitive sequences consist of some gene families, telomeric DNA, and many of the transposable elements. The highly repetitive DNA, which is unlikely to be transcribed, can be found within the centromeric regions in some species and elsewhere. Repetitive DNA sequences may be tightly coiled and may stain darkly (heterochromatin). b. The typical bacterial genome is assumed to consist of one circular chromosome, while exceptions are known. Eukaryotic genomes consist of multiple linear chromosomes with a characteristic number for each species. Eukaryotic chromosomes consist of two telomeres and usually, but not always, a single centromere.

Telomeres are essential for protecting the ends of linear chromosomes and allowing a way of completing chromosome replication without resulting in the shortening of the linear chromosome (see telomerase in Chapter 12). Since bacterial chromosomes are usually circular, telomeres are not needed. The centromere is the portion of the chromosome where microtubules attach during mitosis and meiosis. It often consists of highly repetitive DNA although a specific sequence is not essential to its function. The bacterial chromosome doesn't possess a centromere, although it does have a mechanism to ensure the accurate segregation of daughter chromosomes to daughter cells during cell division. The details of this mechanism are still not completely understood. c. The bacterial chromosome consists of mainly genes with some transposable elements. The DNA sequences between these structures are usually quite short so that the gene density (number of genes per a particular number of nucleotides of DNA) is high. In contrast, eukaryotic chromosomes are often "gene sparse" with few genes per the same number of nucleotides and many sequences not involved in forming genes. Many eukaryotic genes are part of gene families whereas bacterial genes are usually unique with the main exception being the rRNA genes. d. Both types of chromosomes interact with a variety of proteins. With eukaryotes, the nucleosome with its set of histone proteins is a basic structural unit of the chromosome, involved in both the packaging of the DNA and the regulation of gene activity. Additional proteins also interact with eukaryotic DNA to provide additional gene regulation and higher levels of chromosome organization. Bacterial chromosomes also interact with certain proteins, although not with histones, and such proteins are likely involved in gene regulation and chromosome structure and organization, but the details are not yet known.

73. Bison were already experiencing a population bottleneck at the end of the nineteenth century, when the number of breeding adults was low. Their genetic diversity was reduced at that time, so selecting only pure animals returns the population to bottleneck numbers. Selecting bison by the mitochondrial DNA test only might mean including bison that have cattle genes inherited from paternal ancestors.

74. Satellite DNA, also called highly repetitive DNA, is one of three major classes of DNA sequences found in eukaryotic genomes. This class of DNA is present in hundreds of thousands to millions of copies. Highly repetitive fractions of DNA exhibit rapid reassociation kinetics because they are characterized by having numerous (up to millions) copies of identical sequence repeats; thus, once melted, complementary partner strands are readily available and can rapidly re-anneal.

Copyright Macmillan Learning. Powered by Cognero.

Page 33

Name:

Class:

Date:

Chap 11_7e 75. Strain A has a recessive nuclear mutation that causes the petite phenotype. Crossing with a wild type gives a heterozygous diploid. When sporulated, half the haploid spores have the normal allele and half have the recessive allele that causes the petite phenotype.

Strain B has a mitochondrial mutation. Crossing with a wild type gives a heteroplasmic diploid. When sporulated, spores randomly receive either normal mitochondria, mitochondria with the mutant petite allele, or a mixture.

Those with normal mitochondria or a mixture will be non-petite, but the exact proportion of wild-type non-petites is random.

76. It underestimates the number of bison with cattle ancestry because it detects only bison with maternal cattle ancestry.

77. (1) The phenotype cannot be X-linked dominant because II-4 doesn't have the same phenotype as I-1.

(2) The phenotype cannot be X-linked recessive. III-3 and III-4 don't have the same phenotype as II-5.

(3) The phenotype cannot be Y-linked. II-2 doesn't have the same phenotype as I-1; III-3 and III-4 don't have the same phenotype as the II-6.

(4) The phenotype cannot be mitochondrial. II-3 and II-5 don't have the same phenotype as I-2. III-3 and III-4 don't have the same phenotype as II-5.

(5) The phenotype cannot be autosomal recessive. III-3 and III-4 don't have the same phenotype as II-5.

(6) The phenotype could be autosomal dominant if II-5 and II-6 are heterozygotes and III-3 and III-4 are homozygous recessive.

78. If X-linked recessive, 3/4 of the F2 will be wild type; 2/3 of the wild type will be female and 1/3 will be male; 1/4 of the F2 will be mutant, and all will be male. If mitochondrial, all F2, male and female, will be wild type.

79. No, it would not be stable because (1) it would not contain a eukaryotic-specific origin of replication, and therefore it could not replicate in a eukaryotic cell, and (2) bacterial chromosomes don't have centromeres, so they can't segregate in eukaryotic cells (no kinetochores, no spindle fibers, etc.). The bacterial chromosome would be lost and eventually degraded.

80. The core nucleosome from wild-type flies wraps approximately 145 bp of DNA; therefore, if the mutant histone is predicted to wrap an extra 15 bp, the mutant nucleosomes should wrap 160 bp.

81. (1) Unique sequences (1–10 copies per genome) (2) Moderately repetitive sequences (10–100,000 copies per genome; repeats average 150–300 base pairs) (3) Highly repetitive sequences (100,000 millions of copies per genome; repeats usually, but not always, 10 base pairs or fewer) 82. If X-linked recessive, 1/2 the F1 will be wild-type females and 1/2 will be mutant males. In the F2, 1/4 will be wildtype males, 1/4 will be wild-type females, 1/4 will be mutant males, and 1/4 will be mutant females. If mitochondrial, all F1 and all F2 will be mutant.

83. Bacterial cells typically contain a single, circular DNA molecule associated with various proteins to make a compact structure called a nucleoid. Bacteria do not contain histones. Chromosome packing is thought to organize the genetic material into discrete looped domains containing genes that can be selectively activated by relaxing the degree of supercoiling within given domains.

Copyright Macmillan Learning. Powered by Cognero.

Page 34

Name:

Class:

Date:

Chap 11_7e 84. a. The most likely explanation is heteroplasmy for mtDNA molecules in the cells of Jake and his mother. When a female is heteroplasmic for mtDNA, the proportion of mutant mtDNA that gets passed to an offspring is random.

Therefore, it is likely that Jake has a higher proportion of mutant mtDNA molecules in his cells compared to his mother, producing a more severe syndrome. b. Counseling in this situation is highly problematic. This chapter cited the case of a 20-year-old man who carried the MERRF mutation in 85% of his mtDNA and was normal and his cousin who carried it in 96% of his mitochondria and was severely affected. There appears to be a sharp threshold in the percentage of abnormal mtDNA between having the syndrome and not having it. Later in the chapter the author describes a bottleneck during which the mtDNA within the cells is reduced to just a few copies. During this bottleneck, the percentage of abnormal mtDNA can increase or decrease randomly and significantly, making it impossible to predict with confidence the percentage of abnormal mtDNA that will be passed to the next generation. Therefore, it is impossible to predict with confidence the probability that the next child will be affected.

85. Chromatin is composed of repeating nucleosomes that consist of about 200 base pairs of DNA wrapped around a histone octamer complex. DNA wrapped around the octamer core is protected from endonuclease degradation.

However, because the linker DNA between adjacent nucleosomes is exposed, it can be cleaved by endonucleases. Thus, chromatin digested by endonucleases generates 200-base-pair-long fragments of DNA. 86. The plant is heteroplasmic for a mutation in the chloroplast genome that affects chloroplast development and/or photosynthesis. The yellow spots are cells that, by replicative segregation, have received only mutant chloroplast genomes. The green areas are cells that are either homoplasmic for normal chloroplasts or heteroplasmic.

Copyright Macmillan Learning. Powered by Cognero.

Page 35

Name:

Class:

Date:

Chap 12_7e Indicate the answer choice that best completes the statement or answers the question.

The nuclear genome of a single human cell (i.e., the entire diploid complement) contains about 6.6 billion (6.6 × 109) base pairs of DNA. If synthesis at each replication fork occurs at an average rate of 50 nucleotides per second, all the DNA is replicated in 5 minutes. Assume that replication is initiated simultaneously at all origins.

1. How many origins of replication exist in a human diploid genome? a. 220,000 b. 440,000 c. 880,000 d. 2.64 × 107 e. 1.32 × 108 Use the figure below to answer the following questions.

2. Which of the results (A through F) would give a clear experimental conclusion for the semiconservative model? a. A and D b. B and E c. C and F d. D and E e. E and F

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 12_7e 3. Telomerase uses _____ to synthesize new DNA. a. exonuclease activity b. a licensing factor c. strand invasion d. a DNA template e. an RNA template 4. You learn that a Mars lander has retrieved a bacterial sample from the polar ice caps. You obtain a sample of these bacteria and perform the same kind of experiment that Meselson and Stahl did to determine how the Mars bacteria replicate their DNA. Based on the following equilibrium density gradient centrifugation results, what type of replication would you propose for these new bacteria?

a. conservative b. semiconservative c. dispersive d. semiconservative or dispersive e. conservative or dispersive 5. Which activity is NOT associated with DNA polymerases? a. ability to attach a DNA nucleotide to the 3′ end of previously incorporated DNA nucleotide b. ability to excise a newly incorporated nucleotide that does not match the template strand c. ability to "read" a template strand 3′ to 5′ and synthesize a complementary strand d. ability to synthesize a DNA from scratch without a primer e. ability to synthesize new DNA in a 5′ to 3′ direction 6. During DNA replication, the synthesis of the new strand requires the addition of a new dNTP to the 3'-OH group of the growing nucleotide strand by DNA polymerase. Which of the following provides the energy needed for this step? a. the template strand b. the energy of pentose sugar c. 5' end of the growing DNA d. cleavage of two phosphate groups from the dNTP e. The reaction does not require an energy source.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 12_7e 7. The diagram shown below is one half of a replication bubble. As the DNA template strands unwind toward the right, new strands of DNA get synthesized. Which of the following represents a lagging strand?

a. A b. B c. C d. D e. There is no lagging strand in this diagram.

8. Which of the following statements is TRUE of DNA polymerases of eukaryotic cells? a. The same DNA polymerase replicates mitochondrial, chloroplast, and nuclear DNA. b. There are only two different DNA polymerases that function in the process of replication. c. Some DNA polymerases have the ability to function in DNA repair mechanisms. d. All eukaryotic DNA polymerases have 3′ → 5′ exonuclease activity. e. Leading strand synthesis and lagging strand synthesis are performed by the same type of DNA polymerase.

9. What is the function of DNA ligase? a. connects Okazaki fragments by sealing nicks in the sugar–phosphate backbone b. unwinds the double helix by breaking the hydrogen bonding between the two strands at the replication fork c. reduces the torsional strain that builds up ahead of the replication fork as a result of unwinding d. binds to oriC and causes a short section of DNA to unwind e. prevents the formation of secondary structures within single-stranded DNA 10. Okazaki fragments are found in all of the following EXCEPT in a. the leading strand. b. the lagging strand. c. eukaryotic DNA. d. bacterial DNA. e. linear replication models.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 12_7e 11. What is the function of single-strand-binding proteins? a. connect Okazaki fragments by sealing nicks in the sugar–phosphate backbone b. unwind the double helix by breaking the hydrogen bonding between the two strands at the replication fork c. reduce the torsional strain that builds up ahead of the replication fork as a result of unwinding d. bind to oriC and cause a short section of DNA to unwind e. prevent the formation of secondary structures within single-stranded DNA 12. For which of the following is the "end-replication problem" relevant? a. circular DNA b. linear chromosomes c. the centromere region of a chromosome d. rolling-circle model of replication e. theta model of replication 13. Which of the following is a protein that facilitates the termination of replication in E. coli? a. telomerase b. DNA gyrase c. Tus d. primase e. topoisomerase

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 12_7e Use the figure below to answer the following questions.

14. Which of the results (A through F) would give a clear experimental comparison to distinguish between the semiconservative model and dispersive model? a. A and D b. B and E c. C and F d. D and E e. E and F 15. In the diagram below, which letter indicates the 5' end of the leading strand?

a. A b. B c. C d. D e. C and D

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 12_7e The nuclear genome of a single human cell (i.e., the entire diploid complement) contains about 6.6 billion (6.6 × 109) base pairs of DNA. If synthesis at each replication fork occurs at an average rate of 50 nucleotides per second, all the DNA is replicated in 5 minutes. Assume that replication is initiated simultaneously at all origins.

16. Assuming that the origins are approximately equally distributed across the chromosomes, what is the average number of origins per human chromosome? a. 4783 b. 19,130 c. 4.6 × 106 d. 1.21 × 109 e. 2.9 × 1010 17. All DNA polymerases synthesize new DNA by adding nucleotides to the _____ of the growing DNA chain. a. 3′ OH b. 5′ OH c. 3′ phosphate d. 5′ phosphate e. nitrogenous base 18. You are studying a new virus with a DNA genome of 12 Kb. It can synthesize DNA at a rate of 400 nucleotides per second. If the virus uses rolling-circle replication, how long will it take to replicate its genome? a. 7.5 seconds b. 15 seconds c. 30 seconds d. 1 minute e. 2 minutes 19. What type of bonds does DNA ligase create between adjacent nucleotides? a. hydrogen b. phosphodiester c. ionic d. metallic e. ribonucleotide

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 12_7e 20. If a deletion occurs in a gene that encodes DNA polymerase I and no functional DNA polymerase I is produced, what will be the MOST likely consequence of this mutation? a. The DNA strands would contain pieces of RNA. b. The DNA would not exist in a supercoiled state. c. There would be no DNA replication on the leading or lagging strands. d. There would be no RNA primers laid down. e. The DNA will not be able to unwind to initiate replication.

21. Which of the following enzymes do NOT aid in the unwinding of DNA for replication? a. helicase b. single-stranded binding proteins c. primase d. gyrase e. topoisomerase 22. What is the function of DNA gyrase? a. connects Okazaki fragments by sealing nicks in the sugar–phosphate backbone b. unwinds the double helix by breaking the hydrogen bonding between the two strands at the replication fork c. reduces the torsional strain that builds up ahead of the replication fork as a result of unwinding d. binds to oriC and causes a short section of DNA to unwind e. prevents the formation of secondary structures within single-stranded DNA 23. Telomerase activity is MOST likely to be found in which cells in humans? a. red blood cells b. muscle cells c. neurons d. germ line e. liver cells 24. If ribonucleotides were depleted from a cell during S phase, how would DNA synthesis be affected? (Ignore energetic considerations.) a. There would be no effect because ribonucleotides are used in RNA synthesis, not DNA synthesis. b. DNA synthesis would continue but at a slower rate. c. There would only be an effect during M phase, not in S phase. d. DNA synthesis would not be affected because ribonucleotides are only used during the process of transcription. e. Replication would cease because ribonucleotides are required to initiate DNA synthesis.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 12_7e 25. Which of the following statements BEST describes heteroduplex DNA? a. A single-stranded DNA molecule of one chromosome pairs with a single-stranded DNA molecule of another chromosome. b. A single-stranded DNA molecule of one chromosome pairs with a single-stranded RNA molecule of another chromosome. c. Heteroduplex DNA consists of sequences from two different species that are brought together through homologous recombination. d. Heteroduplex DNA consists of an RNA primer and newly synthesized DNA on the lagging strand. e. In heteroduplex DNA newly synthesized DNA has yet to be reassembled into nucleosomes.

26. The origin-recognition complex, ORC, loads the enzyme helicase onto eukaryotic dsDNA during G1 of the cell cycle. What is the consequence of this? a. Unwinding of dsDNA for replication begins in G1 in eukaryotes. b. Unwinding of dsDNA for replication occurs in S phase because the helicase loaded by ORC can't act until an initiator protein first melts the DNA in S phase. c. Unwinding of dsDNA for replication occurs in S phase because the helicase loaded by ORC isn't activated until S phase. d. Unwinding of dsDNA for replication occurs in S phase because a second helicase is required that is only loaded onto the dsDNA in G1. e. None of the above statements are correct.

27. Which of the following statements is TRUE regarding nucleosome formation during DNA replication? a. Nucleosomes are only reassembled on the lagging strand. b. Nucleosome assembly consists entirely of newly synthesized histones. c. Nucleosome assembly occurs at a faster rate in prokaryotes than in eukaryotes. d. The addition of newly synthesized histones is a part of nucleosome assembly. e. Nucleosome assembly does not occur during semiconservative replication.

28. Which of the following does NOT utilize bidirectional replication? a. theta model b. rolling-circle model c. linear model d. eukaryotes e. bacteria

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 12_7e 29. DNA synthesis during replication is initiated from a. a free 5′ OH. b. DNA primers. c. RNA primers. d. telomerase. e. DNA polymerase I. 30. What type of synthesis occurs on the leading strand? a. conservative b. dispersive c. continuous d. discontinuous e. recombinant 31. DNA primase requires a _____ template and _____ nucleotides to initiate primer synthesis. a. DNA; DNA b. RNA; RNA c. DNA; RNA d. RNA; DNA e. leading strand; DNA 32. Suppose Meselson and Stahl had obtained the following results in their experiment. These results would be consistent with which model of replication?

a. conservative replication only b. semiconservative and conservative replication c. semiconservative replication only d. dispersive replication only e. semiconservative and dispersive replication

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 12_7e 33. Which of the following typically only have one origin of replication? a. humans b. prokaryotes c. eukaryotes d. linear model of replication e. plants 34. Meier-Gorlin syndrome results from flaws in DNA replication. Which step in DNA replication is compromised in Meier-Gorlin syndrome? a. licensing of replication origins b. DNA polymerase activity c. helicase activity d. leading strand synthesis e. lagging strand synthesis 35. Which of the following enzyme and function pairs is INCORRECTLY matched? a. DNA gyrase; making and resealing the break to remove the torque as the DNA unwinds b. DNA ligase; sealing nicks in the sugar–phosphate backbone of newly synthesized DNA c. DNA helicase; rewinding and reforming the DNA double helix as the replication terminates d. DNA polymerase III; elongating a new nucleotide strand from the 3'-OH provided by the primers e. Initiator protein; binding to replication origin and separating DNA to initiate replication 36. The proofreading function of DNA polymerases involves _____ activity. a. 5′ → 3′ exonuclease b. 3′ → 5′ exonuclease c. 5′ → 3′ telomerase d. 3′ → 5′ telomerase e. 5′ → 3′ gyrase 37. What would be a likely result of expressing telomerase in somatic cells? a. premature aging b. cancer c. lower rates of replication d. immortality of gametes e. early termination of replication

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 12_7e 38. DNA polymerase I and DNA polymerase III both have _____ activity but only DNA polymerase I has _____ activity. a. 5′ → 3′ exonuclease; 3′ → 5′ exonuclease b. 5′ → 3′ polymerase; 3′ → 5′ polymerase c. 3′ → 5′ polymerase; 5′ → 3′ polymerase d. 3′ → 5′ exonuclease; 5′ → 3′ exonuclease e. 5′ → 3′ polymerase; 3′ → 5′ exonuclease 39. _____ are tandemly repeated DNA sequences located at the ends of eukaryotic chromosomes. a. Replication bubbles b. Telomeres c. Nucleosomes d. Licensing factors e. Holliday junctions 40. Suppose that some cells are grown in culture in the presence of radioactive nucleotides for many generations so that both strands of every DNA molecule include radioactive nucleotides. The cells are then harvested and placed in new medium with nucleotides that are not radioactive so that newly synthesized DNA will not be radioactive. What proportion of DNA molecules will contain radioactivity after two rounds of replication? a. 0 b. 1/8 c. 1/4 d. 1/3 e. 1/2

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 12_7e Use the figure below to answer the following questions.

41. Which of the results (A through F) would give a clear experimental clue for the conservative model? a. A and D b. B and E c. C and F d. D and E e. E and F 42. Which one of the following statements is NOT true for all E. coli DNA polymerases? a. They require a primer to initiate synthesis. b. They use dNTPs to synthesize new DNA. c. They produce newly synthesized strands that are complementary and antiparallel to the template strands. d. They possess 5′ → 3′ exonuclease activity. e. They synthesize in the 5′ → 3′ direction by adding nucleotides to a 3′-OH group.

43. Which of the following eukaryotic DNA polymerases and their role are INCORRECTLY paired? a. δ (delta); lagging strand synthesis of nuclear DNA b. ε (epsilon); translesion DNA synthesis c. η (eta); translesion DNA synthesis d. γ (gamma); replication and repair of mitochondrial DNA e. θ (theta); DNA repair

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 12_7e 44. Which of the following is a necessary step in the Holliday model of recombination? a. DNA primase generates an RNA primer. b. The RNA template of telomerase binds to the telomere. c. Topoisomerases aid in supercoiling. d. DNA polymerase α initiates DNA synthesis. e. A single-strand break occurs in the DNA molecule.

45. The Holliday model describes which of the following processes? a. semiconservative replication b. homologous recombination c. end replication d. RNA primer synthesis e. rolling-circle replication 46. Meselson and Stahl showed that DNA is replicated by a a. conservative system. b. semiconservative system. c. dispersive system. d. semidispersive system. e. conservative system in prokaryotes and dispersive system in eukaryotes.

47. Approximately 6.6 Gb of DNA is present in a single-celled human zygote and must be replicated faithfully at each cell division. What is the approximate size of the haploid human genome? a. 1 Mb b. 3.2. Mb c. 3.2 Gb d. 6.6 Gb e. 12.8 Gb 48. DNA polymerases require all of the following for DNA replication EXCEPT a. a DNA template. b. a primer. c. free 3′ OH. d. 3′ to 5′ polymerase activity. e. dNTPs.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 12_7e 49. You are studying a new virus with a DNA genome of 12 Kb. It can synthesize DNA at a rate of 400 nucleotides per second. If the virus uses theta replication, how long will it take to replicate its genome? a. 7.5 seconds b. 15 seconds c. 30 seconds d. 1 minute e. 2 minutes Indicate one or more answer choices that best complete the statement or answer the question.

50. The radioactive labeling experiment similar to Meselson and Stahl's experiment was performed to determine what happens to histone proteins in eukaryotic DNA replication. Preexisting and newly synthesized histone proteins were labeled with radioactive isotopes of different atomic mass. Select the statement(s) that are consistent with the diagram shown. (Select all that apply.)

a. The newly synthesized histone and preexisting histones exist separately and associate with a different pool of DNA. b. The newly synthesized histone and preexisting histones mix and form smeared bands, representing heterogeneous mixture. c. The newly synthesized histone only associates with newly synthesized DNA while the preexisting histone associates with the old DNA. d. The newly synthesized histone and preexisting histones tend to mix together and reassociate with DNA after each round of the replication. e. The preexisting histones tend to associate with newly synthesized DNA while the newly synthesized histone associates exclusively with the old DNA.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 12_7e 51. Which statement CORRECTLY describes why the DNA synthesis is discontinuous on one template strand while continuous on the other? (Select all that apply.) a. It is due to the antiparallel nature of the DNA molecule. b. It is due to the fact that the DNA polymerase is not active enough. c. It is due to the nature of the DNA polymerase which can only add new nucleotides at the 3' end. d. It is due to the difficulty in unwinding the DNA template for replication. e. It is due to the semiconservative nature of DNA replication.

52. Which of the following statements explaining the mechanism of action of telomerase is correct? (Select all that apply.) a. The extension of telomeres via telomerase occurs at the 3' overhang with a G-rich sequence. b. Telomerase carries a short single-stranded DNA component that is used as a template to extend the telomere. c. The extension of telomeres via telomerase occurs at the 5' end, where DNA polymerase cannot add a new deoxynucleotide after the removal of the very last RNA primer. d. The extension at the 3' overhang is followed by the extension of 5' complementary strand via a currently unknown mechanism. e. Telomerase finishes complementary base pairing of the 5' complementary strand.

53. Explain the need for and mechanism of licensing of DNA replication.

54. Draw a replication fork and indicate all key components and orientations, including lagging and leading strands, DNA helicase, RNA primer, and DNA gyrase.

55. List and describe the key events of DNA replication in their relative order for E. coli.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 12_7e 56. Discuss the main differences in the initiation of recombination proposed by the Holliday model and the doublestrand-break model.

57. You have discovered a special dye that reveals the position of recombination sites on meiotic chromosomes.

You use this dye to count the number of recombination sites and then compare this to the number of genetic exchanges that you can detect by looking at the segregation of markers across the genome. You find many more recombination sites as compared to genetic exchanges. Explain this result.

58. Why do eukaryotic chromosomes have multiple origins of replication, whereas prokaryotic chromosomes typically have only one origin?

59. You discover a drug that specifically inhibits DNA synthesis. You apply the drug to yeast cells after S phase but before prophase of meiosis. You find a drastic reduction in recombination in the meiotic products. How can you explain this?

60. Summarize the similarities and differences in rolling-circle replication, theta replication, and linear eukaryotic replication.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 12_7e 61. Explain the effect on DNA replication of mutations that destroy each of the following activities in DNA polymerase. Also, for each kind of mutation, how might you detect the effect in an in vitro replication reaction? a. 5′ to 3′ polymerase b. 5′ to 3′ exonuclease c. 3′ to 5′ exonuclease

62. What does the term "noncrossover recombinant" mean with respect to homologous recombination?

Nalidixic acid, a first-generation quinolone, was introduced in the 1960s. It is an antibiotic that inhibits topoisomerase activity.

63. Why is this antibiotic safe for human treatment?

64. Levofloxacin, norfloxacin, and ciprofloxacin are examples of second-generation quinolones that also treat bacterial infections by inhibiting bacterial topoisomerase II activity. Why was the development of these second-generation drugs necessary?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 12_7e 65. In the diagram below, the arrow indicates the direction of movement of a replication fork.

a. Which strand, top or bottom, is the leading strand? Explain. b. On which strand, top or bottom, would you expect to find Okazaki fragments? Explain.

66. Why did the Meselson and Stahl experiment require two rounds of replication?

67. What does "proofreading" refer to with regard to DNA replication?

68. Are Okazaki fragments formed on the leading strand during DNA replication? Explain your answer.

69. Which different DNA polymerases are found in eukaryotic cells? What are their functions?

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 12_7e 70. (a) Explain the significance of telomerase in cells of the germ line. (b) Explain the significance of telomerase in cancer cells.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 12_7e Answer Key 1. a 2. a 3. e 4. c 5. d 6. d 7. b 8. c 9. a 10. a 11. e 12. b 13. c 14. d 15. c 16. a 17. a 18. c 19. b 20. a 21. c 22. c 23. d 24. e 25. a 26. c Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 12_7e 27. d 28. b 29. c 30. c 31. c 32. e 33. b 34. a 35. c 36. b 37. b 38. d 39. b 40. e 41. c 42. d 43. b 44. e 45. b 46. b 47. c 48. d 49. b 50. b, d 51. a, c 52. a, d

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 12_7e 53. In order for the large genomes of eukaryotes to be replicated in a timely manner, replication occurs from thousands of origins scattered along the chromosomes. However, the use of multiple origins creates a problem of timing of replication: the entire genome must be replicated once, and only once, each cell cycle. To accomplish this, initiation of replication begins with "licensing" of replication origins early in the cell cycle. Licensing is accomplished by binding of a complex called MCM to the origin early in the cell cycle. After replication has begun at an origin during S phase, a protein called Geminin prevents MCM from binding to DNA and reinitiating replication at that origin. Thus, replication can occur from each origin only once each cell cycle. Geminin degrades after mitosis, allowing MCM to bind again and relicense each origin for the next round of replication.

54.

55. (1) Initiator proteins recognize and bind DNA at specific A/T-rich sites, called origins of replication, and denature (or "melt") the double strands.

(2) Single-strand DNA-binding proteins associate with the denatured DNA to stabilize the single strands and form an initial replication bubble.

(3) DNA helicase binds to the double-strand DNA ahead of the replication bubble and begins breaking the interbase hydrogen bonds.

(4) DNA primase synthesizes RNA primers using the single-strand DNA templates.

(5) DNA gyrase (a topoisomerase) associates with double-strand DNA in front of DNA helicase and breaks/rejoins DNA strands to relieve supercoils generated in the double helix as the replication fork progresses.

(6) DNA polymerase III extends the RNA primers. Continuous DNA polymerization occurs for the leading strand; discontinuous DNA polymerization occurs for the lagging strand. Okazaki fragments (short RNA/DNA hybrid sequences) are formed on the lagging strand.

(7) DNA polymerase I "repairs" the RNA primers and replaces RNA with DNA. (8) For the lagging strand only, DNA ligase seals the "nicks" (breaks in the phosphodiester backbone) remaining between adjacent nucleotides to produce a new, continuous DNA daughter strand.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 12_7e 56. (1) Holliday model: Recombination is initiated with single-stranded breaks at identical positions in homologous chromosomes. The free ends then invade the other (homologous) DNA molecules, and the broken ends join and displace the original complementary strands.

(2) Double-stranded-break model: Recombination is initiated with double-stranded breaks in one of the two aligned chromosomes. Nucleotides on either side of the breaks are degraded, and the partially degraded strand invades the other, intact strand and displaces the homologous strand as it is elongated.

57. Recombination, whether by the Holliday model or by the double-strand-break model, involves Holliday junctions as intermediates. These junctions can be resolved into either crossovers (exchanges) or noncrossovers. The excess of recombination sites must be due to noncrossovers.

58. The large genome size of a typical eukaryotic chromosome, compared to that of a prokaryotic chromosome, necessitates multiple origins of replication so that the genetic material can be copied in a reasonable amount of time.

Also replication is much slower in eukaryotes (the average is about 10–100 nucleotides/second) than in prokaryotes (the average is about 1000 nucleotides/second). For example, the human haploid genome complement is estimated to be about 3.3 × 109 nucleotide base pairs. Divide this number by 23 (the haploid number of human chromosomes) to get an average number of 3.3 × 109/23 = 1.4 × 108, or about 140 million nucleotides (nt) for a single human chromosome. Even at 100 nucleotides/second, it would take (140,000,000 nt) × (100 nt/sec)–1 = (1,400,000 sec) × (1 min/60 sec) × (60 min/hr)–1 × (1 day/24 hr) = 16.2 days to replicate one eukaryotic chromosome if it contained only one origin.

59. According to the double-strand-break model of recombination, after strand invasion DNA synthesis occurs at the 3′ ends of the DNA molecule that experienced the break. An inhibitor of DNA synthesis would block this and interrupt recombination.

60. (1) Rolling-circle replication • Initiated by cleavage in a single DNA strand • Uncleaved strand used as template • New nucleotides added to 3′ end of cleaved strand • Linear single-strand DNA produced; subsequently circularized (2) Theta replication • Two replication forks • Initiated by DNA denaturation at single origin • Single, expanding replication bubble • Bidirectional replication • Circular DNA molecule produced (3) Linear eukaryotic replication • Two replication forks • Initiated by DNA denaturation at multiple origins • Multiple, expanding replication bubbles • Bidirectional replication • Linear DNA molecule produced

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 12_7e 61. a. If the 5′ to 3′ polymerase activity is missing, no dNTPs will be incorporated into DNA. This could be detected using methods such as density gradient centrifugation or electrophoresis, which would allow one to see that no new DNA polymers are being synthesized. b. 5′ to 3′ exonuclease is needed for DNA polymerase to remove RNA primers from the 5′ ends of Okazaki fragments. If the activity is missing, the primers will not be removed from newly synthesized DNA. This could be detected by including radiolabeled ribonucleotides in the reaction mix and seeing that the radioactivity remains with DNA polymers at the end of the reaction. c. 3′ to 5′ exonuclease is needed for proofreading. It allows incorrectly incorporated nucleotides to be removed before proceeding to the next nucleotide. Lack of this activity would decrease the accuracy of replication and allow many more replication errors to occur. This could be detected by sequencing the products of replication and determining the number of errors.

62. The terms "crossover" and "noncrossover" refer to the arrangement of DNA near to, but not participating in, a homologous recombination event. In the process of recombination, two double-stranded DNA molecules join at a point of four-stranded DNA called the Holliday junction. When the Holliday junction is cleaved and DNA is rejoined into two separate, double-stranded DNA molecules, the composition of the resulting DNA molecules in adjacent regions—for example, a flanking pair of genes with one proximal and one distal—depends on which of two possible orientations the DNA strands are cleaved in. In one cleavage orientation, the segments of DNA that flank the region involved in recombination will be exchanged and emerge in a new, recombinant composition, like that discussed for single crossovers in meiosis in Chapter 5. Such a recombinant is therefore said to be of the crossover type. However, if the Holliday junction is resolved by cleavage in the other orientation, then all DNA segments that flank each side of the recombination region will emerge with the same composition of flanking DNA that was present before the recombination event. Recombinant DNA strands are still formed because a short segment of DNA has been exchanged, but the exchange is limited to the site of the initial DNA break(s) and the range of Holliday-junction migration. Such a DNA strand is, therefore, called a noncrossover recombinant.

63. The activity of these drugs is specific to type II bacterial topoisomerases (DNA gyrases). These drugs would not interact with the mammalian topoisomerases due to structural constraints.

64. Mutations in bacterial DNA gyrases to quinolones have arisen requiring the development of novel drugs to treat the same bacteria.

65. a. The bottom strand, which has 5′ to 3′ synthesis in the same direction as movement of the replication fork, is the leading strand. b. Okazaki fragments are found on the top strand, which experiences discontinuous synthesis because the direction of synthesis is opposite that of the movement of the replication fork.

66. The first round of replication, which produced a single band of intermediate density, was enough to rule out the conservative model of replication. However, a second round was necessary to show that now two bands of equal intensity were observed, consistent with semiconservative replication and inconsistent with dispersive replication.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 12_7e 67. Proofreading is the ability of DNA polymerases to back up, remove a newly added nucleotide, and replace it.

Proofreading enables the cell to correct mistakes if the wrong base is added during 5′ to 3′ DNA polymerization. If a noncomplementary base is incorporated, then base pairing with the template is disrupted, leading to altered conformation of the DNA polymerase on the DNA. These conformational perturbations stall the polymerase, which allows its 3′ to 5′ enzymatic activity to remove the mispaired base. Removal restores a normal conformation so that the DNA polymerase resumes 5′ to 3′ activity, incorporates the correct nucleotide in the vacant site, and continues replicating the template.

68. No, they are not. Okazaki fragments form because DNA polymerase adds nucleotides in only one direction (5′ to 3′), yet it must copy two DNA templates—the leading strand and the lagging strand—that are antiparallel. Once an RNA primer is synthesized 5′ to 3′ on the leading strand, DNA synthesis may begin in the 5′ to 3′ direction along the exposed template. The replication fork moves in the same physical direction that the DNA polymerase on the leading strand does, so exposed template is continuously supplied to the polymerase, and it may continue uninterrupted without a need to form Okazaki fragments. However, for the lagging strand, when the RNA primers are synthesized 5′ to 3′ and the polymerase begins synthesizing DNA, the polymerase moves in a direction that is physically opposite to the overall movement of the replication fork along the template. Therefore, additional RNA primers must be added at interrupted intervals as replication fork movement exposes new template DNA behind the polymerase.

69. (1) DNA polymerase α: initiation of nuclear DNA synthesis and DNA repair (2) DNA polymerase δ: completion of replication on lagging strands (5′ → 3′ polymerase activity) (3) DNA polymerase γ: replication and repair of mitochondrial DNA (4) DNA polymerase ε: leading strand synthesis (5) Other eukaryotic DNA polymerases (ζ, η, and θ) allow DNA replication to proceed past damaged regions of DNA (DNA lesions) or to play various roles in DNA repair processes.

70. (a) Telomerase allows for the ends of chromosomes to be replicated. In the absence of telomerase, the ends of chromosomes are clipped after each round of replication, causing the telomeres to be eventually lost and the ends of the chromosomes to become unstable. In order for the chromosomes to be maintained for use in the next generation, telomerase must be expressed in cells of the germ line. (b) Most somatic cells normally do not express telomerase. Because of this, somatic cells normally can divide only a limited number of times before the telomeres are completely removed and the chromosomes become unstable.

However, in most cancers, mutations occur that allow telomerase to be expressed. Because the telomeres are restored after replication, these cells are able to divide an unlimited number of times, allowing cancer cells to reproduce themselves without limit and allowing tumors to grow to large size and cancer to spread within the body.

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 13_7e Indicate the answer choice that best completes the statement or answers the question.

1. Whereas the nucleotide strand used for transcription is termed the _____, the nontranscribed strand is called the _____. a. promoter; terminator b. terminator; promoter c. transcription apparatus; TATA box d. template strand; nontemplate strand e. nontemplate strand; template strand 2. What would be the consequence of a mutation in the gene that encodes sigma factor for bacterial transcription? a. The initiation of transcription may begin at random points of the DNA template. b. The RNA polymerase may not be released from the DNA template. c. Transcription may be delayed indefinitely, which may kill the cell. d. The RNA transcript may not be able to dissociate from the DNA template. e. The transcription may end prematurely, resulting in shorter mRNA transcript.

3. Which of the following RNA molecules is required for the process of translation? a. crRNA b. tRNA c. snRNA d. snoRNA e. siRNA 4. An in vitro transcription system that contains a bacterial gene does not initiate transcription. What is one possible problem? a. Histones that were on the DNA when it was isolated from E. coli are blocking access to the template. b. There is a mutation in the inverted repeat sequence that prevents a hairpin secondary structure from forming. c. There is a mutation at –10, where a promoter consensus sequence is located. d. Rho factor has not been added. e. TATA-binding protein (TBP) has not been added.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 13_7e 5. Which of the following statements is NOT true? a. Both DNA and RNA are synthesized in a 5′-to-3′ direction. b. During RNA synthesis, the DNA template strand is read in a 3′-to-5′ direction. c. During RNA synthesis, new nucleotides are added to the 3′ end of the growing RNA molecule. d. RNA polymerase has 5′-to-3′ polymerase activity. e. RNA molecules have the same 5′-to-3′ orientation as the DNA template strands to which they are complementary.

6. In prokaryotes, a group of genes that are usually transcribed following a specific stimulus is organized as an "operon" under a single shared promoter (Chapter 16; see the figure below). The result of the transcription of the genes in the presence of a stimulus would generate a:

a. crRNA. b. miRNA. c. polycistronic mRNA. d. rRNA. e. siRNA. 7. If the DNA strand 5'–GTACCGTC–3' were used as a template, what would be the sequence of the transcribed RNA? a. 5′–GUACCGUC–3′ b. 5′–GACGGTAC–3′ c. 5′–CAUGGCAG–3′ d. 5′–GACGGUAC–3′ e. 5′–GUCGGUAC–3′ 8. Which of the following statements is FALSE regarding TFIID? a. It contains a TATA-binding protein. b. It aids in initiation of transcription. c. It binds to the core promoter. d. It binds to the TATA box. e. It is a transcriptional activator protein.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 13_7e 9. Where are promoters usually located? a. upstream of the start site b. downstream of the start site c. near nucleotide +25 d. near the hairpin loop e. downstream of the terminator 10. For a given gene, both strands of a DNA are used as a template when which of the following molecules is synthesized? a. RNA b. DNA c. both RNA and DNA d. neither RNA nor DNA 11. In eukaryotes, which RNA polymerase transcribes the genes that encode proteins? a. RNA polymerase I b. RNA polymerase II c. RNA polymerase III d. RNA polymerase IV e. RNA polymerase V 12. In transcription, to which end of the elongating strand are nucleotides always added? a. 3′ b. 5′ c. 3′ in prokaryotes and 5′ in eukaryotes d. It depends on which RNA polymerase is being used. e. It depends on which DNA strand is being used as the template.

13. Which of the following classes of RNA is NOT naturally found in eukaryotes? a. mRNAs b. snRNAs c. miRNAs d. crRNAs e. piRNAs

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 13_7e 14. An in vitro transcription system that contains a bacterial gene initiates transcription from random points on the DNA. Which of the following proteins MOST likely is missing from the reaction? a. sigma factor b. rho factor c. RNA polymerase II d. TATA-binding protein (TBP) e. TFIID 15. An in vitro transcription system transcribes a bacterial gene but terminates inefficiently. What is one possible problem? a. There is a mutation in the –10 consensus sequence, which is required for efficient termination. b. Rho factor has not been added. c. Sigma factor has not been added. d. A hairpin secondary structure has formed at the 3′ end of the mRNA, interfering with termination. e. Histones were added prematurely and interfered with termination.

16. Which of the following is not necessary for RNA polymerase to recognize the promoter of a bacterial gene? a. sigma factor b. origin of replication c. –10 consensus sequence d. –35 consensus sequence 17. Which of the following statements is TRUE regarding transcription in most organisms? a. All genes are transcribed from the same strand of DNA. b. Both DNA strands are used to transcribe a single gene. c. Different genes may be transcribed from different strands of DNA. d. The DNA template strand is used to encode double-stranded RNA. e. The DNA nontemplate strand is used to encode single-stranded RNA.

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 13_7e 18. What is the RNA sequence transcribed from the DNA shown below?

a. 5'–CCGTATAATGA–3' b. 3'–GGCATATTACT–5' c. 5'–GGCAUAUUACA–3' d. 5'–CCGUAUAAUGA–3' e. 3'–GGCAUAUUACU–5' 19. When does sigma factor normally dissociate from RNA polymerase? a. after transcription has terminated b. after the process of initiation c. after the addition of nucleosomes d. after the binding of rho e. following the addition of nucleosomes 20. Which process is illustrated in the diagram below?

a. transcription b. translation c. RNA processing d. replication e. nucleosome assembly

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 13_7e 21. Choose the BEST consensus sequence for the following set of nucleotide sequences.

TAAGACGCCATGA AAAGTCGCAATCA AAAGTTCCGTTCA AGAGTTGCTATCA AGAGTTGCAATCA a. YACGTRGCATG/CA b. AYTRTRGCATGA c. AAAGTNGANTCA d. ARAGTYGCNTCA e. ACTNCGYTGARA 22. Over time, DNA replaced RNA as the primary carrier of genetic information, and the chemical stability of DNA is believed to be the key reason for this. Which attribute of DNA is the reason behind its chemical stability? a. DNA lacks a free hydroxyl group on the 2′-carbon atom of its sugar. b. Unlike RNA, DNA is usually double stranded. c. DNA does not usually form hairpin loops. d. One of the two pyrimidines found in DNA does not involve uracil. e. DNA contains thymines, which make it more chemically stable.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 13_7e 23. The following diagram represents the transcription unit and corresponding RNA transcripts as "Christmastree-like" structures captured in the electron micrograph.

Which of the following statements CORRECTLY describes the why the transcription unit looks like a "Christmas tree" with different length of tree branches along the DNA "tree trunk"? a. Depending on where the transcription begins on the DNA template, different lengths of transcripts get generated. b. Each position on the DNA dictates the specific length of the RNA transcript generated from that particular position. c. The transcript randomly grows from the DNA without any particular reason. d. A single DNA can be transcribed many times continuously, and the RNA transcript gets longer as the transcription apparatus moves down the DNA. e. There is no known reason for the transcription unit to assume the "Christmas-tree-like" shape.

24. In which of the following organisms would transcription be the LEAST similar to archaea? a. E. coli b. yeast c. plants d. mice e. humans 25. In eukaryotic cells, where does the basal transcription apparatus bind? a. core promoter b. regulatory promoter c. terminator d. enhancer e. ribozyme

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 13_7e 26. What is the function of eukaryotic RNA polymerase I? a. transcription of rRNA genes b. transcription of mRNA genes c. transcription of tRNA genes d. transcription of snRNAs e. initiation of transcription (but not elongation) 27. Which of the following molecules is synthesized using nucleotides containing the bases adenine, guanine, cytosine, and uracil? a. RNA only b. DNA only c. both RNA and DNA d. neither RNA nor DNA 28. Which one of the following statements regarding eukaryotic transcription is NOT true? a. Eukaryotic transcription involves a core promoter and a regulatory promoter. b. There is no one generic promoter. c. A group of genes is transcribed into a polycistronic RNA. d. Chromatin remodeling is necessary before certain genes are transcribed. e. There are several different types of RNA polymerase.

29. _____ probably began the evolution of life on Earth. a. DNA b. RNA promoters c. DNA polymerases d. RNA polymerases e. Ribozymes 30. The TATA-binding protein (TBP) binds to the TATA box sequence in eukaryotic promoters. What is its function in transcriptional initiation? a. It blocks access of RNA polymerase to the promoter until removed by general transcription factors. b. It is the subunit of prokaryotic RNA polymerase that is required to recognize promoters. c. It modifies histones so that nucleosomes can be removed from DNA for transcription. d. It bends and partly unwinds DNA at a promoter. e. It creates a phosphodiester bond between the nucleotides.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 13_7e 31. What types of bonds are created between nucleotides during the process of transcription? a. ionic b. oxygen c. phosphodiester d. hydrogen e. both phosphodiester and hydrogen 32. In a transcription reaction, two phosphate groups are cleaved from the incoming: a. deoxyribonucleoside diphosphate. b. deoxyribonucleoside triphosphate. c. ribonucleoside diphosphate. d. ribonucleoside triphosphate. e. ribozyme.

33. Which nonstandard Watson–Crick base pair might be found in a RNA molecule with secondary structure? a. A–G b. A–U c. A–C d. C–U e. G–U 34. Which of the following molecules is synthesized using triphosphate nucleotides as a substrate for a polymerase enzyme that forms phosphodiester bonds? a. RNA only b. DNA only c. both RNA and DNA d. neither RNA nor DNA 35. What may be the consequence of a mutation in the gene that encodes eukaryotic Rat1 exonuclease for eukaryotic transcription mediated by RNA polymerase II? a. The transcription may end prematurely, resulting in shorter mRNA transcript. b. The transcription may not be properly terminated, and RNA polymerase II may not be released. c. The transcription may be delayed indefinitely, which may kill the cell. d. The RNA transcript may not be able to dissociate from the DNA template. e. The transcription may not terminate and result in much longer RNA.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 13_7e 36. Which of the following features of the rho protein is primarily responsible for its ability to cause termination of transcription? a. recognizing unstructured RNA b. helicase activity c. migrating behind RNA polymerase d. RNA-binding activity e. polymerase activity 37. The polymerase that synthesizes which of the following molecules uses DNA as a template and synthesizes new strands from 5′ to 3′? a. RNA polymerase only b. DNA polymerase only c. both RNA polymerase and DNA polymerase d. neither RNA polymerase nor DNA polymerase 38. Prokaryotic promoters contain the sequence TATAAT at a position _____ from the transcription start. a. +1 b. –1 c. –10 d. –25 e. –35 39. In eukaryotes, tRNAs are transcribed in: a. the nucleus and function in the nucleus. b. the nucleus but function in the cytoplasm. c. the cytoplasm and function in the cytoplasm. d. both the nucleus and the cytoplasm and function in the cytoplasm. e. the cytoplasm and function in the nucleus.

40. The enzyme involved in DNA replication that most closely resembles RNA polymerase is: a. DNA polymerase I. b. DNA polymerase III. c. primase. d. telomerase. e. helicase.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 13_7e 41. Which of the following molecules is made of nucleotides joined by phosphodiester bonds that connect the 2′ OH to the 5′ phosphate? a. RNA only b. DNA only c. both RNA and DNA d. neither RNA nor DNA 42. In prokaryotes, rho-independent transcription termination depends on a secondary structure formed in: a. the RNA polymerase that is transcribing the gene. b. the DNA template. c. the RNA that is being transcribed. d. a protein factor that binds to RNA polymerase. e. a protein factor that binds to the RNA that is being transcribed.

43. The following diagram represents a transcription unit.

What would likely happen to transcription of this transcription unit if a certain consensus sequence within region A is removed via deletion? a. The mutation in region A would result in truncated RNA transcript that is shorter than expected RNA. b. The mutation in region A does not affect the transcription as the transcript is copied only from regions B and C. c. The transcript may get copied in reverse direction as the orientation of region A is important for the direction of the transcription. d. The transcript may not be produced as the consensus sequence within region A may be important for the initiation of the transcription. e. The transcription occurs normally as there are other consensus sequences that would compensate for the loss of one.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 13_7e 44. If the sequence of an RNA molecule is 5'–GGCAUCGACG–3', what is the sequence of the nontemplate strand of DNA? a. 5'–GGCATCGACG–3' b. 3'–GGCATCGACG–5' c. 5'–CCGTAGCTGC–3' d. 3'–CCGTAGCTGC–5' e. 3'–CGTCGATGCC–5' 45. Which of the following statements about RNA polymerase is NOT true? a. RNA polymerase adds a ribonucleotide to the 3' end of a growing RNA molecule. b. RNA polymerase binds to a promoter to initiate transcription. c. During transcription of a gene, RNA polymerase reads only one strand of DNA. d. RNA polymerase reads a template strand of DNA 5' to 3'. e. RNA polymerase has many subunits.

46. Which of the following is a sequence of DNA where transcription is initiated? a. hairpin loop b. TBP c. initiator d. sigma factor e. promoter 47. Mutations in which of the following regions upstream of the RNA coding sequence are LEAST likely to affect the transcription of a gene? a. –10 consensus sequence within the promoter region b. –35 consensus sequence within the promoter region c. the region between –10 and –35 consensus sequences d. the upstream elements at –40 to –60 position e. All mutations on the sequences upstream of a gene will have deleterious effects on gene transcription.

48. What is the function of general transcription factors? a. They are DNA sequences to which RNA polymerase binds. b. They direct nucleosome assembly. c. They bind to regulatory promoters to increase the rate of transcription. d. They bind to enhancers to allow minimal levels of transcription. e. They are a part of the basal transcription apparatus.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 13_7e 49. Which of the following types of RNA gets translated? a. rRNA b. mRNA c. tRNA d. miRNA e. rRNA, mRNA, and tRNA all get translated.

50. Which of the following statements is TRUE regarding the termination of transcription? a. In some organisms, transcription terminates thousands of nucleotides past the coding sequence. b. Transcription typically terminates precisely at the hairpin loop terminator sequence. c. In prokaryotes, transcription terminates as soon as rho has bound to the RNA. d. In yeast, transcription terminates as soon as Rat1 has bound to the RNA. 51. Which of the following is NOT required for transcription? a. ribonucleotides b. RNA primers c. DNA template d. RNA polymerase e. promoter 52. In prokaryotic RNA polymerases, the holoenzyme consists of the core enzyme and the: a. rho factor. b. TFIID. c. TBP. d. omega subunit. e. sigma factor.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 13_7e Indicate one or more answer choices that best complete the statement or answer the question.

53. The following diagram represents a transcription unit.

Which region(s), A, B, and/or C, would be transcribed into an RNA transcript? (Select all that apply.) a. A b. B c. C 54. The following diagram represents the transcription unit and corresponding RNA transcripts as "Christmastree-like" structures captured in the electron micrograph.

Which of the following statements CORRECTLY describes the orientation of the DNA and RNA transcript?

(Select all that apply.) a. A represents the 5' end while B represents the 3' end of the DNA template undergoing transcription. b. A represents the 3' end while B represents the 5' end of the DNA template undergoing transcription. c. C represents the 5' end while D represents the 3' end of the RNA transcript. d. C represents the 3' end while D represents the 5' end of the RNA transcript. e. The specific orientation of the DNA template and RNA transcript cannot be determined based on the given information.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 13_7e 55. Mutations occurred in one of the inverted repeat sequences within the rho-independent terminator sequence of a bacterium. What would likely be the consequence of this mutation? (Select all that apply.) a. The transcription may end prematurely. b. The transcription may not be initiated at all. c. The transcription may be delayed. d. The RNA transcript may not be able to dissociate from the DNA template. e. The transcription may not terminate and result in much longer RNA. 56. Compare rho-dependent and rho-independent transcription termination in terms of their use of RNA secondary structures.

57. In bacteria the processes of DNA replication and transcription occur simultaneously. What molecular strategies exist in bacteria to avoid and/or deal with "head-on" and "rear-end" collisions between DNA and RNA polymerases moving along the same DNA molecule?

58. Two sequences of a prokaryotic promoter are shown below. Sequence 1 is the wild-type sequence, and sequence 2 is a mutant sequence (mutation shown in bold) that results in increased transcription of the gene.

Explain why sequence 2 shows this phenotype.

59. Describe three ways that eukaryotic transcription initiation is different from prokaryotic transcription initiation.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 13_7e 60. The discovery of ribozymes led to the theory that the evolution of life on Earth began with an "RNA world."

Describe the chemical properties and functions of RNA that would allow it to be the basis of the first selfreplicating systems.

61. Describe how the activities and requirements of prokaryotic RNA polymerase and eukaryotic RNA polymerase II are both similar and different.

62. The discovery of ribozymes led to the theory that the evolution of life on Earth began with an "RNA world."

Describe how the current cellular roles of RNA support this theory.

63. How are enhancers different from promoters?

64. If you wanted to express a eukaryotic gene in E. coli, would the normal promoter be sufficient for transcription? Why or why not?

65. Organisms belonging to the archaea group possess a TATA-binding protein (TBP), but they are structurally very similar to members of the eubacteria (i.e., contain no nucleus, are single celled). What does this conundrum mean regarding the evolutionary relationships among the archaea, eubacteria, and eukaryotes?

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 13_7e 66. Explain why eukaryotic transcripts from the same RNA-polymerase-II-transcribed gene may vary in the sequences at the 3′ end.

67. You analyze a newly cloned eukaryotic gene by comparing the DNA sequence of the gene to the sequence of the transcribed RNA and find that it does not contain recognizable promoter sequences upstream of its coding sequence. How can you explain this result, and how might you confirm your hypothesis?

68. If you remove the TATA box and place it immediately upstream of a transcription start site of a eukaryotic gene and subsequently transcription of the mRNA is assayed, will you still achieve transcription from the same start site?

69. The discovery of ribozymes led to the theory that the evolution of life on Earth began with an "RNA world."

What properties of proteins and DNA make them more suitable than RNA as enzymes and the cell's genetic material?

70. What would you add to an in vitro transcription system that contains a Drosophila gene for glyceraldehyde 3-phosphate dehydrogenase, an enzyme in glycolysis, in order to get basal transcription?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 13_7e 71. An in vitro system to detect transcription initiation functions in protein extracts from either eukaryotic or archaebacterial cells is added. Initiation does not occur if extracts from eubacterial cells are added. What does the protein extract contain? Explain the results using the different sources of protein extracts.

72. What would you add to an in vitro transcription system that contains an E. coli gene for glyceraldehyde 3phosphate dehydrogenase, an enzyme in glycolysis, in order to get transcription that begins from the normal transcription start site?

73. Compare and contrast the operation of promoters and enhancers in affecting transcription of genes in eukaryotes. Include a discussion of how they operate and from what positions with respect to the transcription initiation site and the roles of each in bringing about and regulating transcription.

74. If you were asked to isolate total RNA from two unknown samples and then were required to identify if the RNA was from prokaryotes or eukaryotes, what aspects regarding the classes of RNA present would help you distinguish one from the other?

75. On the DNA strands shown below, two RNA polymerase enzymes are using the top strand as a template. In the boxes, label the 5′ and 3′ ends of the DNA molecules and the RNA molecules being made. With arrows, indicate the directions, left to right or right to left, that the polymerases are moving.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 13_7e 76. A new mutation in Saccharomyces cerevisiae, a eukaryotic yeast, causes the cells to be unable to produce the amino acid histidine. Specifically, they cannot catalyze the first reaction in the histidine biosynthesis pathway. When examined closely, they are producing a completely wild-type enzyme for this reaction but at greatly reduced levels. Explain the mutation in terms of transcription

77. The following diagram represents DNA that is part of the RNA-coding sequence of a transcription unit.

The sequence of the RNA product of transcription is 5′–UGGCAUCGCCUAU–3′. Label the diagram of the DNA molecule to indicate which strand serves as the template strand. Also label the 5′ and 3′ ends of both strands.

78. Diagram an active transcription unit as it would appear at high magnification. On the diagram indicate each of the following: (a) the DNA molecule (b) the RNA molecule that was initiated earliest (c) the 5′ end of an RNA molecule (d) the 5′ end of the template strand of DNA (e) the approximate location of the promoter (f) a molecule of RNA polymerase (g) an arrow indicating the direction of transcription

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 13_7e 79. An RNA molecule has the following percentage of bases: A = 15%, U = 30%, C = 20%, and G = 35%. a. Is this RNA single stranded or double stranded? How can you tell based on the base composition? b. What would be the percentage of bases in the template strand of the DNA that encodes this RNA molecule?

80. What are the three different RNA polymerases found in all eukaryotes, and what types of genes do they transcribe?

81. Transcription of a gene using what is normally the nontemplate strand can be accomplished using standard genetic engineering techniques to move the promoter. The resulting RNA is called "antisense RNA." Antisense RNA is often used to prevent production of a protein. Explain how antisense RNA would interact with RNA normally made by the gene and how this would interfere with protein production.

82. Histone acetyltransferases add acetyl groups to lysines, which are positively charged amino acids. How might this affect the association of a nucleosome with DNA?

83. Your goal is to achieve a high level of transcription—a level similar to its in vivo levels—of a eukaryotic gene in an in vitro transcription system. You add the DNA template for this gene, which contains the following, to the in vitro transcription system: a TATA box sequence located –25 to –30 bp upstream of the transcription start site, the sequence for the gene, and terminator sequences. How would you interpret the result if you do not get high levels of RNA transcription, and what corrective measures would you take to achieve your goal?

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 13_7e Answer Key 1. d 2. a 3. b 4. c 5. e 6. c 7. d 8. e 9. a 10. b 11. b 12. a 13. d 14. a 15. b 16. b 17. c 18. d 19. b 20. a 21. d 22. a 23. d 24. a 25. a 26. a Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 13_7e 27. a 28. c 29. e 30. d 31. c 32. d 33. e 34. c 35. b 36. b 37. c 38. c 39. b 40. c 41. d 42. c 43. d 44. a 45. d 46. e 47. c 48. e 49. b 50. a 51. b 52. e 53. b, c 54. b, c Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 13_7e 55. d, e 56. (1) Rho-dependent transcription termination relies on a lack of RNA secondary structure so that rho can bind the RNA, move up to the RNA polymerase, and remove the RNA from the DNA template.

(2) Rho-independent transcription termination relies on a hairpin in the transcribed RNA that causes RNA polymerase to pause, facilitating termination.

57. Head-on collisions are largely avoided because mRNAs for highly expressed genes are transcribed in the same direction as the replication. Rear-end collisions cannot be avoided entirely, but when collisions occur, they can be dealt with by restarting stalled replication forks, removing blocked transcription complexes, and repairing DNA breaks and mutations that arise.

58. The mutation changes the –10 Pribnow box so that it exactly fits the consensus sequence for RNA polymerase binding.

59. (1) There are many different eukaryotic RNA polymerases. Each recognizes the promoter of a different type of gene.

(2) Eukaryotic RNA polymerases require general transcription factors to associate with a promoter.

(3) Eukaryotic promoters do not have the same –10 and –35 consensus sequences that prokaryotic promoters have.

(4) Eukaryotic RNA polymerases do not have the polymerase subunits that are required for promoter recognition.

(5) Eukaryotic transcription initiation requires altering or removing nucleosomes at the gene to be transcribed.

60. (1) As a nucleic acid, RNA can serve as a template for self-replication: An RNA can be copied into a complementary strand that is a template for generating more of the original RNA. (2) As a nucleic acid, RNA can carry genetic information in its base sequence.

(3) RNA could have performed the reactions required of a self-replicating system because RNA ribozymes have catalytic activity.

61. Both transcribe mRNA in a 5′ to 3′ direction, use rNTPs as a substrate, use a single DNA strand as a template, and do not require a primer to begin RNA synthesis. They associate with different proteins and bind to different promoter sequences to begin transcription: the prokaryotic enzyme requires the sigma subunit and uses the –10 and –35 sequences; the eukaryotic enzyme requires TBP and a whole host of additional general transcription factors. Additionally, the eukaryotic enzyme uses several different binding sequences like the TATA box, TFIIB recognition element, and so forth. The eukaryotic enzyme may also interact with enhancer sequences.

62. (1) RNA is an intermediate between DNA, the permanent genetic information, and proteins.

(2) RNA has a role as a primer for DNA replication.

(3) RNA is required for protein synthesis as mRNA, rRNA, and tRNA. 63. (1) Unlike for promoters, the distance between enhancers and the gene they influence is not critical.

(2) Enhancers may affect several genes in their vicinity.

(3) Enhancers may be downstream or upstream of the genes whose transcription they affect.

64. No. Prokaryotic and eukaryotic promoters have different sequences and different proteins that associate with them.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 13_7e 65. It suggests that the archaea and eukaryotes are more closely related to each other than the archaea are to the eubacteria.

66. (1) The 3′ end is the termination end. Termination on polymerase-II-transcribed genes can occur at multiple sites.

(2) RNA polymerase transcribes until the 3′ end is processed by a cleavage complex. The complex may prevent termination until the processing sequence is transcribed and the complex can cleave the 3′ end of the new RNA. 67. This gene is most likely a polymerase-III-transcribed gene that contains an internal promoter. In order to confirm this hypothesis, look for known polymerase III promoter sequences within the coding sequence.

68. No. The TATA box needs to be present a certain number of nucleotides upstream of the transcription start site to allow enough space for the assembly of TATA-binding protein and other transcription factors on the core promoter.

The RNA polymerase can then be placed appropriately over the transcription start site.

69. (1) Proteins are made of 20 different amino acids; RNA is made of relatively uniform nucleotides with only four different bases. Because they can be more chemically diverse, proteins are better suited to catalyzing a variety of cellular reactions.

(2) DNA is double stranded, with a structure that protects the molecule from degradation. The lack of 2′ OH makes the molecule less susceptible to self-hydrolysis. Because it is more stable, DNA is a better molecule for the cell's genetic material.

70. (1) RNA polymerase II (2) basal transcription factors (TFIIA, TFIIB, TFIID, TFIIE, TFIIF, and TFIIH) (3) ribonucleoside triphosphates with A, C, G, and U bases (rNTPs) 71. The extracts that work contain a TATA-binding protein (TBP). Eukaryotic and archaebacterial cells use a TBP (and its associated transcription factors, TFIIB); eubacterial cells do not.

72. (1) RNA polymerase (2) Sigma factor (3) Ribonucleoside triphosphates with A, C, G, and U bases (rNTPs)

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 13_7e 73. Both promoters and enhancers are DNA sequences that regulate transcription of adjacent DNA sequences. Both are needed for a high level of transcription of a gene. Both serve as binding sites for proteins that are involved in bringing about and regulating transcription.

Promoters are found just upstream of the transcription start site and need to be in a specific orientation with respect to the transcription start site. Eukaryotic promoters typically include a core promoter located just upstream of the transcription start site, as well as a regulatory promoter located immediately upstream of the core promoter.

The core promoter allows for binding of the basal transcription apparatus (including RNA polymerase). The core promoter by itself usually can only bring about a low level of transcription. The regulatory promoter allows for the binding of a variety of additional transcription factors that can interact with basal transcription factors at the core promoter to bring about regulated gene expression at a higher level. The activity of promoters is constrained in that they can only operate from a specific position and in a specific orientation with respect to the transcription start site.

Enhancers are DNA sequences that can be found near or far from and upstream or downstream of the transcription initiation site. Regulatory proteins bind to enhancers and interact with the transcription apparatus at a promoter to bring about a higher level of transcription. Because of the ability of enhancers to act at various distances and positions with respect to the promoter, a large number of enhancers can be found on a single gene, allowing the gene to be responsive to a variety of different signals and conditions in the cell.

74. RNA from prokaryotes will contain mRNA, tRNA, and rRNA. In addition to these three types of RNA, eukaryotic samples will contain pre-mRNA, snRNA, snoRNA, scRNA, miRNA, and siRNA. If the eukaryotic sample happens to be from mammalian testes, it will also contain piRNA.

75.

76. The mutation is either in the regulatory promoter for the gene or in an enhancer, reducing transcription.

77.

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 13_7e 78. An acceptable answer would be:

79. a. This RNA molecule must be single stranded, like most RNA molecules. If the molecule were double stranded, it would consist of complementary base pairs so that A = U and C = G, which is not the case. b. The template strand of DNA would be complementary to this RNA molecule, so the composition would be T = 15%, A = 30%, G = 20%, and C = 35%.

80. (1) RNA polymerase I transcribes large rRNA. (2) RNA polymerase II transcribes pre-mRNA, some snRNA, snoRNAs, and some miRNAs. (3) RNA polymerase III transcribes tRNA, small rRNA, some snRNAs, and some miRNA. 81. Antisense RNA would bind to the normal RNA from a gene because it would be antiparallel and complementary.

The double-stranded RNA would not be able to be translated, so the protein encoded by the gene would not be produced.

82. (1) DNA is negatively charged.

(2) Acetylating lysines would neutralize the positive charge, which would allow them to form ionic bonds with DNA. (3) Nucleosomes that contain histones with a reduced ability to bond with DNA would be less tightly associated with DNA. 83. Two causes contributed to the low level of transcription and need to be corrected: (a) The template did not include regulatory promoter regions including enhancers. (b) The regulatory proteins that bind to the enhancers and activate high levels of transcription are absent in the in vitro system.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 14_7e Indicate the answer choice that best completes the statement or answers the question.

1. Which of the following is found in the primary product of transcription but not in a mature mRNA molecule? a. start codon b. promoter c. exons d. introns e. stop codon 2. The sequence below represents a pre-mRNA. What would happen if the G in the 5' splice site were mutated to a C? mRNA: 5′–ACUGGACAGGUAAGAAUACAACACAGUCGGCACCACG–3′ a. The U2 snRNA would not be able to bind to the branch point because it could not recognize it. b. The spliceosome complex would be degraded because it could no longer recognize the 5' splice site. c. The U1 snRNA would not be able to bind complementarily to the 5' splice site. d. Splicing would still occur appropriately because the G is not essential at the 5' splice site. e. The 5' cap would not be able to be added because it requires the 5' splice site to be functional.

3. Guide RNAs are needed in: a. transcription. b. translation. c. RNA interference. d. RNA editing. e. RNA splicing.

4. Which of the following statements about ribosomes and ribosomal RNA is NOT true? a. Ribosomes typically contain about 80% of the total cellular RNA. b. Ribosomal RNA is processed in both prokaryotes and eukaryotes. c. In eukaryotes, genes for rRNA are usually present within tandem repeats. d. Each ribosomal RNA component is encoded by a separate gene. e. In eukaryotes, the rRNA transcripts are processed further by snoRNAs within the nucleus.

5. Which of the following intron types requires spliceosomes for removal? a. group I intron b. group II intron c. group III intron d. nuclear pre-mRNA e. tRNA Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 14_7e 6. Which of the following statements CORRECTLY describes the concept of alternative splicing? a. Eukaryotic gene and protein sequences are precisely colinear. b. With the rare exception of RNA editing, every nucleotide contained in a processed mRNA molecule is derived from exon sequences. c. Every other intron is removed in an alternate manner to generate a functional mRNA transcript. d. Only a subset of the same mRNA transcripts is specifically selected for splicing in the nucleus. e. Multiple protein products are often produced from single eukaryotic genes.

7. Which of the following observations supports the notion that the gene is simply a set of DNA sequences that are transcribed into a single RNA molecule that encodes a single polypeptide? a. Alternative splicing—a single gene can yield multiple mRNA and protein products. b. A single ribosomal RNA transcript can liberate several RNA molecules via further processing. c. RNAs can be the functional product of a gene without being translated into a protein product. d. Within a protein-coding region, each codon represents a specific amino acid that will be linked to form a polypeptide. e. Regulatory elements are part of a gene that regulate timing, degree, and specificity of gene expression but are not transcribed.

8. Which of the following statements about group I and group II introns is NOT true? a. Both group I and II introns form elaborate and characteristic secondary structures with loops. b. The splicing mechanism of group II introns is similar to that of spliceosome-mediated nuclear premRNA splicing. c. The length of group I and group II introns is much longer than the exons within the structures. d. Group I and group II introns are exclusively found in mitochondrial and chloroplast encoded genes. e. Both group I and group II introns are found in bacterial genes.

9. Which class of RNA is MOST abundant in cells? a. mRNA b. tRNA c. rRNA d. snRNA e. miRNA 10. Which of the following elements would NOT be found in an mRNA molecule? a. protein-coding region b. 3′ untranslated region c. 5′ untranslated region d. promoter e. start and stop codons Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 14_7e 11. Given the figure below, within which of the following would the Shine–Dalgarno sequence be located? The sequences in red and light gray are included in the mature mRNA. The sequences in light gray make up the protein-coding sequence of the gene.

a. promoter b. exon 1 c. exon 4 d. intron 1 e. A Shine–Dalgarno sequence is not present in this figure.

12. The sequence below represents a pre-mRNA. Which of the following represents the sequence of the spliced mRNA that would result from this pre-mRNA sequence? mRNA: 5′–ACUGGACAGGUAAGAAUACAACACAGUCGGCACCACG–3′ a. 5′–ACUGGACAGUCGGCACCACG –3′ b. 5′–GUAAGAAUACAAC–3′ c. 5′–UGACCUGUCAGCCGUGGUGC–3′ d. 5′–ACUGGACAGGUAAGAAUACAACACAGUCGGCACCACG–3′ e. 5′–AGAAUACAACACAGUCGGCACCACG–3′ 13. Which of the following statements BEST explains why only pre-mRNA receives a 5′ cap? a. The enzyme that initiates the capping step is known to associate with RNA polymerase II, which generates pre-mRNAs. b. Only pre-mRNAs contain proper sequences for the cap to be added on. c. The tail of the pre-mRNA can recruit the right combination of enzymes for capping. d. Nuclear pore complexes only recognize pre-mRNAs and allow them out to the cytoplasm for the capping process to begin. e. rRNA and tRNAs do not exit the nucleus to receive the cap via enzymes in the cytoplasm.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 14_7e 14. Scientists once believed that each gene can encode a single polypeptide. We now know that _____ and _____ allow a single gene to encode more than one polypeptide. a. transcription; translation b. polyadenylation; RNA transport c. DNA methylation; chromatin condensation d. alternative processing; RNA editing e. gene silencing; RNA interference 15. The 5' cap in an mRNA plays a role in translation initiation. Which of the following could be a plausible mechanism by which a 5' cap could enhance initiation? a. The 5' cap could assist in bringing together the snRNPs for spliceosome assembly. b. The 5' cap could recruit proteins that would help to assemble the ribosomes. c. The 5' cap could assist in the identification of stop codons within the mRNA. d. The 5' cap could serve as a marker for the ribosome to locate promoters. e. The 5' cap could assist in the unwinding of the DNA to allow ribosome access to the DNA. 16. Which of the following statements regarding gene structure is TRUE? a. The amino acid sequence of a polypeptide can be precisely predicted by the nucleotide sequence of the gene that encodes it. b. The number of introns found in the genes of an organism is determined exclusively by species to which it belongs. c. The number of exons and introns generally correlates to the complexity of the organism. d. Intron cleavage and exon splicing are both mediated exclusively by protein enzymes. e. The number of exons is always less than the number of introns in a gene.

17. Suppose a mutation occurred that prevented a eukaryotic pre-mRNA from receiving a 5' cap. What would be an expected result? a. Transcription would continue past the end of the gene coding sequence resulting in a longer pre-mRNA transcript. b. Translation would continue past the end of the gene coding sequence resulting in a longer pre-mRNA transcript. c. Transcription would not occur as the transcription factors would not be able to bind to the promoter. d. Translation would not occur as the ribosome would not be able to bind to the mRNA. e. Replication would not occur as DNA polymerase would not be able to bind to the DNA at the origin of replication.

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 14_7e 18. What do group I and group II introns have in common? a. Both are found in mitochondrial genes. b. Both are found in bacteriophages. c. Both are known to be self-splicing introns. d. Both are found in protein-coding genes of chloroplasts. e. Both are frequently found in eukaryotic genes.

19. The 5′ cap on an mRNA is important for all the processes listed below except for the _____ of an mRNA molecule. a. transcription b. intron removal c. stability d. initiation of translation e. ribosomal interaction 20. Which of the following nitrogenous bases is frequently modified enzymatically to become a rare type of nitrogenous base in tRNA? a. adenine b. uracil c. thymine d. cytosine e. guanine 21. Given the figure below, within which of the following would the 5' untranslated region be located? The sequences in red and light gray are included in the mature mRNA. The sequences in light gray make up the protein-coding sequence of the gene.

a. promoter b. exon 1 c. exon 4 d. intron 1 e. A 5' untranslated region would not be present in this figure.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 14_7e 22. Which of the following intron types is present only in eukaryotes? a. nuclear pre-mRNA b. group I intron c. group II intron d. tRNA e. group III intron 23. A drug that destroyed small nucleolar RNA (snoRNA) would inhibit which process? a. RNA splicing of pre-mRNA b. translation c. assembly of the nucleosome d. replication e. transcription 24. To which part on a tRNA would an amino acid attach during tRNA charging? a. 3′ acceptor arm b. anticodon arm c. TψC arm d. DHU arm e. extra arm 25. Which of the following statements about bacterial mRNA transcripts is TRUE? a. Unlike eukaryotes, bacterial mRNA transcripts do not typically contain untranslated regions. b. The Shine–Dalgarno sequence associates with an RNA component of ribosomes. c. Transcription and translation take place sequentially in bacterial cells. d. Most of bacterial genes contain a large number of introns and a small number of exons. e. The 5′ end and 3′ end of mRNA transcripts are modified in bacteria.

26. A scientist is studying a gene known as the ABC gene in bacteria. Into a bacterial cell, she inserts an miRNA that is complementary to a portion of the ABC mRNA found in the cell. What result would you predict? a. There will be an increase in the amount of ABC protein made. b. There will be a decrease in the amount of ABC protein made. c. There will be an increase in the transcription of the ABC gene. d. There will be a decrease in the transcription of the ABC gene. e. No change will result from the insertion of the miRNA.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 14_7e 27. The 5′ and 3′ untranslated regions (UTRs) of processed mRNA molecules are derived from: a. exons. b. introns. c. promoters. d. terminators. e. the protein-coding region.

28. Which of the following small RNA types is unique to prokaryotes? a. siRNA b. crRNA c. miRNA d. piRNA e. lncRNA 29. Which of the following phenomena is NOT affected by the presence of alternative splicing? a. speciation b. development c. organismal complexity d. tissue specificity e. RNA interference 30. Below is a list of steps of intron removal and splicing during pre-mRNA processing. Please select the choice that lists the steps in the CORRECT sequential order.

1. Attachment of snRNP U1 to the 5′ splice site 2. Transcription of the DNA template into the pre-mRNA molecule 3. Release of lariat structure 4. Splicing together of exons 5. Transesterification reaction at the branch point adenine a. 1, 2, 3, 4, 5 b. 4, 1, 3, 5, 2 c. 2, 1, 5, 3, 4 d. 3, 5, 1, 2, 4 e. 5, 3, 4, 1, 2

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 14_7e 31. Which of the following classes of RNAs is unique to eukaryotes? a. messenger RNA (mRNA) b. ribosomal RNA (rRNA) c. transfer RNA (tRNA) d. small nuclear RNAs (snRNAs) e. CRISPR RNAs (crRNAs) 32. A microbiologist isolated a mutant strain of E. coli that is extremely susceptible to bacteriophage infection, by a wide range of bacteriophages. Which of the following genes might she explore as a possible site of the mutation that results in this phenotype? a. a gene that encodes a small nucleolar RNA b. a gene that encodes a Cas protein c. a gene that encodes a Piwi-interacting RNA d. a gene that encodes either a miRNA or a siRNA e. a gene that encodes an immune RNA 33. What is the similarity between miRNAs, siRNAs, and piRNAs? a. All three types originate from transposons or viruses and are found in all organisms. b. They all target and degrade the gene from which they were transcribed. c. All three are generated from a single-stranded RNA that gets cleaved. d. All three can influence chromatin structure, which, in turn, can influence gene expression. e. All three associate with Piwi proteins in order to mediate RNA degradation.

34. A key modification in the 3′ end of eukaryotic mRNA is the addition of 50 to 250 adenine nucleotides, forming a poly(A) tail. Which of the following is NOT a function of the poly(A) tail? a. The stability of mRNA transcripts in the cytoplasm is affected by the poly(A) tail. b. The poly(A) tail facilitates the attachment of the ribosome to the mRNA. c. The poly(A) tail is important for proper nuclear export of the mRNA. d. The poly(A) tail at the 3′ end translates to a long stretch of repeated amino acids. e. Multiple proteins will recognize and bind to the poly(A) tail in the cytoplasm.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 14_7e 35. The human gene encoding calcitonin contains six exons and five introns and is located on chromosome 11. The pre-mRNA transcript from this gene can generate either calcitonin or calcitonin gene-related peptide (CGRP) in a tissue-specific manner. Calcitonin produced from the thyroid gland is 32 amino acids long and functions to regulate the calcium while CGRP, which contains 37 amino acids, is produced by the brain cells and involved in transmission of pain. Which of the following processes makes production of two functionally and structurally different proteins from the same gene possible? a. self-spicing introns b. differential transcription c. alternative replication d. 5′ capping and polyadenylation e. alternative RNA processing 36. How many introns are present on a gene that consists of four exons and is known to have introns? a. two b. three c. four d. five e. The number cannot be determined from the information provided.

37. The spliceosome is a large, ribonucleoprotein complex located in the: a. cytoplasm. b. endoplasmic reticulum. c. Golgi apparatus. d. nucleus. e. nucleolus.

38. Which of the following spliceosomal components specifically recognizes and binds to the branch point of the intron during pre-mRNA splicing? a. U1 b. U2 c. U5 d. U6 e. spliceosomal proteins

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 14_7e 39. Which of the following consensus sequences are NOT found in nuclear introns? a. GU at the 5′ splice site at the beginning of the intron b. AG at the 3′ splice site at the end of the intron c. CCA at the 3′ site downstream of the branch point d. A at the lariat branch point site e. 3′ CAGG consensus sequence at the 3′ splice site 40. The information needed during RNA editing comes MOST directly from: a. pre-mRNA. b. mRNA. c. rRNA. d. tRNA. e. guide RNA. 41. Anticodons are found in _____ molecules. a. mRNA b. tRNA c. rRNA d. snRNA e. miRNA 42. Which mechanism allows for the production of polypeptides that are not entirely encoded by DNA? a. regulated transcription b. RNA interference c. alternative RNA processing d. RNA editing e. colinearity 43. Which of the following statements CORRECTLY describes the facts about introns? a. The number of introns is always less than the number of exons in a gene. b. Introns are degraded in the cytoplasm. c. All eukaryotic genes contain an intron. d. Mitochondrial and chloroplast genes do not contain introns. e. Introns do not contain sequence-specific information.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 14_7e 44. During the posttranscriptional processing of pre-mRNA, a 5′ cap is added to an mRNA in a step-by-step manner. Which of the following reasons prevents the 5′ capping process, involving methylation, from occurring on a DNA strand? a. lack of an OH group on the 2′ carbon of the deoxyribose b. lack of an OH group on the 3′ carbon of the deoxyribose c. lack of a uracil nitrogenous base on the DNA strand d. lack of GTP hydrolysis associated with DNA transcription e. lack of a H on the 4′ carbon of the deoxyribose 45. When studying a plant's protein production, a scientist found two different proteins. The first one contained amino acids from exons 1, 2, 3, 4, 5, and 6, while the second one only contained amino acids from exons from 1, 2, and 3. Which of the following is MOST likely responsible for this difference? a. a mutation in the gene that encodes an miRNA b. posttranslational modification c. RNA editing d. alternative RNA processing e. a mutation in the gene that encodes a snoRNA 46. What would be the MOST likely effect of mutating the consensus sequence found at the 5' splice site of an intron? a. A shorter than normal protein would be produced. b. Replication would be inhibited. c. A longer than normal mRNA would be produced. d. A longer than normal DNA would be produced. e. Transcription would terminate prematurely.

47. Which of the following rRNA components originates from a separate gene transcript rather than as a cleaved product of a long single precursor rRNA transcript? a. prokaryotic 16S rRNA b. prokaryotic 23S rRNA c. eukaryotic 18S rRNA d. eukaryotic 5.8S rRNA e. eukaryotic 5S rRNA

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 14_7e 48. Below is a list of steps of eukaryotic pre-mRNA processing. Please select the choice that lists the steps in the CORRECT sequential order.

1. Recognition and binding the 3′ AAUAAA sequence by specific protein factors 2. Cleavage at the poly(A) site 3. Addition of the 5′ cap 4. Export to the cytoplasm 5. Addition of the poly(A) tail a. 3, 1, 2, 5, 4 b. 2, 3, 4, 5, 1 c. 4, 2, 3, 1, 5 d. 1, 3, 5, 4, 1 e. 5, 4, 1, 3, 2 49. Suppose an organism ingests a drug that disassembles its spliceosomes, rendering them nonfunctional. Which of the following would be seen MOST immediately in this organism? a. All translation would stop. b. tRNA bases would no longer be modified into rare bases. c. Introns would not be removed from the pre-mRNA. d. mRNA would not be able to bind the 5' cap. e. rRNA would no longer be appropriately processed.

50. If a splice site were mutated so that splicing did not take place, what would be the effect on the amino acid sequence? a. It would be shorter than normal. b. It would be longer than normal. c. It would be the same length but would encode a different protein. d. It would be unable to fold into its correct structure. e. It depends on the mutant mRNA sequence.

51. A scientist is studying a gene known as the XYZ gene in eukaryotes. Into a eukaryotic cell, she inserts an miRNA that is complementary to a portion of the XYZ mRNA found in the cell. What result would you predict? a. There will be an increase in the amount of XYZ protein made. b. There will be a decrease in the amount of XYZ protein made. c. There will be an increase in the transcription of the XYZ gene. d. There will be a decrease in the transcription of the XYZ gene. e. No change will result from the insertion of the miRNA.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 14_7e 52. During the posttranscriptional processing of pre-mRNA, a 5′ cap is added to an mRNA. Arrange the following steps of the capping process in the CORRECT order.

1. Addition of a guanine nucleotide via a 5′–5′ bond 2. Removal of a phosphate from a ribonucleotide triphosphate at the 5′ head of the pre-mRNA 3. Methylation at the 2′ position of the sugar in the second and the third nucleotides 4. Methylation at position 7 of the terminal guanine base a. 1, 2, 3, 4 b. 2, 4, 1, 3 c. 4, 1, 3, 2 d. 2, 1, 4, 3 e. 3, 2, 4, 1 53. The calcitonin gene can encode either the hormone calcitonin or a protein called calcitonin-gene-related peptide depending on which 3' cleavage site is used. In the thyroid gland, cleavage and polyadenylation occur after the fourth exon leading to calcitonin production. However, in the brain, the exact same transcript is cleaved after the sixth exon yielding calcitonin-gene-related peptide. This is an example of: a. multiple capping. b. alternative RNA processing. c. polyadenylation. d. environmental influence. e. mutation.

54. Below is a list of steps in the processing of ribosomal RNAs. Please select the choice that lists the steps in the CORRECT sequential order.

1. Methyl groups added to specific bases and the 2′-carbon atom of some ribose sugars 2. Transcription of the rRNA precursors from DNA 3. Cleavage of precursor rRNA 4. Individual rRNA molecules ready for ribosome assembly 5. Trimming of precursor rRNA a. 3, 1, 2, 5, 4 b. 2, 3, 4, 5, 1 c. 4, 2, 3, 1, 5 d. 1, 3, 5, 4, 1 e. 2, 1, 3, 5, 4

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 14_7e 55. Which mechanism allows for more than one polypeptide to be encoded by a single gene? a. regulated transcription b. RNA interference c. alternative RNA processing d. self-splicing of introns e. RNA methylation 56. Cystic fibrosis is caused by a mutation in the CFTR gene. The normal CFTR gene comprises approximately 190,000 nucleotides and produces an mRNA of 6128 nucleotides in length. What is a possible explanation for the difference in the sizes of the gene and the mRNA? a. The promoter of the CFTR gene is likely silenced by miRNAs prior to mRNA production. b. The CFTR gene likely has many introns that are excised prior to translation of the CFTR protein. c. The 5' cap and the poly(A) tail get removed prior to translation of the CFTR protein. d. Methylation of the 5' cap silences portions of the gene, preventing those regions from being transcribed into mRNA. e. The mutation that causes cystic fibrosis creates a new terminator sequence, resulting in a shorter mRNA. 57. Which of the following regulatory RNA types is different from the rest in terms of its length? a. siRNA b. crRNA c. miRNA d. piRNA e. lncRNA 58. siRNAs and miRNAs function in which of the following processes? a. transcription b. translation c. RNA interference d. RNA editing e. RNA splicing

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 14_7e Indicate one or more answer choices that best complete the statement or answer the question.

59. Which of the following statements about internal modifications of pre-mRNAs is TRUE? (Select all that apply.) a. Internal modifications are found in eukaryotic mRNAs. b. The addition of a methyl group to the nitrogen at position 6 of adenine (m6 A) is a rare internal modification. c. m6 A modifications are most often found near start codons and in 5′ UTRs. d. m6 A affects mRNA stability, splicing and translation. e. The addition of a methyl group to the nitrogen at position 3 of guanine (m3 G) is an example of an internal modification.

60. Which of the following statements about RNA is INCORRECT? (Select all that apply.) a. Posttranscriptional modifications to RNAs only occur in eukaryotes. b. Secondary structure is the highest level of structure that RNA can adopt. c. Modification of eukaryotic pre-mRNAs involves the addition of a 5′ cap, splicing out of introns, and the addition of a 3′ poly(A) tail. d. tRNAs are involved in translation. e. rRNAs are involved in translation.

61. Which of the following statements about BOTH small nuclear RNAs (snRNAs) and small nucleolar RNAs is TRUE ? (Select all that apply.) a. Both are critical for ribosome assembly. b. Both involve splicing together of exons. c. Both associate with proteins to carry out modifications of other RNA molecules. d. Both are found in prokaryotes and eukaryotes. e. Both are required for expression of functional gene products.

62. Which of the following statements about long noncoding RNAs (lncRNAs) is TRUE? (Select all that apply.) a. All regulate gene expression. b. They do not contain open reading frames (ORFs). c. They are involved in X-chromosome inactivation in humans. d. Both are found in prokaryotes and eukaryotes. e. Some may be the result of imprecise transcription.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 14_7e 63. Which of the following processes supports the observation that the amino acid sequence of a protein may not be the same as that encoded by its gene? (Select all that apply.) a. RNA editing b. alternative splicing c. multiple 3′ cleavage sites d. 5′ capping e. errors that occurred during transcription 64. What is a pulse–chase experiment? Explain how this technique can be used to follow the products of a shortterm event.

65. Devise a strategy to prove that splicing occurs in the nucleus.

66. Compare and contrast the mechanisms of splicing for nuclear pre-mRNA introns, group I introns, group II introns, and tRNA introns.

67. Explain how RNA editing violates a general principle of molecular genetics.

68. What attributes make miRNA a potent agent for silencing gene expression by allowing it to silence a large number of genes simultaneously?

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 14_7e 69. For a time, a gene was defined as a sequence of DNA that encodes a polypeptide. Critique this outdated notion of what a gene is and propose a more comprehensive definition.

70. In your own words, list a comprehensive definition for "gene" at the molecular level.

71. What are guide RNAs (gRNAs), and what do they do?

72. Use the pre-mRNA sequence shown below to answer the following questions. mRNA: 5′ ACUGGACAGGUAAGAAUACAACACAGUCGGCACCACG 3′ a. Underline the 5′ and 3′ splice sites, then write the sequence of the spliced mRNA in the space provided. b. Predict what would happen if the G in the 5′ splice site were mutated to a C. c. We learned in this chapter that the 5′ cap in an mRNA plays a role in translation initiation. What do you think is one plausible mechanism by which a 5′ cap can enhance initiation? How can you experimentally demonstrate that a 5′ cap is important for this process?

73. Draw the tRNA cloverleaf secondary structure, labeling the various loops and stems. Indicate the functions of the acceptor stem and the anticodon. How are modified bases produced in tRNAs?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 14_7e 74. Discuss the features of C. elegans that make it an important model system for studies of animal genetics and development. Include in your answer some unique features of C. elegans compared to other model systems.

75. Devise and describe a strategy to conduct a pulse–chase experiment to determine if newly transcribed eukaryotic RNA actually moves from the nucleus to the cytoplasm.

76. Describe the evidence that suggested that RNA is the carrier of genetic information from the nucleus to the ribosome.

77. Critique the idea that the information needed to encode a particular polypeptide always resides within a single gene.

78. What is an snRNP, and what role does it play in the cell?

79. In 1958 Watson and Crick proposed that genes are proteins are colinear meaning that the number of nucleotides in a gene is directly proportional to the number of amino acids in the sequence of the protein encoded by that gene. Describe (a) which part(s) of the concept of collinearity are still useful and (b) where the concept of collinearity has limitations.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 14_7e Answer Key 1. d 2. c 3. d 4. d 5. d 6. e 7. d 8. d 9. c 10. d 11. e 12. a 13. a 14. d 15. b 16. c 17. d 18. c 19. a 20. b 21. b 22. a 23. b 24. a 25. b 26. e Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 14_7e 27. a 28. b 29. e 30. c 31. d 32. b 33. d 34. d 35. e 36. b 37. d 38. b 39. c 40. e 41. b 42. d 43. a 44. a 45. d 46. c 47. e 48. a 49. c 50. e 51. b 52. d 53. b 54. e Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 14_7e 55. c 56. b 57. e 58. c 59. a, d 60. a, b 61. c, e 62. b, c, e 63. a, b, c, e 64. A pulse–chase experiment involves transiently introducing labeled precursors into an active biological system ("pulse"), then removing the labeled precursor and adding unlabeled (or "cold") precursors to "chase" the label so that it can be monitored. By monitoring the movement of the label in the system, questions regarding the kinetics, stability, and location of labeled cellular components can be answered.

65. Collect cells and fractionate them into nuclear and cytoplasmic fractions. Isolate the proteins from each of the fractions. Add a pre-mRNA (made in vitro) with introns to each of the protein fractions. Monitor the progressive decrease in the size of pre-mRNA in the nuclear fraction by separating the RNA products on a gel, which can resolve size differences between pre-mRNA (longer) and spliced mRNA (shorter). The cytoplasmic fraction will be incapable of splicing, and therefore, the pre-mRNA remains intact over a period of time without undergoing splicing. The difference in the ability to splice suggests that the nuclear fraction contains the machinery essential for splicing.

66. (1) Pre-mRNA introns: Spliceosome-mediated intron excision and exon splicing. Introns form lariat structures prior to their release.

(2) Group I introns: All group I introns fold into a common secondary structure containing nine looped stems, which are necessary for transesterification and splicing. Group I introns self-splice exclusively and thus do not require spliceosome-related components.

(3) Group II introns: The splicing mechanism of group II introns is very similar to the mechanism used by nuclear genes, involving snRNPs and two transesterifications, which generate a lariat structure. However, all group II introns form a specific secondary structure and self-splice.

(4) tRNA introns: For both prokaryotes and eukaryotes, different tRNAs are processed differently, so there is no standard processing pathway to speak of. The splicing process for all tRNAs is not spliceosome mediated.

Instead, generally speaking, tRNA introns are excised by endonucleases and released for degradation, and then the two ends of the tRNA molecule—which are held in place by intramolecular bonding within the secondary tRNA structure—are ligated together to form the complete, processed tRNA molecule.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 14_7e 67. Except for RNA viruses, genetic information resides exclusively within the nucleotide sequence of DNA. This fundamental principle is violated by RNA editing, a process whereby the coding sequence of an mRNA molecule is altered posttranscriptionally. This results in the translation of a polypeptide having an amino acid sequence that differs from the coding sequence contained in the gene.

68. The pre-miRNA is cleaved into one or more smaller RNA molecules with a hairpin. Dicer binds to this hairpin structure and removes the terminal loop. One of the miRNA strands is incorporated into the RISC; the other strand is released and degraded. The RISC attaches to a complementary sequence on the mRNA, usually in the 3′ untranslated region of the mRNA. The region of close complementarity, called the seed region, is quite short, usually only about seven nucleotides long. Because the seed sequence is so short, each miRNA may potentially pair with sequences on hundreds of different mRNAs. Thus, miRNAs potentially can silence a large number of genes simultaneously.

69. There are many problems with this simple concept of a gene. First, many genes encode functional RNA molecules rather than polypeptides. These functional RNAs (tRNA, rRNA, snRNA, etc.) produce RNA molecules that have some function other than encoding polypeptides. Second, many genes (perhaps most in at least some species) can encode more than one polypeptide through mechanisms such as alternative processing and RNA editing. Third, all genes contain 5′ and 3′ untranslated regions, and many genes (especially in eukaryotes) contain introns that are transcribed but do not encode polypeptides. Finally, transcription of a DNA sequence is regulated by upstream and downstream sequences that are not transcribed. Since these sequences are important to the normal expression and function of the gene product, they probably should be considered to be part of the gene.

A more comprehensive definition of a gene would include a transcribed region that produces a functional RNA or an mRNA that encodes one or more polypeptides, along with adjacent DNA sequences that are involved in regulating expression of the gene. This definition is not perfect given that a single regulatory element can regulate more than one transcription unit, making it part of more than one gene by this definition. Also, this definition does not account for the fact that some genetic information carried on a particular RNA molecule might not be provided by the transcription unit but rather by other genes that control alternative RNA processing and RNA editing.

70. A gene is a sequence of DNA, including all structural and regulatory sequences, that encodes information dictating synthesis of one or more discrete polypeptides or RNA molecules.

71. Guide RNAs play a crucial role in some RNA editing strategies by containing sequences that are partly complementary to segments of specific pre-edited mRNA molecules. After the gRNA associates with the corresponding mRNA, the mRNA is cleaved and specific nucleotides are added, deleted, or altered depending on the template provided by the gRNA. After the mRNA is modified, its ends are ligated back together, and the intact, now altered, mRNA can be translated.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 14_7e 72. a. Pre-mRNA: 5′ ACUGGACAGGUAAGAAUACAACACAGUCGGCACCACG 3′ spliced mRNA: 5′ ACUGGACAG/UCGGCACCACG 3′ ( / indicates an exon–intron junction) b. The mRNA would not be spliced properly, if at all, because the U1 snRNA would not be able to bind complementarily to the 5′ splice site, due to the mutation. A different site with a close consensus sequence might be found, resulting in a completely different mRNA sequence. Ultimately, the protein encoded by this gene would be different from the normal protein and probably would not have the normal function. c. One possibility is that the 5′ cap could serve as a recruiting site for proteins to bind and help assemble the ribosomes to facilitate translation initiation. In an in vitro translation system, translation initiation efficiency of an mRNA with and without cap can be measured by monitoring the levels of proteins synthesized. If the hypothesis that 5′ cap enhances translation initiation efficiency is correct, then in this experiment more protein will be produced from the mRNA with a cap.

73. See Figure 14.20 for tRNA structure.

Acceptor stem Acceptor arm Anticodon arm DHU arm TψC arm The 3′ end of all tRNAs (prokaryotic and eukaryotic) contains the trinucleotide sequence 5′–CCA–3′ (the acceptor stem). Amino acids are covalently attached to the 3′ adenines in the acceptor stems of their cognate tRNAs by highly specialized enzymes called aminoacyl tRNA synthetases.

Anticodons are trinucleotide sequences located at the bottom of the tRNA cloverleaf structure that interact, through specific complementary base pairing, with given anticodons contained in the genetic code carried by mRNA. Codon–anticodon specificity alone dictates which amino acids are incorporated into a given polypeptide during ribosome-mediated translation.

Transfer RNAs contain rare bases, in addition to the standard four bases found in RNA. The rare bases in tRNA are initially transcribed as standard bases, but they are posttranscriptionally modified through chemical changes mediated by specific tRNA-modifying enzymes.

74. C. elegans is a small nematode worm that has served well as a model system for studying animal genetics and development. It has all of the common benefits of other model organisms, including small size (1 mm), short generation time (3 days), fecundity (each female lays 200–1000 eggs), and ease of lab culture. Like Mendel's peas and some other genetic model systems, C. elegans is capable of self-fertilization or crossing. The genome is relatively small and has been completely sequenced. Approximately 25% of the genes in C. elegans are also found in the human genome. C. elegans is especially useful in the study of development due to its transparent body, which makes it easier to observe all stages of development, and due to its regular pattern of cell divisions that produces the adult body. RNA interference allows for particular genes to be turned off in order to assess gene functions. Transgenic technology is also available to allow for further assessment of gene functions. C. elegans has been especially useful in the study of development, apoptosis, genetic control of behavior, and aging.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 14_7e 75. Uracil is present only in RNA (and not DNA). Therefore, add labeled (e.g., using radioactivity) uracil to the growth Moderate of an actively growing eukaryotic cell culture. The uracil will be taken into the cells and incorporated into newly synthesized RNA. After a short time, gently spin down the cells, wash away the old (labeled) growth Moderate, and transfer the cells to new media containing "cold" (unlabeled) uracil. At various time intervals, take a small sample of cells from the culture and fractionate (separate) the cells into nuclear and cytosolic components.

Isolate RNA from each component and fractionate the RNA using gel electrophoresis. Monitor all soluble samples and gels for radioactivity. At first, radioactivity will be present only in the nuclear fraction, but over time you will measure increasing radioactivity in the cytosolic component, which suggests that RNA is moving from the nucleus to the cytoplasm. To verify this assumption and confirm that the radioactivity measured is not simply unincorporated labeled uracil diffusing out of the nucleus, perform autoradiography on the RNA gels. The presence of high-molecular-weight radioactivity will confirm incorporation into RNA, and comparing the time-course results will confirm RNA movement. Note that RNA movement can alternatively be monitored using fluorescent labels, coupled with sophisticated confocal microscope-imaging techniques, which optically section intact cells.

76. Using radioactively labeled uracil, Gros and coworkers performed pulse–chase experiments demonstrating that labeled phage RNA, which was synthesized in the nucleus, later associated with the ribosomes and was distinct from them. This RNA component was called messenger RNA. 77. A major exception to this idea is RNA editing. For those transcripts that are edited, some of the information needed to encode the polypeptide is carried in the corresponding structural gene, and some of the information is carried in the gene that produces the guide RNA molecule. In some mRNAs that have been studied, more than 60% of the sequence is determined by RNA editing.

Another exception is alternative RNA processing. Although all of the information needed to encode the various polypeptide products of a particular gene might be contained within the gene (unless RNA editing is at work), which particular product is made within a particular cell is determined by the processing mechanisms, the components of which are encoded by a variety of other genes.

78. A small nuclear ribonucleoprotein is a complex of small nuclear RNAs (snRNAs)—U1, U2, and so on—and proteins that catalyze the transesterification reactions and splicing of exons during eukaryotic mRNA processing.

79. a) For a fully processed, mature mRNA, the number of nucleotides in the open reading frame (ORF)/protein-coding region does tend to correspond directly to the number of amino acids that will be incorporated into the protein translated from that mRNA. (The number of amino acids is equal to the number of nucleotides divided by 3. b) The concept of colinearity does not take into account the fact that there are often noncoding sequences in genes including sequences at the start (5′ UTR), middle (introns), and end (3′ UTR) of genes and their mRNA products that make the gene longer than anticipated by colinearity.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 15_7e Indicate the answer choice that best completes the statement or answers the question.

1. Mechanisms that exist to detect and deal with errors in mRNA in order to ensure the accurate transfer of genetic information are collectively referred to as: a. mRNA surveillance. b. proofreading function. c. RNA interference. d. alternative processing. e. RNA transition.

2. Which of the following statements about the formation of the peptide bond between amino acids is INCORRECT? a. The formation of peptide bonds results in the formation of a water molecule. b. The amino group of the first amino acid and the carboxyl group of the second amino acid are involved in forming a peptide bond. c. The carboxyl group of the first amino acid reacts with the amino group of the second amino group to form a peptide bond. d. A polypeptide formed by a series of peptide bonds will result in two distinct free ends, one with a free amino group and the other with a free carboxyl group. e. The number of peptide bonds formed in a polypeptide varies from protein to protein.

3. Which one of the following codons codes for a different amino acid from the rest? a. CUU b. CUC c. UUA d. UUU e. CUA

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 15_7e

4. After the peptide bond forms, what will happen next? a. tRNA A will be carrying the polypeptide, and it will shift to the P site. b. tRNA A will be carrying the polypeptide, and it will shift to the A site. c. tRNA B will be carrying the polypeptide, and it will shift to the P site. d. tRNA B will be carrying the polypeptide, and it will shift to the A site. e. The ribosome disassembles to release the tRNAs and to allow new tRNA to enter.

5. What would be the consequences of a two-nucleotide-deletion mutation in the middle of the first exon of a protein-coding gene? a. The result would be no change in the protein, since deletions are very rare. b. A reading frameshift would occur. All codons (and thus amino acids) downstream of the codon in which the deletion occurred would be different, including the stop signal. c. Some codons might be altered since there are changes in the bases of the DNA, but the stop codon would remain the same and the protein would still be made. d. Since exons are not coding regions, no changes in the protein would occur. e. No change in the reading frame would occur due to the degeneracy of the genetic code.

6. Gal4 is a protein that acts as transcriptional activator in the model organism S. cerevisiae. A Gal4 monomer folds from a single polypeptide chain and has several domains with different activities including DNA-binding, dimerization, and trans-activation. At which level of protein structure does a domain belong? a. between primary and secondary b. between secondary and tertiary c. between tertiary and quaternary d. above quaternary e. none of the above

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 15_7e 7. The genetic code is NOT universal for: a. prokaryotes, which use a different genetic code than eukaryotes. b. a few mitochondrial genes, which substitute one sense codon for another. c. viruses, which use an entirely different genetic code. d. archaebacteria, which have their own genetic code. e. animal species whose cells are more advanced and complex.

8. Which molecule allows the release of mRNA from a stalled ribosome? a. miRNA b. snoRNA c. incRNA d. tmRNA e. siRNA 9. An example of a motif is the helix–turn–helix motif, which can have DNA-binding activity. At which level of protein structure does a motif belong? a. between primary and secondary b. between secondary and tertiary c. between tertiary and quaternary d. above quaternary e. none of the above 10. Which of the following events is NOT part of prokaryotic translation initiation? a. IF-3 separates ribosome subunits so that a small subunit can bind mRNA through base pairing of the 16S rRNA and the Shine–Dalgarno sequence on the mRNA. b. An initiator tRNAformylmet binds the initiation codon, with the help of IF-1 and IF-2 complexed with GTP. The tRNAformylmet is positioned in the P site. c. A peptide bond is formed by the peptidyl transferase activity of the large subunit rRNA. The polypeptide chain on the tRNA in the P site is transferred to the amino acid on the tRNA in the A site. d. IF-3 dissociates, allowing a large subunit to bind the 30S initiation complex.

11. An mRNA has the codon 5′ UAC 3′. What tRNA anticodon will bind to it? a. 5′ AUG 3′ b. 5′ GUA 3′ c. 5′ ATC 3′ d. 5′ CTA 3′ e. 5′ CAU 3′

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 15_7e 12. A tRNA anticodon is 5′ GCU 3′. What amino acid does it carry? a. Ala b. Arg c. Ser d. Pro e. Thr 13. Codons that specify the same amino acid are said to be: a. wobbly. b. isoaccepting. c. hypothetical. d. synonymous. e. anonymous.

14. What is the minimum number of different aminoacyl-tRNA synthetases required by a cell? a. 64, one for each codon b. 61, one for each sense codon c. 30, one for each different tRNA d. 50, one for each different tRNA e. 20, one for each amino acid

15. The next step in the translation of this mRNA will be the formation of a peptide bond between which two of the numbered amino acids? a. amino acid 2 and amino acid 3 b. amino acid 2 and amino acid 4 c. amino acid 1 and amino acid 3 d. amino acid 1 and amino acid 2 e. amino acid 3 and amino acid 4

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 15_7e 16. Translating an mRNA requires two other types of RNA: a. tRNA and mRNA. b. tRNA and miRNA. c. tRNA and rRNA. d. rRNA and siRNA. e. snRNA and snoRNA. 17. Which amino acid is coded by the stop codons in most organisms? a. met b. pro c. trp d. cys e. none 18. Which of the following statements describes the "wobble" rules CORRECTLY? a. There is a flexible pairing between tRNA and amino acid as there are more tRNAs than the number of amino acids. b. The number of the genetic code exceeds the number of amino acids available in the cell. c. There are multiple tRNAs that may bind to the same amino acids. d. There are multiple codons that may code for the same amino acids. e. The third base pairing between the tRNA and mRNA is relaxed.

19. A bacterial protein is encoded by the following mRNA sequence: 5′-AUGGUGCUCAUGCCCTAA-3′. The second methionine codon (AUG) in this mRNA sequence will: a. serve as the initiation codon. b. encode N-formylmethionine. c. encode methionine that will eventually be removed. d. encode unformylated methionine. e. be skipped as the translation progresses.

20. Which of the following statements about protein folding and posttranslational modifications of proteins is CORRECT? a. All nascent polypeptide chains have the intrinsic ability to fold into the active conformation based on the primary structure. b. Only eukaryotic proteins undergo alterations following translation. c. Amino acids within a protein may be modified by molecular chaperones. d. Signal sequence of a protein helps direct a protein to a specific location within the cell. e. Attachment of a protein called ubiquitin directs proteins to enter into the nucleus.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 15_7e 21. Which of the following is NOT required during the process of tRNA charging? a. amino acid b. tRNA c. GTP d. ATP e. aminoacyl-tRNA synthetase An auxotrophic E. coli strain requires adenine to grow because of a mutation in a gene for an adenine synthesis enzyme. The following shows part of the wild-type and mutant alleles of the gene, including the start codon. The bottom strand is the template for transcription.

22. How does the mutation affect the enzyme? a. The protein won't be synthesized and translation will stop at the third codon. b. The protein will not be changed because the mutation did not change the amino acid. c. The protein will be changed because the mutation altered the amino acid. d. The protein will not be changed because the mRNA was not changed.

23. During initiation, the _____ subunit is the first part of the ribosome to associate with the mRNA. a. small b. large c. intermediary d. secondary e. tertiary Refer to the following sequence: 5′ ...GGAGCUCGUUGUAUU... 3′ 24. Which amino acids does this sequence code for, if the reading frame is as shown, starting from the correct end? a. Gly-Ala-Arg-Cys-Ile... b. Pro-Arg-Ala-Thr-Stop c. Met-Asn-Glu-Leu… d. Glu-Leu-Val-Val-Phe… e. Leu-Glu-Gln-His-Asn…

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 15_7e 25. During elongation, an incoming charged tRNA enters at the _____ site of the ribosome. a. peptidyl (P) b. aminoacyl (A) c. exit (E) d. Shine–Dalgarno e. Kozak 26. What is the function of peptidyl transferase activity? a. It charges tRNAs. b. It acetylates the end of a protein after translation. c. It cleaves the polypeptide from the last tRNA during termination. d. It moves ribosomes along mRNA during translation. e. It forms peptide bonds.

27. There are _____ different codons, which encode 20 amino acids and three stop codons. a. 16 b. 20 c. 23 d. 61 e. 64

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 15_7e An auxotrophic E. coli strain requires adenine to grow because of a mutation in a gene for an adenine synthesis enzyme. The following shows part of the wild-type and mutant alleles of the gene, including the start codon. The bottom strand is the template for transcription.

28. E. coli strains that have both the original ade1- mutant allele shown above and a mutant allele for a gene that encodes a lysine tRNA are able to make some normal ade1 enzyme. The mutant lysine tRNA allele makes a tRNA that has one base that is different from the wild-type tRNA. Why might the mutant lysine tRNA allele affect overall cell growth? a. The mutant tRNAlysine will occasionally cause translation to continue beyond the normal stop codon on mRNAs that normally use UAG as a stop codon. b. The premature stop codon from the ade1– allele will always cause shortened proteins. c. The mutant tRNA will prevent the translational machinery from recognizing the normal stop codon on the mRNA and so will result in incorrect proteins. d. The mutant tRNAlysine will recognize the stop codon on the mutant ade1– allele and cause the protein to be terminated. e. The mutant tRNAlysine will incorporate a lysine at every UAG stop codon.

29. Which of the following statements about translation initiation in eukaryotes is ALWAYS true? a. The initiation codon is AUG. b. Binding of the mRNA occurs at the 5′ cap. c. Ribosomes are involved. d. Open reading frames (ORFs) only exist in the coding region of mRNA. e. All of the above are correct.

30. The amino acid sequence of a polypeptide is referred to as the _____ sequence of the polypeptide. a. primary b. secondary c. tertiary d. quaternary

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 15_7e Refer to the following sequence: 5′ ...GGAGCUCGUUGUAUU... 3′ 31. The nucleotide sequence 5′ ...GGAGCUCGUUGUAUU... 3′ is changed to 5′ GGAGACUCGUUGUAUU 3′. Why does or why doesn't the amino acid sequence change? a. The reading frame changes after the mutation (the addition of an A in the fifth position) and so the amino acid sequence is modified after that point. b. The reading frame, starting at the 5′ end of this sequence, would be modified because of this change and so the entire amino acid would be different. c. The amino acid is not changed since the coding sequence was not changed at the 5′ position. d. A premature stop codon caused by this change would result in a truncated polypeptide. e. Just one amino acid would be changed in the resulting polypeptide.

32. A yeast strain was exposed to a chemical mutagen. As expected, exposure to a mutagen resulted in a DNA sequence change in an essential gene you examined. Yet this mutation did not result in any lethal phenotype.

Which of the following answers would NOT explain this apparent discrepancy? a. The DNA sequence change occurred in a synonymous nucleotide position of an amino acid, and as a result the protein sequence remained unaltered. b. The protein sequence was not affected, and no lethal phenotype manifested. c. The DNA sequence change resulted in an amino acid change, but that amino acid change had no negative effect on protein function (i.e., was a neutral change). d. The DNA sequence change occurred in a nonsynonymous nucleotide position of an amino acid, and as a result the protein sequence remained unaltered. e. All mutations that cause a change in DNA sequence result in a lethal phenotype.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 15_7e

33. If the bottom strand of the DNA is the template, the tRNA anticodon sequence for the first RNA codon, left to right or 5′ to 3′, is: a. GGA. b. AUG. c. CAC. d. UCC. e. CCU. Refer to the following sequence: 5′ ...GGAGCUCGUUGUAUU... 3′ 34. The sequence 5′ ...GGAGCUCGUUGUAUU... 3′ is changed to 5′… GGAGACUCGUUGUAUU… 3′.

What would be the effect on the amino acid sequence? a. There would be no effect on the amino acid sequence. b. This is an insertion mutation, so there would be a premature stop codon. c. The amino acid sequence would be Asn-Thr-Thr-Thr-Leu. d. The amino acid sequence would be Thr-Ser-Tyr-Leu-Asn. e. The amino acid sequence would be Gly-Asp-Ser-Leu-Tyr.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 15_7e 35. Which of the following mRNA codons will bind to the tRNA anticodon 5′ GCU 3′, considering wobble–base pairing rules. a. 5′ AGU 3′ and 5′ AGC 3′ b. 5′ UGA 3′ and 5′ CGA 3′ c. 5′ AGC 3′ d. 5′ CGA 3′ e. 5′ AGU 3′, 5′ AGC 3′, 5′ AGA 3′, and 5′ AGG 3′ 36. The genetic code is said to be "degenerate" because: a. there are more codons than amino acids. b. there are more amino acids than codons. c. different organisms use different codons to encode the same amino acid. d. some codons specify more than one amino acid. e. there are more tRNAs than amino acids.

An auxotrophic E. coli strain requires adenine to grow because of a mutation in a gene for an adenine synthesis enzyme. The following shows part of the wild-type and mutant alleles of the gene, including the start codon. The bottom strand is the template for transcription.

37. How does this mutation change the mRNA? a. The third codon in the mRNA is changed from AAG (lysine) to UAG (stop). b. The third codon in the mRNA is changed from GCA (alanine) to GCU (alanine). c. The third codon in the mRNA is changed from CAA (glutamine) to CUA (leucine). d. The third codon in the mRNA is changed from CGU (arginine) to CGA (arginine). e. There is no change in the mRNA.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 15_7e

38. Process 2, illustrated above, represents: a. replication. b. transcription. c. translation. d. RNA processing. e. RNA interference.

39. Which of the following is observed in prokaryotes but not in eukaryotes? a. UGG is an example of a stop codon only found in prokaryotes. b. An mRNA can be translated by only one ribosome at a time in prokaryotes. c. The 5' end of a prokaryotic mRNA can be translated while the 3' end is still being transcribed. d. Translation does not require any protein factors in prokaryotes. e. In prokaryotes, ribosomes move along an mRNA in the 3' to 5' direction.

40. Which of the following does NOT enhance the binding of the ribosome to the 5´ end of the mRNA? a. 5′ cap b. 3′ poly(A) tail c. cap-binding proteins d. poly(A) proteins e. enhancer Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 15_7e 41. An mRNA has the stop codon 5′ UAA 3′. What tRNA anticodon will bind to it? a. 5′ ATT 3′ b. 5′ AUC 3′ c. 5′ ACU 3′ d. 5′ UUA 3′ e. none Refer to the following sequence: 5′ ...GGAGCUCGUUGUAUU... 3′ 42. This sequence is RNA because: a. it is single stranded. b. it contains U (uracil) and no T (thymine). c. it runs in a 5′ to 3′ direction. d. it codes for amino acids . e. it is a small molecule.

43. When codons that code for the same amino acid differ in their _____, a single tRNA might bind both of them through wobble base pairing. a. 5′ base b. middle base c. 3′ base 44. The function of aminoacyl-tRNA synthetases is to: a. transcribe tRNA genes. b. match tRNA anticodons and mRNA codons at the ribosome. c. attach appropriate amino acids to corresponding tRNAs. d. form the peptide bond between amino acids at the ribosome. e. synthesize and transport amino acids to the ribosomes.

45. For any sequence of nucleotides, how many reading frames are possible? a. 1 b. 2 c. 3 d. 5 e. 10

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 15_7e

46. If the bottom strand of the DNA from the diagram above serves as the template strand, the RNA sequence, left to right 5′ to 3′, is: a. AUAGGCAGU. b. UCCCAGGUG. c. CACCUGGGA. d. AGGGUCCAC. e. GACAUUAGA. 47. If the bottom strand of the DNA serves as the template, the amino acid sequence of the protein produced from the RNA would be: a. Met-Leu-Ser. b. Arg-Val-His. c. Thr-Ile-Phe. d. Pro-Gly-Trp. e. Lys-Val-His.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 15_7e 48. Which of the following mechanisms specifically allows detection and rapid degradation of mRNA containing a premature termination codon? a. RNA interference b. no-go decay c. nonsense-mediated mRNA decay d. transfer-messenger RNA mediated ribosomal removal e. nonstop mRNA decay 49. Which of the following statements about translation is CORRECT? a. A special tRNA that does not have an attached amino acid binds to stop codons to terminate translation. b. The first three bases at the 5′ end of an mRNA are the AUG at which translation begins. c. The codon for methionine appears only at the beginning of the mRNA for a protein, not in the middle or in the end. d. In eukaryotes, the 5′ cap and the 3′ poly(A) tail are involved in translation initiation. e. Ribosomes move along an mRNA in the 3′ to 5′ direction.

50. Which of the following statements about proteins is INCORRECT? a. All proteins are made up of some combination of 20 essential amino acids. b. Like nucleic acids, polypeptides have polarity. c. A single polypeptide has primary, secondary, and tertiary structures. d. α-helix and β-pleated sheets do not require a specific sequence of amino acids to form. e. Some proteins contain more than one polypeptide chain.

51. Which of the following statements does NOT describe the events in prokaryotic translation elongation? a. The nucleotides in the Shine–Dalgarno sequence of the mRNA pair with their complementary nucleotides in the 16S rRNA. b. A ribosome with a growing peptide attached to a tRNA in the P site accepts a charged tRNA with the next amino acid into the A site. The charged tRNA enters as a complex with EF-Tu and GTP. c. If the anticodon of the charged tRNA matches the codon, GTP is cleaved and EF-Tu exits and is regenerated to EF-Tu-GTP by EF-Ts. d. A peptide bond is formed by the peptidyl transferase activity of the large subunit rRNA. The polypeptide chain on the tRNA in the P site is transferred to the amino acid on the tRNA in the A site. e. The ribosome translocates toward the 3′ end of the mRNA with the aid of EF-G and GTP hydrolysis.

The empty tRNA that was in the P site moves to the E site and exits. The tRNA with the polypeptide that was in the A site moves to the P site.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 15_7e 52. During initiation of translation: a. the initiator tRNAmet binds to the A site of a ribosome. b. specific rRNA base pairs with a sequence in mRNA to position a ribosome at the start codon. c. IF-3 must be recruited to the 30S ribosome in order for the 70S initiation complex to assemble. d. there is no energy expenditure as the tRNA binding to mRNA is via complementary base pairing. e. both 70S and 30S ribosome subunits must simultaneously recognize an mRNA to bind.

Indicate one or more answer choices that best complete the statement or answer the question.

53. Translation in prokaryotes and eukaryotes has some similarities but differs in important ways. Which of the following statements about similarities and differences in prokaryotic and eukaryotic translation are TRUE? (Select all that apply.) a. Initiation in prokaryotes and eukaryotes begins with a formylmethionine. b. Prokaryotic and eukaryotic ribosomes are different sizes. c. Prokaryotic ribosomes are sensitive to antibiotics that do not affect eukaryotic ribosomes. d. Prokaryotic and eukaryotic large subunits contain two rRNAs. e. Prokaryotic mRNAs are short-lived. Eukaryotic mRNAs vary in their half-life. f. Recognition of the start codon in prokaryotes and eukaryotes involves the 5′ cap, the 3′ poly(A) tail, and the Kozak sequence near the start codon. g. Prokaryotic initiation, elongation, and termination factors are different from those in eukaryotes. h. Transcription and translation routinely occur simultaneously in prokaryotes and eukaryotes. i. Elongation involves the formation of a peptide bond between the amino acids that are attached to the tRNAs in the P and A sites of the ribosome in both prokaryotes and eukaryotes.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 15_7e An auxotrophic E. coli strain requires adenine to grow because of a mutation in a gene for an adenine synthesis enzyme. The following shows part of the wild-type and mutant alleles of the gene, including the start codon. The bottom strand is the template for transcription.

54. E. coli strains that have both the original ade1- mutant allele shown above and a mutant allele for a gene that encodes a lysine tRNA are able to make some normal ade1 enzyme. The mutant lysine tRNA allele makes a tRNA that has one base that is different from the wild-type tRNA. Which of the following statements would help explain how the lysine tRNA mutation allows synthesis of the normal ade1 enzyme? (Select all that apply.) a. The mutation in the lysine tRNA gene changes the sequence of the anticodon. b. The mutant tRNA pairs with UAG, a stop codon. c. When the mRNA from the ade1– allele is being translated, a mutant tRNAlysine could incorporate a lysine into the protein, bypassing the mutant UAG. d. The ade1– mutant would always result in a stop signal regardless of the mutant tRNA. e. The mutant tRNA would always match with UAG and prevent normal termination of the polypeptide chain.

55. What was the key technical deficiency that prevented Nirenberg and colleagues from deciphering the genetic code? What were the two important breakthroughs that still allowed them to break the genetic code?

56. Where is the base inosine (I) found, and what is its function?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 15_7e 57. Four different uracil auxotrophs of Neurospora, a eukaryotic mold, are tested for growth on uracil and uracil precursors. The data are shown in the following table. A plus sign (+) means growth.

Mutant 1 Mutant 2 Mutant 3 Mutant 4

Compound A B C + + + – – + + – + – – –

D – – – –

Uracil + + + +

Neurospora can be grown as haploids or diploids. Haploid mutants 1 and 2 are fused to make a diploid. If both 1 and 2 carry recessive mutations, on which compounds will the diploid be able to grow?

58. An auxotrophic E. coli strain requires adenine to grow because of a mutation in a gene for an adenine synthesis enzyme. The following shows part of the wild-type and mutant alleles of the gene, including the start codon. The bottom strand is the template for transcription.

E. coli strains that have both the original ade1– mutant allele shown above and a mutant allele for a gene that encodes a lysine tRNA are able to make some normal ade1 enzyme. The mutant lysine tRNA allele makes tRNA that has one base that is different from the wild-type tRNA. Explain how the lysine tRNA mutation allows synthesis of the normal ade1 enzyme.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 15_7e 59. An auxotrophic E. coli strain requires adenine to grow because of a mutation in a gene for an adenine synthesis enzyme. The following shows part of the wild-type and mutant alleles of the gene, including the start codon. The bottom strand is the template for transcription.

Explain why the mutant lysine tRNA allele from the previous question might affect overall cell growth.

60. A random sequence of synthetic RNA is made using equal proportions of A and C. When it is translated, which amino acids will be incorporated into the protein?

61. The genetic code uses three bases to encode one amino acid. Why can't the code use only two bases to encode each amino acid?

62. Which amino acids are encoded, if the reading frame is as shown below, starting from the correct end?

5′ ...GGAGCUCGUUGUAUU... 3′

63. Describe the events in prokaryotic translation initiation.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 15_7e 64. E. coli only has a single tryptophan tRNA gene. Why, given this arrangement, might it be difficult to identify mutants that alter the anticodon of this tRNA gene?

65. List three differences between prokaryotic and eukaryotic translation.

66. Describe the events in prokaryotic translation elongation.

67. This DNA sequence represents an open reading frame (ORF) of a transcriptional unit. Transcribe and then translate this gene.

5′ ATGGGAGCTCGTTGTATTTGA 3′ 3′ TACCCTCGAGCAACATAAACT 5′

68. What would be the effect on the following amino acid sequence if the sequence were changed to 5′ GGAGACUCGUUGUAUU 3′? Explain why the amino acid sequence changes.

5′ ...GGAGCUCGUUGUAUU... 3′

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 15_7e 69. Although the genetic code is nearly universal, some variations do exist. In vertebrate mitochondria, UGA codes for Trp (instead of termination), AUA codes for Met (instead of Ile), and AGA and AGG are stop codons (instead of coding for Arg). Translate the following coding strand DNA sequences using both the standard code and the vertebrate mitochondrial code. a. 5′ ATGGCCATAAGATGA 3′ b. 5′ATGGGGGATCGCTAA 3′ c. 5′ATGTGATGGCATCTTATAAATTGATAA 3′

70. Write the anticodon, with correct polarity, of all tRNAs that will bind to the mRNA codon 5′ UCG 3′, considering wobble–base pairing rules.

71. Four different uracil auxotrophs of Neurospora, a eukaryotic mold, are tested for growth on uracil and uracil precursors. The data are shown in the following table. A plus sign (+) means growth.

Mutant 1 Mutant 2 Mutant 3 Mutant 4

Compound A B C + + + – – + + – + – – –

D – – – –

Uracil + + + +

Diagram the uracil pathway, showing the step at which each mutant is blocked.

72. Write the codon, with correct polarity, of all mRNA codons that will bind to the tRNA anticodon 5′ GCU 3′, considering wobble–base pairing rules.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 15_7e 73. The antibiotic streptomycin binds to the 30S subunit in bacteria and interferes with translation. Streptomycin resistance (SmR) can occur through mutation in the rpsL gene, which encodes the ribosomal protein S12. The antibiotic can no longer bind to the altered protein, so translation can occur without interference from the antibiotic. However, in the absence of streptomycin, many SmR strains of bacteria grow less well than the nonmutant streptomycin-sensitive (SmS) bacteria. a. Propose a model for why SmR bacteria grow better than SmS bacteria in the presence of streptomycin but not as well in the absence of streptomycin. b. During translation, an incorrect amino acid is occasionally incorporated into the polypeptide chain—that is, the amino acid added does not correspond to the codon at that location in the mRNA. You develop an assay that can assess this rate of incorrect amino acid incorporation and thereby estimate the accuracy of translation in bacteria. You find that SmR bacteria are more accurate than SmS bacteria in this experiment. How might this observation affect your model above?

74. An auxotrophic E. coli strain requires adenine to grow because of a mutation in a gene for an adenine synthesis enzyme. The following shows part of the wild-type and mutant alleles of the gene, including the start codon. The bottom strand is the template for transcription.

How does this mutation change the mRNA, and how does it affect the enzyme?

75. Prokaryotic ribosomes are inhibited by antibiotics like tetracyclines. As eukaryotes, humans are unaffected by these antibiotics. However, ribosomes within human mitochondria more closely resemble prokaryotic ribosomes than eukaryotic ribosomes. Why aren't tetracyclines more toxic to humans if they can, in theory, inhibit mitochondrial ribosomes?

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 15_7e 76. You identify a novel deep-sea-vent (DSV) bacteria and wish to explore its molecular biology. You find that you can introduce DNA sequences into the bacteria and it will transcribe and translate those sequences.

However, you notice some curious differences between the expected proteins and the actual proteins produced. a. A 9-bp DNA sequence that in E. coli encodes a short peptide of Met-Trp-Phe instead produces a peptide of Met-Cys-Val-Trp-Gly-Val-Phe in the DSV bacteria. What is the sequence of the coding strand of the short DNA fragment? Propose an explanation for the longer peptide produced by the DSV bacteria. b. Curiously, a 9-bp sequence that in E. coli produces a peptide of Met-Thr-Asn does not produce any peptide in the DSV bacteria. What is the sequence of the coding strand of this DNA fragment? Propose an explanation for the absence of any peptide.

77. What was Beadle and Tatum's definition of a gene? Give a modern definition and explain how it differs from that of Beadle and Tatum.

78. What would be the consequences of a two-nucleotide-deletion mutation in the middle of the first exon of a protein-coding gene?

79. The wild-type sequence of a protein is ... Asp-Ile-Cys-Trp.… The sequence of a mutant form of the protein is ... Asp-Ile-Tyr-Trp.... Write all possible DNA sequences for the wild-type and mutant alleles of the gene for this protein; underline the changed nucleotide in the mutant allele.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 15_7e 80. Is the following sequence RNA or DNA? How can you tell?

5′ ...GGAGCUCGUUGUAUU... 3′

81. A yeast strain was exposed to a chemical mutagen. As expected, exposure to a mutagen resulted in a DNA sequence change in an essential gene you examined. Yet this mutation did not result in any lethal phenotype.

How do you explain this apparent discrepancy?

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 15_7e Answer Key 1. a 2. b 3. d 4. c 5. b 6. c 7. b 8. d 9. b 10. c 11. b 12. b 13. d 14. e 15. a 16. c 17. e 18. e 19. d 20. d 21. c 22. a 23. a 24. a 25. b 26. e Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 15_7e 27. e 28. a 29. c 30. a 31. a 32. e 33. e 34. e 35. a 36. a 37. a 38. c 39. c 40. e 41. e 42. b 43. c 44. c 45. c 46. d 47. b 48. c 49. d 50. a 51. a 52. b 53. b, c, e, g, i 54. a, b, c Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 15_7e 55. To solve the genetic code, all they had to do was to generate an mRNA of a particular sequence and then in vitro translate it to make a corresponding protein. Then, if they could have sequenced both the mRNA and protein, they would have been able to break the code. However, in the early 1960s, RNA and protein sequencing technology did not exist.

First, Nirenberg and colleagues exploited an important aspect of an enzyme called polynucleotide phosphorylase, which allows an enzyme to make an RNA polymer without the need of a template (an essential requirement for RNA polymerase). Using this enzyme, Nirenberg and colleagues made a variety of synthetic RNAs to solve some of the codons. However, because of the limitation in making a variety of random polymers with this approach, it was still not possible to solve many other codons.

They overcame this problem by using ribosome-bound tRNAs. They found that a very short sequence of mRNA —even one consisting of a single codon—would bind to a ribosome. By synthesizing a variety of very short mRNAs, they then isolated the tRNAs that were bound to the mRNA and ribosomes and determined which amino acids were present on the bound tRNAs. 56. Inosine is found in the anticodon of tRNA at first position. It is involved in wobble base pairing with the third position of the codon and is capable of pairing with A, U, or C. It permits tRNAs to bind with some synonymous codons.

57. All compounds: D, B, A, C, and uracil 58. The mutation in the lysine tRNA gene changes the sequence of the anticodon. The anticodon now pairs with UAG, a stop codon. When the mRNA from the ade1– allele is being translated, if a mutant tRNA lysine enters the ribosome at the UAG, the lysine carried by the tRNA will be incorporated into the protein, bypassing the mutant UAG. 59. The mutant tRNAlysine will occasionally cause translation to continue beyond the normal stop codon on mRNAs that normally use UAG as a stop codon. The presence of some longer versions of normal proteins may affect cell growth.

60. pro, lys, thr, gln, his, thr, and asn 61. With four bases in RNA, there are only 16 combinations of two bases. There are 20 different amino acids, so this would not be enough to encode all amino acids.

62. Gly-Ala-Arg-Cys-Ile 63. a. IF-3 separates ribosome subunits so that a small subunit can bind mRNA through base pairing of the 16S rRNA and the Shine–Dalgarno sequence on the mRNA. b. An initiator tRNAformylmet binds the initiation codon, with the help of IF-1 and IF-2 complexed with GTP. The tRNAformylmet is positioned in the P site. c. IF-3 dissociates, it allows a large subunit to bind the 30S initiation complex.

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 15_7e 64. In addition to having only the single tryptophan tRNA gene, there is only a single tryptophan codon in the genetic code. Any changes to the tryptophan tRNA anticodon would make the bacteria unable to recognize the tryptophan codon in mRNA. Therefore, any proteins that ought to contain the tryptophan amino acid will be unable to be translated successfully, which would be lethal to the organism. Other amino acids have multiple codons that code for them and, in many cases, multiple copies of the tRNA gene. In this event, a mutation in the anticodon of a single tRNA gene would not prevent the incorporation of these amino acids, and so mutants in these tRNA genes can be identified (see Chapter 18), unlike the case for the tryptophan tRNA gene.

65. (1) Initiation in prokaryotes begins with a formyl-methionine. In eukaryotes, it begins with methionine.

(2) Prokaryotic and eukaryotic ribosomes are different sizes.

(3) Prokaryotic ribosomes are sensitive to antibiotics that do not affect eukaryotic ribosomes.

(4) Prokaryotic large subunits contain two rRNAs, while eukaryotic large subunits contain three.

(5) Prokaryotic mRNAs are short-lived. Eukaryotic mRNAs vary in their half-life.

(6) Recognition of the start codon is different. Prokaryotes have an rRNA that binds to the mRNA. Eukaryotic recognition of the start codon involves the 5′ cap, the 3′ poly(A) tail, and the Kozak sequence near the start codon.

(7) Prokaryotic initiation, elongation, and termination factors are different.

(8) Transcription and translation routinely occur simultaneously in prokaryotes but are not routine in eukaryotes.

66. a. A ribosome with a growing peptide attached to a tRNA in the P site accepts a charged tRNA with the next amino acid into the A site. The charged tRNA enters as a complex with EF-Tu and GTP. b. If the anticodon of the charged tRNA matches the codon, GTP is cleaved, and EF-Tu exits and is regenerated to EF-Tu-GTP by EF-Ts. c. A peptide bond is formed by the peptidyl transferase activity of the large subunit rRNA. The polypeptide chain on the tRNA in the P site is transferred to the amino acid on the tRNA in the A site. d. The ribosome translocates toward the 3′ end of the mRNA with the aid of EF-G and GTP hydrolysis. The empty tRNA that was in the P site moves to the E site and exits. The tRNA with the polypeptide that was in the A site moves to the P site.

67. (a) Transcribed mRNA sequence: 5′ AUGGGAGCUCGUUGUAUUUGA 3′ (b) Translated protein sequence: Met-Gly-Ala-Arg-Cys-Ile (c) Which amino acid is at the C-terminus of this peptide? Ile 68. The amino acid sequence would be Gly-Asp-Ser-Leu-Tyr. The reading frame, or grouping of bases used as codons, changes after the mutation (the addition of an A in the fifth position).

69. a. Standard: Met-Ala-Ile-Arg; Mitochondrial: Met-Ala-Met b. Standard: Met-Gly-Asp-Arg; Mitochondrial: Met-Gly-Asp-Arg c. Standard: Met; Mitochondrial: Met-Trp-Trp-His-Leu-Met-Asn-Trp 70. 5′ CGA 3′ 71.

D

1 →

B

3 →

A

Copyright Macmillan Learning. Powered by Cognero.

2 →

C

4 →

uracil

Page 28

Name:

Class:

Date:

Chap 15_7e 72. 5′ AGU 3′ 5′ AGC 3′ 73. a. In the SmR bacteria, streptomycin cannot bind the ribosome because of the mutation. This observation suggests a change in the shape of the ribosome caused by the mutation. However, the change in the ribosome must not be so severe as to prevent the ribosome from functioning entirely, since the bacteria can still grow. Nevertheless, in the absence of streptomycin, the altered ribosome likely works less well than the normal ribosome, resulting in slower growth than in the SmS strain of bacteria. b. The model above proposed that the mutant ribosome works less well, but the observation of greater accuracy by the mutant seems to contradict that model. However, another parameter in the function of a ribosome (aside from accuracy) would be speed. If the translation rate of the mutant ribosome was slower than that of the normal ribosome, that factor could account for the slower growth of the mutant. Apparently, then, the bacteria must have a trade-off between accuracy and speed—the mutant ribosome is more accurate but at the cost of slower speed.

74. The third codon in the mRNA is changed from AAG (lysine) to UAG (stop). So the protein won't be synthesized, and translation will stop at the third codon.

75. Mitochondrial ribosomes are protected within several membranes. Antibiotics need to be transported across the plasma membrane, the mitochondrial outer membrane, and the mitochondrial inner membrane to reach them.

76. a. The DNA sequence must be 5′-ATGTGGTTN-3′ to encode Met-Trp-Phe. These amino acids are present in the longer DSV peptide, too. However, two intervening amino acids separate them. The simplest explanation is that, somehow, the DSV ribosome only shifts one base after reading a codon. Thus, the DSV ribosome translates the codons AUG-UGU-GUG-UGG-GGU-GUU-UUN to produce the longer peptide. b. The DNA sequence is 5′-ATGACNAAY-3′ (where N is any base and Y is either C or T). Consistent with the model above, the second codon the DSV ribosome would encounter is UGA, which is a stop codon. Therefore, no peptide would form.

77. They defined it as the one gene, one enzyme hypothesis. A modern version may be called the one gene, one polypetide hypothesis because we now know that enzymes can be made up of multiple polypeptide chains, each encoded by a gene. However, our modern definition of the gene must also take into account that many genes encode functional RNA products, like rRNA. 78. A reading frameshift would occur. All codons (and thus amino acids) downstream of the codon in which the deletion occurred would be different, including the stop signal.

79. Wild-type allele 5′ GAT/C ATC/A/T TGT/C TGG 3′ 3′ CTA/G TAG/T/A ACA/G ACC 5′ Mutant allele 5′ GAT/C ATC/A/T TAT/C TGG 3′ 3′ CTA/G TAG/T/A ATA/G ACC 5′ 80. It is RNA because it contains U (uracil) and no T (thymine).

Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 15_7e 81. The DNA sequence change occurred in a synonymous nucleotide position of an amino acid, and as a result the protein sequence remained unaltered. Since the protein sequence was not affected, no lethal phenotype manifested.

Another possibility is that the DNA sequence change resulted in an amino acid change, but that amino acid change had no negative effect on protein function (i.e., was a neutral change).

Copyright Macmillan Learning. Powered by Cognero.

Page 30

Name:

Class:

Date:

Chap 16_7e Indicate the answer choice that best completes the statement or answers the question.

1. What is the function of allolactose in regulation of the lac operon? a. inducer b. repressor c. activator d. promoter e. regulatory protein 2. An example of a gene product encoded by a regulatory gene is: a. beta-galactosidase enzyme. b. allolactose. c. repressor protein. d. an operator. e. a terminator.

3. When a structural gene is under negative inducible control, what would be the result of a mutation that eliminated the repressor protein? a. The structural gene would be constitutively expressed due to the lack of negative inducible control. b. The transcription of the structural gene would not be affected, as a repressor is not required. c. The mutation would lead to activation of an activator upon the lack of a repressor protein, which would allow the transcription to continue. d. Since transcription requires a repressor protein, the transcription would be turned off. e. More cAMP would be produced in a cell to compensate for the lack of a repressor protein.

4. A deletion occurs in the trp operon DNA of E. coli and results in the loss of the attenuation region in the 5′ UTR of the RNA. The DNA sequences of the structural genes and the operator/promoter region are not affected by deletion. What effect will this deletion have on the regulation of this mutant trp operon compared to a wild-type operon? a. In the presence of tryptophan, transcription of the structural genes will be enhanced compared with a wild-type operon. b. In the absence of tryptophan, transcription of the structural genes will be reduced compared with a wild-type operon. c. In the presence of tryptophan, the repressor will bind to the operator/promoter region in the mutant operon more strongly than in a wild-type operon. d. In the absence of tryptophan, RNA polymerase will not bind to the operator/promoter region in the mutant operon. e. In the presence of tryptophan, transcription will be initiated at the second structural gene in the mutant operon.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 16_7e 5. Which of the following E. coli strains will synthesize permease in the presence of lactose and absence of glucose? a. lacI– lacP– lacO+ lacZ+ lacY+ b. lacIs lacP– lacO+ lacZ– lacY+/lacI+ lacP+ lacO+ lacZ+ lacY+ c. lacI– lacP+ lacO+ lacZ+ lacY– d. lacI– lacP+ lacOc lacZ– lacY– /lacI+ lacP– lacO+ lacZ– lacY+ e. lacI+ lacP+ lacOc lacZ+ lacY– /lacI– lacP+ lacO+lacZ– lacY+ 6. Which of the following statements about regulation of gene expression is CORRECT? a. An inducible gene is transcribed when a specific substance is absent. b. A gene is any DNA sequence that is transcribed into an mRNA molecule only. c. All genes are transcribed at all times as long as they have a functional promoter. d. The regulation of gene expression is the same in both eukaryotes and prokaryotes. e. The regulation of gene expression is critical for the control of life processes in all organisms.

7. Which of the following secondary structures causes attenuation of structural genes of the trp operon? a. 1+2 loop b. 1+3 loop c. 2+4 loop d. 2+3 loop e. 3+4 loop 8. Which of the following statements about DNA-binding proteins is NOT true? a. Specific amino acids within the motif form hydrogen bonds with DNA. b. These proteins can affect the expression of a gene. c. Most DNA-binding proteins bind dynamically. d. Some of these proteins incorporate a metal ion such as zinc. e. Once bound, most of DNA-binding proteins remain on DNA permanently.

9. In the presence of both lactose and glucose, which of the following occurs with the E. coli lac operon? a. The lacI gene does not produce repressor. b. The cAMP–CAP complex is not available to bind near or at the lac promoter. c. The lacZ and lacY genes are fully expressed. d. Lactose is converted to glucose and galactose. e. Lactose binds to the operator.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 16_7e 10. There are enzymes called aminoacyl-tRNA synthetases that attach tRNAs to the appropriate amino acid.

Assume that the aminoacyl-tRNA synthetase that attaches tryptophan to its tRNA in an E. coli mutant strain is only partially active. It is active enough for the strain that carries it to be viable but is much less active than that of wild-type cells. What would be the expected effect of this mutation on attenuation in the trp operon? a. Pairing of regions 3 and 4 in the 5′ UTR of the RNA will be enhanced so attenuation will increase in the presence of tryptophan. b. Pairing of regions 2 and 3 in the 5′ UTR of the RNA will be decreased so attenuation will increase in the presence of tryptophan. c. Pairing of regions 3 and 4 in the 5′ UTR of the RNA will be decreased so attenuation will decrease in the presence of tryptophan. d. Attenuation will not change but the trp repressor will be more active and the transcription of the trp operon structural genes will be decreased in the presence of tryptophan. e. Attenuation will be decreased in the absence of tryptophan but will be enhanced in the presence of tryptophan.

11. Which of the following is NOT a common DNA-binding motif? a. the helix-loop-helix motif b. the beta sheet–alpha helix motif c. the zinc-finger motif d. the homeodomain motif e. the leucine-zipper motif 12. A lac operon of genotype lacI+ lacP+ lacO+ lacZ+ lacY– will produce β-galactosidase but not permease when: a. lactose is present. b. lactose is absent. c. in the presence or absence of lactose. d. glucose is present. e. glucose and lactose are both present.

13. Which of the following DNA-binding motifs are composed of three alpha helices? a. zinc-finger motif b. leucine-zipper motif c. homeodomain motif d. helix-turn-helix motif e. steroid receptor

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 16_7e 14. E. coli lac operon control by CAP is: a. negative inducible. b. negative repressible. c. positive inducible. d. positive repressible. e. regulated by riboswitches.

15. A mutant E. coli strain, grown under conditions that normally induce the lac operon, produces particularly high amounts of ß-galactosidase. What is a possible genotype of the cells? a. lacI+ lacP+ lacO+ lacZ– lacY+ lacA+ b. lacI+ lacP+ lacOc lacZ+ lacY+ lacA+ c. lacI– lacP+ lacO+ lacZ– lacY+ lacA+ d. lacI+ lacP– lacO+ lacZ+ lacY+ lacA+ e. lacI– lacP– lacO+ lacZ– lacY+ lacA– 16. What would happen to the lac operon in the absence of allolactose? a. The structural genes within the lac operon will be constitutively transcribed. b. The activator protein will be bound to the operator, which will turn on the structural gene behind it. c. The repressor regulator protein binds to the operator and prevents the transcription of the structural gene. d. The catabolite activator protein becomes inactivated and no transcription occurs. e. The cAMP level rises in the absence of allolactose, which in turn inactivates the transcription.

17. RNA molecules that are complementary to particular sequences on mRNA are called: a. complementary RNA. b. sense RNA. c. antisense RNA. d. riboswitches. e. ribozymes.

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 16_7e 18. Suppose that you perform an experiment where you construct a plasmid that carries a copy of the lac operator region (lacO) but no other part of the lac operon. (The lac repressor can bind to single operator regions.) This plasmid is placed in an E. coli cell, which has a normal copy of the lac operon in its chromosome. When this strain is grown, the number of plasmids reaches about 50 copies per bacterial cell.

What is the expected phenotype of such a strain in the absence of glucose? a. The lac operon will be turned on even in the absence of lactose. b. The lac operon will be turned off even in the presence of lactose. c. The lac operon will be regulated normally. d. The lac operon will be initially turned on in the presence of lactose but eventually it will be turned off even though lactose is still present. e. The lac operon will be expressed only if galactose is added to the Moderate.

19. Where would the lac repressor be bound in a (nonmutant) E. coli cell that is growing in low glucose and high lactose? (I = lac repressor gene; Z, Y, A = lac operon structural genes; P = lac promoter; O = lac operator)

a. P b. O c. P and O d. I, P, O e. The repressor would not be bound.

20. Which of the following statements about gene regulation concerning operons is INCORRECT? a. A negative repressible gene is controlled by a regulatory protein that inhibits transcription. b. For a gene under negative repressible control, a small molecule is required to prevent the gene's repressor from binding to DNA. c. For a gene under positive repressible control, the normal state is transcription of a gene, stimulated by a transcriptional activator. d. A regulator gene has its own promoter and is transcribed into an independent mRNA. e. Presence of operons where genes of related functions are clustered is common in bacteria but not in eukaryotes.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 16_7e 21. A mutation occurs in the trp operon DNA of E. coli and results in the change to the two UGG tryptophan codons in the 5′ UTR of the RNA to UAG stop codons. What effect will this mutation have on the regulation of this mutant trp operon compared to a wild-type operon? a. In the presence of tryptophan, transcription of the structural genes will be reduced compared with a wild-type operon. b. In the absence of tryptophan, transcription of the structural genes will be reduced compared with a wild-type operon. c. In the presence of tryptophan, the repressor will bind to the operator/promoter region with the mutant operon more strongly than with a wild-type operon. d. In the absence of tryptophan, RNA polymerase will not bind to the operator/promoter region with the mutant operon. e. In the presence of tryptophan, transcription of the structural genes will be enhanced compared with a wild-type operon.

22. The _____ is a type of regulator protein that binds to a region of DNA in the promoter of a gene called the operator and prevents transcription from taking place. a. inducer b. repressor c. activator d. inactivator e. terminator 23. Which parts of the DNA region shown in the diagram encode proteins? (I = lac repressor gene; Z, Y, A = lac operon structural genes; P = lac promoter; O = lac operator)

a. P b. I, P, O c. P, O, Z, Y, A d. I, Z, Y, A e. I, P, O, Z, Y, A

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 16_7e 24. It is possible for a repressor to negatively regulate the expression of an operon because: a. the repressor induces the expression of the inducer by binding to the promoter that comes before the inducer gene. b. one of the structural genes expressed in the operon negatively regulates the repressor. c. the repressor-binding site overlaps the promoter site of the operon, allowing it to physically block the binding of RNA polymerase. d. the repressor-binding site on the DNA overlaps with the translation start site, thereby preventing the transcription. e. the repressor physically blocks where the activator should be binding on the operator region.

25. When a structural gene is under positive inducible control, what would be the result of a mutation that eliminates the activator protein? a. The structural gene will be constitutively expressed due to the lack of inducible control. b. The transcription of the structural gene will not be affected, as an activator is not required. c. The mutation will lead to activation of a repressor upon the lack of an activator protein, which will block transcription. d. Since transcription will require an activator protein, transcription will be turned off. e. More cAMP will be produced in a cell to compensate for the lack of an activator protein.

26. What is the function of cAMP in regulation of the lac operon? a. It activates a repressor protein. b. It activates an activator protein. c. It inactivates a repressor protein. d. It inactivates an activator protein. e. It causes attenuation.

27. A partial diploid E. coli cell of lacI+ lacP– lacOc lacI+ lacY+ / lacIs lacP+lacO+ lacZ+ lacY– will synthesize: a. both lacZ and lacY functional gene products in the absence of lactose. b. neither lacZ nor lacY functional gene products in the presence of lactose. c. both lacZ and lacY functional gene products in the presence of lactose. d. lacY but not a functional lacZ gene product in the absence of lactose. e. lacZ but not a functional lacY gene product in the absence of lactose.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 16_7e 28. The trp operon is known to operate by both negative repressible regulation of the operator and attenuation.

Which of the following statements does NOT support the reason dual control exists to regulate the operon? a. The repression alone is never complete, and some transcription can be initiated. b. The combined mechanism provides a much finer tuning of tryptophan synthesis regulation. c. Attenuation and repression allow the cell to more sensitively respond to the tryptophan level. d. It is most likely due to the fact that the attenuation is the evolutionary relic, which by accident has remained. e. Repression responds to the cellular levels of tryptophan, while attenuation responds to the number of tRNA charged with tryptophan.

29. A lac operon of genotype lacI+ lacP+ lacO+ lacZ– lacY+ will not produce β-galactosidase but will produce permease when: a. lactose is present. b. lactose is absent. c. in the presence or absence of lactose. d. glucose is present. e. glucose and lactose are present.

30. An operon is controlled by a repressor. When the repressor binds to a small molecule, it binds to DNA near the operon. The operon is constitutively expressed if a mutation prevents the repressor from binding to the small molecule. The type of control illustrated is: a. negative inducible. b. negative repressible. c. positive inducible. d. positive repressible. e. catabolite repression.

31. E. coli lac operon control by lacI is: a. negative inducible. b. negative repressible. c. positive inducible. d. positive repressible. e. attenuation.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 16_7e 32. A mutant E. coli strain, grown under conditions that normally induce the lac operon, does not produce ßgalactosidase. What is a possible genotype of the cells? a. lacI+ lacP+ lacO+ lacZ+ lacY– lacA+ b. lacI+ lacP+ lacOc lacZ+ lacY+ lacA+ c. lacl+ lacP+ lacO+ lacZ+ lacY+ lacA+ d. lacI+ lacP– lacO+ lacZ+ lacY+ lacA+ e. lacI– lacP+ lacO– lacZ+ lacY+ lacA– 33. A partial diploid E. coli cell of lacI+lacP+lacOc lacZ– lacY– / lacI– lacP+ lacO+lacZ– lacY+ genotype will synthesize: a. both lacZ and lacY gene products in the absence of lactose. b. neither lacZ nor lacY gene products in the presence of lactose. c. lacZ but not lacY gene product in the presence of lactose. d. lacY but not lacZ gene product in the absence of lactose. e. lacY but not lacZ gene product in the presence of lactose.

34. Which of the following is generally constitutively transcribed? a. regulatory genes b. structural genes c. operator elements d. promoter elements e. operons 35. RNA-mediated repression is carried out by: a. nonsense RNA. b. sense RNA. c. antisense RNA. d. riboswitches. e. ribozymes.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 16_7e 36. If there are mutations that inactivate lacP and lacI, which of the following statements is true? (I = lac repressor gene; Z, Y, A = lac operon structural genes; P = lac promoter; O = lac operator)

a. These are mutations that are, respectively, cis- and trans-acting on lac operon expression. b. These are mutations that are, respectively, trans- and cis-acting on lac operon expression. c. These will affect the expression of I only. d. These will affect the expression of only Z, Y, and A. e. These mutations will have no effect.

37. Assume that a mutation occurs in the promoter for the lacI regulatory gene and this mutation results in a tenfold increase in the transcription of lacI. What would be the expected consequences of such a mutation? a. The lac structural genes would be fully expressed even in the presence of glucose and absence of lactose. b. The lac operon would be regulated normally. c. The lac structural genes would not be fully induced even in the absence of glucose and presence of lactose. d. The lac structural genes would be partially expressed in the presence of glucose and fully expressed if lactose is also present. e. The lac structural genes would now be fully expressed in the absence of glucose and lactose but not expressed in the absence of glucose and presence of lactose.

38. When binding of the inducer to the repressor causes a conformational change, which then prevents the repressor from binding to DNA, the repressor is called a(n) _____ protein. a. coactivator b. allosteric c. structural d. operating e. responsive 39. A mutant E. coli strain is found that synthesizes β-galactosidase in the absence of glucose whether or not lactose is present. What two mutations might lead to this outcome? a. lacIs mutation or lacOc mutation b. lacP– mutation or lacI– mutation c. lacP– mutation or lacOc mutation d. lacI– mutation or lacOc mutation e. lacP– mutation or lacIs mutation Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 16_7e 40. Proteins with DNA-binding motifs predominantly bind to the _____ of DNA. a. major groove b. minor groove c. paired nitrogenous bases d. phosphate groups e. deoxyribose sugar 41. The formation of 1+2 and 3+4 secondary structures of 5′ UTR region mRNA from the trp operon is triggered when: a. the tryptophan level inside the bacterial cell is extremely low. b. the tryptophan level inside the bacterial cell is high. c. the repressor protein fails to bind to the operator. d. there is a spontaneous mutation introduced into the 5′ UTR. e. the structural gene transcription within the trp operon gets initiated.

42. Which of the following types of eukaryotic gene regulation take place at the level of DNA? a. alternation of chromatin structure b. mRNA processing c. RNA interference d. mRNA stability e. posttranslational modification 43. In the absence of tryptophan, what happens to the genes within the trp operon? a. The regulator without tryptophan-binding prevents the genes from being transcribed. b. The regulator dissociates from the operator, and structural genes get transcribed. c. Lack of tryptophan increases the level of cAMP, which leads to activation of CAP protein and gene expression. d. The active repressor binds to the operator, and genes do not get transcribed. e. The active activator binds to the operator, and transcription of structural gene takes place.

44. A lac operon of genotype lacI– lacP+ lacO+ lacZ+ lacY+ will produce β-galactosidase and permease when: a. lactose is present. b. lactose is absent. c. lactose is present or not. d. glucose is present. e. glucose and lactose are both present.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 16_7e 45. A promoter that affects only genes that are on the same piece of DNA is referred to as a(n) _____-acting promoter. a. cis b. trans c. enhancer d. positive e. negative 46. If a mutation prevents the formation of the antiterminator 2+3 loop in the trp operon, what would be the effect? a. transcription only when tryptophan is absent b. transcription only when tryptophan is present c. constitutive attenuation of transcription d. constitutive transcription e. no effect, as the 2+3 loop has no function 47. The trp operon in E. coli contains a 5′ UTR sequence that is translated into a small polypeptide of 14 amino acids, which includes two tryptophans. If the two trp codons in the 5′UTR of the RNA are changed to serine codons and the resulting cells are starved for tryptophan but not for any other amino acid, what will be the effect of the mutant codons on the operation of the trp operon? a. The trp repressor will be synthesized in greater amounts. b. Expression of the trp structural genes will be less than with a normal trp operon. c. Expression of the trp structural genes will be greater than with a normal trp operon. d. The leader polypeptide will not be synthesized. e. The expression of the operon will not be affected by these mutations.

48. Transcriptional control that acts by regulating the continuation of transcription is called: a. riboswitching. b. antitermination. c. negative control. d. operator mutation. e. attenuation.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 16_7e 49. Which of the following facts about riboswitches is INCORRECT? a. Binding of certain molecules to the riboswitches results in the formation of specific secondary structures of mRNA. b. Certain molecules that bind to riboswitches may act as repressors or inducers of transcription. c. Riboswitches are not only found in bacterial cells but also in archaeal, fungal, or plant cells. d. Riboswitches are typically found in the 3′ UTR of the mRNA structure. e. The secondary structure that forms riboswitches typically contains a base stem and several branching hairpins.

50. An operon is controlled by a repressor. When the repressor binds to a small molecule, it is released from binding to DNA near the operon. The operon is never expressed if a mutation prevents the repressor from binding to the small molecule. The type of control illustrated is: a. negative inducible. b. negative repressible. c. positive inducible. d. positive repressible. e. attenuation.

Indicate one or more answer choices that best complete the statement or answer the question.

51. In bacteria, which of the following modes of regulation of gene expression are mainly used? (Select all that apply.) a. Alteration of chromatin structure b. Regulation of transcription c. Regulation of mRNA processing d. Regulation of translation e. Regulation of posttranslational modifications 52. What are the benefits of regulation of gene expression to a bacterial cell? (Select all that apply.) a. It is more energy efficient. b. It allows for biochemical flexibility to respond to different environmental conditions. c. It allows for selective expression of a subset of genes in the genome to be expressed at any one time. d. It allows transcription to be spatially separated from translation. e. It allows translation to begin before transcription is complete.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 16_7e 53. Fill in the blanks in the "level of transcription" column of this table with (+) for high levels of transcription, (+/–) for moderate levels of transcription, and (–) for minimal levels of transcription of the lac operon. Consider regulation by both the lac repressor and CAP (catabolite activator protein). The strain is wild type with no partial diploidy. The first line is filled in for reference.

Moderate conditions high glucose, no lactose no glucose, high lactose high glucose, high lactose no glucose, no lactose

Level of transcription –

54. (a) Define the term computational biology. (b) Based on their computational biology work on operons, what did Igoshin and Ray conclude about the reason(s) for the evolution of operons in prokaryotes?

55. Aside from the case of lacOc, where the operator is unable to bind to the repressor protein, briefly describe another mutant that could have caused constitutive expression of lac operon.

56. What is the difference between negative control operons that use induction and systems that use repression?

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 16_7e 57. The bacterium Bacillus subtilis can grow on minimal media with a variety of sugars as a carbon source. One such sugar is mannose, metabolized by the products of the man operon. Expression of the operon is controlled by a regulatory protein encoded in a separate gene, manR. Depending on conditions, the regulatory protein may bind at one of two sites in the operon, as follows: (i) When mannose is absent from the cell, the regulatory protein is in a conformation called R1. R1 can bind specifically at an operator site manO. Binding of R1 at manO reduces transcription of the operon fourfold from a basal level of 20 units. (ii) When mannose is present in the cell, it binds to the regulatory protein, causing it to undergo an allosteric transition from conformation R1 to a new conformation, called R2. R2 cannot bind at manO. However, R2 can bind specifically at a different site called the initiator, manI. Binding of R2 at manI increases transcription of the operon twofold from the basal level.

Mutations m1–m3, which affect expression of this operon, were identified. Each mutation affects only a single component of the operon. Levels of operon activity were measured in haploids. They were also measured in partial diploids with an F′ carrying the wild-type alleles of all genes and regulatory elements described above.

For each mutation, describe which component is affected. In addition, explain the observed activity in the haploid and partial diploid in each case.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 16_7e 58. The following table shows several bacterial strain lac operon genotypes (some are partial diploids). a. Fill in the blanks in the “lactose absent” and “lactose present” columns in this table. (+) means significant levels of active β-galactosidase enzyme can be detected. (–) means no significant levels of active βgalactosidase enzyme are present. The first line is filled in for reference.

Strain genotype 1. lacI+ lacP+ lacO+ lacZ+ lacY+ 2. lacI+ lacP+ lacOc lacZ+ lacY+ 3. lacI+ lacP+ lacOc lacZ– lacY+ 4. lacIs lacP+ lacOc lacZ+ lacY+ 5. lacI– lacP+ lacO+ lacZ+l acY+ / lacI+ 6. lacIs lacP+ lacO+ lacZ+ lacY+ / lac I+

Lactose absent –

Lactose present +

b. Using your answers in the previous table, explain the dominance relationships among the three lacI alleles (lacI+, lacI−, lacIs ).

59. Describe one similarity and one difference in how the trp and lac repressor proteins function.

60. The lysC gene in Bacillus subtilis encodes the first enzyme in the metabolic pathway that generates lysine from its precursor molecule. In the presence of high concentrations of lysine, lysC expression is reduced.

Researchers have generated mutant bacteria that produce the LysC enzyme constitutively, even in the presence of lysine. A mutant identified in this manner had two nucleotide changes in the 5′ UTR of the lysC gene itself. These nucleotide changes are thought to affect the function of a riboswitch regulating transcription of lysC. Describe two possible ways in which the mutations might affect the function of this riboswitch.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 16_7e 61. The dotted line in the following graph shows levels of glucose in a culture of wild-type E. coli grown in Moderate that initially contains both glucose and lactose. The solid line shows levels of transcription of the lac operon. Describe what is happening to the culture and the lac operon, referring to the lac repressor, allolactose, cAMP, and CAP (catabolite activator protein).

62. Fill in the blanks in the following table for the transcription status of the lac operon ((+) for transcription and (–) for no transcription) in the absence or presence of lactose. The first line is filled in for reference.

Strain genotype lacI+ lacP+ lacO+ lacZ+ lacY+ lacI+ lacP+ lacOc lacZ+ lacY+ OR lacI– lacP+ lacO+ lacZ+ lacY+ lacI+ lacP– lacO+ lacZ+ lacY+ OR lacIs lacP– lacO+ lacZ+ lacY+

Lactose absent –

Lactose present +

63. The 5′ UTR of the trp operon RNA contains several UGG tryptophan codons. What would be the effect on transcription-attenuation gene regulation if the trp codons were converted to UGC cysteine codons?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 16_7e 64. Fill in the blanks in this table with “yes” or “no” for each condition of lac operon regulation. The strain is wild type with no partial diploidy. The first line is filled in for reference.

65. An E. coli strain of chromosomal genotype lacI– lacP+ lacO+ lacZ+ lacY+ constitutively expresses the genes of the lac operon. The strain is converted to wild-type lac operon regulation by the addition of an extra piece of DNA. What gene or genes are contained on this extra DNA that explain this conversion to wild type?

Include an explanation of cis- or trans-acting factors and how they work.

66. Fill in the blanks in the following table with "yes" or "no" for each condition of trp operon regulation. The strain is wild type, with no partial diploidy.

Moderate

trp repressor trp repressor Antiterminator bound to trp? bound to hairpin formed in operator? mRNA?

Transcription attenuated?

high tryptophan low tryptophan

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 16_7e 67. Explain why glucose-dependent catabolite repression in E.coli is important and how it is possible to achieve this repression without influencing glucose metabolism.

68. You have isolated two mutations linked to the lac operon, which you designate Lac1– and Lac2– , that cause constitutive expression of the operon. You construct strains carrying a lac operon with a mutant lacY gene on an F′ plasmid. You test both ß-galactosidase activity and Lac permease activity in the strains you constructed using the artificial inducer IPTG.

Are Lac1– and Lac2– dominant or recessive? Do they act in cis or in trans? Indicate how you can determine both of these properties for each mutation. What type of lac mutations best fit the properties of Lac1– and of Lac2– ?

69. Draw a diagram of the transcription and translation of the trp operon under high tryptophan conditions. In addition to the trp operon, show RNA polymerase; the 5′ UTR RNA with regions 1, 2, 3, 4, and the string of uracils; and a ribosome.

70. In the experiments described in the text, Jacob and Monad deciphered that the lacI+ gene product can function in trans and regulate the lac operon either on the plasmid or on the chromosome. What genotype of the partial diploid bacterial strain was key to their experiments and led them to such a conclusion? Explain why this strain helped them to reach that conclusion.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 16_7e 71. Imagine the following scenario: You take the regulatory region of the trp operon (including the promoter, operator, and 5′ UTR) and attach it upstream of the structural genes of the lac operon. You then introduce this artificial construct into a mutant strain in which its own lac operon is completely nonfunctional. Indicate the level of ß-galactosidase activity in each of the following cases and explain why you expect that level of activity: a. No tryptophan, no lactose b. High tryptophan, high lactose

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 16_7e Answer Key 1. a 2. c 3. a 4. a 5. e 6. e 7. e 8. e 9. b 10. c 11. b 12. a 13. c 14. c 15. b 16. c 17. c 18. b 19. e 20. b 21. e 22. b 23. d 24. c 25. d 26. b Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 16_7e 27. b 28. d 29. a 30. b 31. a 32. d 33. e 34. b 35. e 36. a 37. c 38. b 39. d 40. a 41. b 42. a 43. b 44. c 45. a 46. c 47. b 48. e 49. d 50. a 51. b, d 52. a, b, c

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 16_7e 53.

Moderate conditions high glucose, no lactose no glucose, high lactose high glucose, high lactose no glucose, no lactose

Level of transcription – + +/– –

54. a) Computational biology can be called "in silico" biology. It does not involve wet bench experiments but rather complex math, mathematical modelling, programming, and computer algorithms to examine problems in biology. b) Through computational biology, Igoshin and Ray concluded operons evolved to reduce biochemical noise.

(Biochemical noise refers to random fluctuations in gene expression that lead to less than optimal levels of multiple gene products working in a common process.) In an operon, a few genes are transcribed from a single promoter under common regulatory control. Igoshin and Ray studied multiple natural operons in E. coli and found these operons had groupings of genes whose products showed an overall decrease in biochemical noise than expected from groupings of genes at random.

55. A mutation (cis acting) in the promoter sequence of lacI could abolish lacI RNA and protein synthesis. In the absence of lacI protein, lac operon will be constitutively expressed.

56. (1) Induction: Stimulates expression of a gene in response to a specific substrate Repression: Prevents expression of a gene when a specific product is present (2) Induction: Transcription is normally off and is stimulated in response to a specific substrate.

Repression: Transcription is normally on and is shut off in response to a specific product.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 16_7e 57. The basal level of activity is 20 units of activity, but in the absence of the mannose substrate, activity is reduced fourfold, to 5 units. In the presence of mannose, the basal level of activity is doubled to 40 units. In the partial diploid, the repressed level of expression is 10 units (5 units × 2) because there are two copies of the operon, and the induced level is 80 units (40 units × 2). m1: In the haploid, there is neither repression nor activation of the basal level of activity, as indicated by the 20 units of activity both in the presence and absence of substrate. This phenotype is corrected in trans by a transacting factor from the F′. The manR regulatory protein is the only trans-acting factor described in the system.

Because the partial diploid is phenotypically wild type, the added manR protein must be sufficient for normal activity. Therefore, m1 is a mutant of the manR gene, such that the gene product can neither activate nor repress expression, and introduction of a wild-type version restores both activation and repression. m2: In the haploid, repression is normal in the absence of mannose, but there is no activation beyond the basal level with the addition of mannose. In the partial diploid, the defect is not corrected in trans, such that the addition of mannose only allows for a basal 20 units of activity from the chromosomal operon and the full 40 units of activity from the F′ operon. Because the defect is cis-acting, affecting the ability to activate transcription, the mutation is likely in the manI initiator element such that the R2 form of the regulatory protein cannot bind and increase transcription. m3: In the haploid, there is no repression in the absence of mannose, but activation remains normal. In the partial diploid, there is no correction of the defect as levels of uninduced activity appear to be the sum of the regulated F′ (5 units) and the unregulated chromosomal copy (20 units). This mutation is thus in a cis-acting element affecting the ability to repress transcription. Therefore, m3 is a mutation in the manO operator site.

58. a.

Strain genotype 1. lacI+ lacP+ lacO+ lacZ+ lacY+ 2. lacI+ lacP+ lacOc lacZ+ lacY+ 3. lacI+ lacP+ lacOc lacZ– lacY+ 4. lacIs lacP+ lacOc lacZ+ lacY+ 5. lacI– lacP+ lacO+ lacZ+ lacY+ / lacI+ 6. lacIs lacP+ lacO+ lacZ+ lacY+/ lac I+

Lactose absent – + – + – –

Lactose present + + – + + –

b. lacIs is (trans) dominant to both lacI+ and lacI−; lacI+ is (trans) dominant to lacI−. When both lacI+ and lacI

− are present in the cell, we see the lacI+ phenotype (see strain 5). When lacIs is present in the cell, we see the

lacIs phenotype (see strain 6).

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 16_7e 59.

Similarity

Difference

Both bind operator sequences in DNA. When bound, transcription is repressed.

Both are encoded by genes not included in the operons they regulate.

When the lac repressor binds the inducer (lactose), it cannot bind to DNA. When the trp repressor binds the corepressor (tryptophan), it does bind to DNA.

60. The nucleotide changes could change two aspects of the riboswitch: (1) they could disrupt the secondary structure of the mRNA, thereby preventing proper formation of a terminator structure and causing constitutive transcription regardless of whether lysine is present or not; or (2) they could prevent binding of lysine (or another small regulatory molecule) to the mRNA, again preventing terminator formation and allowing constitutive transcription.

61. When glucose and lactose are present, E. coli will prefer to use glucose. When lactose is high, allolactose binds the lac repressor, removing it from the operator and allowing transcription of the lac operon. However, when glucose is high, cAMP is low, so CAP does not activate transcription of the lac operon to its highest levels. As glucose is used up, shown by the dotted line going down, cAMP levels increase, cAMP binds CAP, and transcription of the lac operon is activated to its highest levels.

62.

Strain genotype lacI+ lacP+ lacO+ lacZ+ lacY+ lacI+ lacP+ lacOc lacZ+ lacY+ OR lacI– lacP+ lacO+ lacZ+ lacY+ lacI+ lacP– lacO+ lacZ+ lacY+ OR lacIs lacP– lacO+ lacZ+ lacY+

Lactose absent – +

Lactose present + +

–

–

63. Translation through the 5′ UTR would no longer be dependent on the availability of the amino acid tryptophan.

(However, it would now be dependent on levels of cysteine.) Regardless of levels of tryptophan, translation would continue through the 5′ UTR, the attenuator hairpin would form, and the antiterminator loop would not form. The entire trp operon would not be transcribed (unless levels of cysteine were low).

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 16_7e

64.

65. The extra piece of DNA contains a wild-type lacI+ gene. It encodes a wild-type trans-acting lac repressor that can diffuse throughout the cell and bind to the chromosomal operator of the lac operon and restore wild-type regulation.

66.

Moderate

high tryptophan low tryptophan

trp repressor trp repressor Antiterminator bound to trp? bound to hairpin formed in operator? mRNA? yes yes no no no yes

Transcription attenuated? yes no

67. First, energy conservation is the key aspect that drives catabolite repression. Given the availability of carbon sources, whichever carbon sources will result in the least amount of energy expended to obtain their stored energy as well as any carbon sources that will result in net positive energy will be used by the E. coli. Second, the genes that are essential to utilize glucose are not dependent on the cAMP–CAP complex to promote their transcription.

This nondependency allows glucose to shut off transcription of other operons that are required to utilize alternate carbon sources.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 16_7e 68. Because only the bacterial chromosome has a functional lacY gene and only the F′ has a functional lacZ gene, we can think of permease activity (from lacY+) and ß-galactosidase activity (from lacZ+) as reporters for lac operon expression from the bacterial chromosome and the F′, respectively. That is, permease activity indicates whether the operon on the bacterial chromosome is regulated, constitutive, or uninducible; and ß-galactosidase activity indicates whether the operon on the F′ is regulated, constitutive, or uninducible.

Lac1– is a dominant, cis-acting mutation. It is dominant because permease activity is still constitutive in the partial diploid with one wild-type allele (the F′ should contain a Lac1+ allele) and the chromosomal Lac1– allele. Lac1– is cis-acting because it can only confer constitutive expression on the reporter in cis (on the same piece of DNA, lacY+), while the reporter in trans (on a physically separate piece of DNA, lacZ+) is regulated appropriately.

Lac2– is a dominant, trans-acting mutation. It is dominant because permease activity is still constitutive in the partial diploid with the wild-type allele on the F′ and the chromosomal Lac2– mutation. Lac2– is trans-acting because it can confer constitutive expression on both the reporter in cis (lacY+) and the reporter in trans (lacZ+).

Lac1– best fits the properties of lacOc mutations (cis-acting dominant). A mutation in the operator sequence prevents binding of the LacI repressor, causing constitutive expression, but only for genes on the same chromosome. A trans-acting factor is usually a protein, whereas a cis-acting factor is usually a DNA sequence.

Therefore, the Lac2– mutation appears to be a mutation affecting a protein. Besides the structural genes, the only protein component of the lac operon is the LacI repressor protein. However, most of the time, mutations in lacI are recessive, because a functional copy of lacI on an F′ can substitute for a lacI mutation in the bacterial chromosome. For this allele to be dominant, we have to propose that the mutant form of LacI protein generated by the Lac2– mutation has to actually interfere with the function of the normal LacI protein produced from the F′. A mutant allele that can interfere with the function of a normal allele is typically referred to as a dominant negative, and mutations of this kind that have been identified in the lac operon are known as lacI–d mutations.

69.

70. A partial diploid strain with the genotype lacI+ lacZ– /lacI– lacZ+ was central to their efforts. This strain functioned normally, synthesizing β-galactosidase only when lactose was present. In this strain, the functional lacZ+ gene was not physically linked to the functional lacI+ gene. Despite that, since a lacI+ gene could regulate a lacZ+ gene located on a different DNA molecule, they concluded that a functional lacZ+ can act in trans to regulate the lac operon that is present in the same chromosome.

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 16_7e 71. a. In the artificial construct, the normal regulatory apparatus of the lac operon is completely absent. Instead, the lacZ and lacY genes are now associated with the regulatory proteins and sequences of the trp operon. Therefore, the level of ß-galactosidase activity will be entirely dependent on tryptophan levels and entirely independent of lactose levels. In the absence of tryptophan, the trp repressor protein will be unbound, and the antiterminator loop will form in the 5′ UTR due to ribosome stalling. Therefore, expression will be high, and ß-galactosidase activity will be high. b. When tryptophan levels are high, tryptophan will bind to the trp repressor and act as a corepressor. Therefore, expression will be low. In addition, high tryptophan levels lead to read-through in the 5′ UTR, formation of the attenuation loop, and termination of transcription. As a result, lacZ will not be transcribed, ß-galactosidase will not be generated, and activity levels will be low.

Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 17_7e Indicate the answer choice that best completes the statement or answers the question.

1. A boundary element is also known as a(n): a. insulator. b. repressor. c. enhancer. d. coactivator. e. mediator.

2. Which of the following molecules is capable of targeting chromatin-remodeling complexes to specific DNA sequences to modify chromatin structure and activate gene expression? a. transcriptional repressor b. enhancer c. DNAse I d. transcriptional activator e. histone deacetylase 3. When regions around genes become sensitive to the enzyme DNase I, this is an indication that those regions are: a. becoming transcriptionally active. b. becoming more condensed. c. binding to the single-strand binding proteins. d. destabilizing and transcriptionally inactive. e. becoming highly methylated by a methylase.

4. RITS consists of: a. siRNAs and proteins. b. miRNAs and proteins. c. RISCs and mRNAs. d. methyl groups and histone proteins. e. DNA, histone proteins, and mRNAs.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 17_7e 5. A scientist is using crosslinked chromatin immunoprecipitation (XChIP) to determine the DNA sequences to which a protein of interest binds. However, when she attempts to sequence the DNA in the last step of the procedure, she is unable to obtain any DNA to sequence. Which of the following is a likely reason she is getting these results? a. The antibody she is using to precipitate the protein–DNA complex is not specific enough and binds the protein in question and five other proteins. b. The protein and DNA are not properly crosslinked. c. The antibody she is using is not specific to the DNA sequence to which the protein binds. d. The DNA is too tightly complexed with histone proteins to allow sequencing. e. The cells were lysed during the precipitation stage of the process.

6. Which of the following CANNOT be determined by chromatin immunoprecipitation (ChIP)? a. the types of modification present on the DNA-binding proteins b. the location of modified histones that activate or repress transcription c. the position of transcription factors and associated regulators on the chromosome d. identification of active promoters on a genome-wide level e. the exact amount of protein expressed from a gene 7. siRNAs and miRNAs are produced by the: a. cleavage of RISCs by endonucleases. b. cleavage of functional mRNA within the cytoplasm. c. cleavage of pre-mRNA in the nucleus. d. cutting and processing of double-stranded RNA by Dicer enzymes. e. cutting and processing of double-stranded RNA by Slicer enzymes.

8. Which of the following sequences or molecules is LEAST relevant to the assembly of the basal transcription apparatus for transcription? a. core promoter b. general transcription factors c. TATA box d. RNA polymerase e. enhancer

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 17_7e 9. Which of the following statements about eukaryotic gene regulation is TRUE? a. Transcriptional activator proteins bind to the DNA in a nonspecific manner. b. Eukaryotic enhancers are a part of the basal transcription apparatus. c. The eukaryotic regulatory promoters are highly conserved with the same consensus sequences throughout the genome. d. Mediators are protein complexes involved in regulating transcription rates. e. The transcriptional repressors always bind to the insulator elements.

10. mRNAs are degraded by enzymes called: a. DNAse I. b. ribozymes. c. heat-shock proteins. d. silencers. e. ribonucleases.

11. Which of the following statements about CpG islands is CORRECT? a. CpG islands are commonly found at the 3′ UTR regions. b. The CpG island methylation is universal across both prokaryotes and eukaryotes. c. Methylated CpG islands are associated with long-term gene repression. d. Transcriptionally active DNA has a higher frequency of methylated CpG. e. There is an association between DNA methylation at the CpG island and acetylation of histone via recruitment of acetylases.

12. Which of the following mechanisms does NOT involve siRNA- and miRNA-based gene regulation? a. cleavage of mRNA b. inhibition of translation c. posttranslational modification d. degradation of mRNA e. transcriptional silencing 13. A scientist created transgenic C. elegans worms that expressed a gene that made the worms glow green under fluorescent light. Then she injected double-stranded RNA complementary to this gene. Which of the following results was MOST likely to be seen? a. The worms glowed green under fluorescent light. b. The worms were not able to glow green under fluorescent light. c. The injection of double-stranded RNA was lethal and the worms died. d. The worms had an increased life span. e. The CpG islands near the promoter of the gene became demethylated.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 17_7e 14. A genetics student identified a male fly (Drosophila) that had an XX genotype. Which of the following is the MOST accurate plausible explanation? a. The fly could only have a mutation in its sex-lethal (Sxl) gene. b. The fly could only have a mutation in its transformer (tra) gene. c. The fly could only have a mutation in its double-sex (dsx) gene. d. The fly could have a mutation in either its sex-lethal (Sxl) or transformer (tra) gene. e. The fly could have a mutation in its sex-lethal (Sxl), transformer (tra), or double-sex (dsx) gene.

15. Which of the following statements about histones and gene expression is CORRECT? a. In a general sense, highly condensed DNA bound with histone proteins represses gene expression. b. Acetylation involves the addition of acetyl groups to histone proteins, and it usually results in repression of transcription. c. Addition of methyl groups to the tails of histone proteins always results in activation of transcription. d. Histone code refers to the modification that takes place on the globular domain of the octamer histone core. e. Phosphorylation of cytosines generally leads to increase in transcription..

16. In Arabidopsis, FLD (a deacetylase enzyme) stimulates flowering. Which of the following statements is TRUE? a. FLD deacetylates histones that bind to regions of the FLC gene and stimulates its transcription. b. FLD deacetylates histones surrounding the FLD gene, causing suppression of FLD transcription. c. FLD deacetylates histones that bind to the FLC gene, causing repression of FLC transcription. d. FLD causes repression of FLC translation.

17. When siRNAs are present, the rate of mRNA degradation _____ and the rate of protein production _____. a. increases; increases b. increases; decreases c. decreases; decreases d. decreases; increases e. stays constant; decreases 18. Which of the following eukaryotic gene regulatory mechanisms acts after transcription has been initiated? a. changes in chromatin structure b. changes in transcriptional regulator proteins c. assembly of the basal transcription apparatus d. factors that increase RNA polymerase stalling e. association of transcriptional coactivators

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 17_7e 19. If a deacetylase inhibitor were injected into a mouse brain, which of the following would you expect to occur? a. A general increase in gene expression would occur. b. A general decrease in gene expression would occur. c. A general increase or decrease in gene expression, depending on the gene, would occur. d. Acetyl groups would be removed from histones. e. Methyl groups would be removed from the DNA. 20. Choose the CORRECT order of procedures in performing chromatin immunoprecipitation (ChIP).

1. DNA sequencing and fragment identification 2. Removal of crosslinking and separation of DNA and proteins 3. Crosslinking proteins and chromosomes via UV 4. Antibody incubation for immunoprecipitation 5. Protein degradation via proteases 6. Cell lysis and chromosome fragmentation a. 1, 3, 2, 4, 5, 6 b. 2, 6, 5, 1, 3, 4 c. 4, 3, 2, 5, 1, 5 d. 3, 6, 4, 2, 5, 1 e. 5, 1, 3, 6, 4, 2 21. Which of the following statements about chromatin-remodeling complexes is INCORRECT? a. They can reposition nucleosomes on DNA. b. They can cause conformational changes in DNA. c. They can cause conformational changes in nucleosomes. d. They can alter the chemical structure of histones. e. They can interact directly with transcription factors.

22. Suppose a scientist discovered a mutant strain of C. elegans worms that had lost the ability to regulate gene expression via RNA interference mechanisms. Which of the following might she predict is a possible source of this defect? a. There is a mutation in the gene that encodes the Dicer enzyme that makes it nonfunctional. b. There is a mutation that results in an overexpression of siRNAs. c. There is a mutation in the gene that encodes the Slicer enzyme that allows it to cut RNA more efficiently. d. There is a mutation in a gene that encodes an acetylase enzyme that prevents histone acetylation. e. There is a mutation in a gene that encodes an enzyme that selects the correct splice site for alternative splicing.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 17_7e 23. Proteins that affect the structure of DNA bound to histones without altering histone chemical structure are called: a. RISCs. b. chromatin-remodeling complexes. c. splicing complexes. d. MRE activator proteins. e. dicers.

24. What is the consequence of methylation of DNA sequences called CpG islands? a. active transcription b. transcription repression c. recruitment of chromosome-remodeling complex d. transcriptional stalling e. insulator sequence formation 25. A scientist has discovered a new gene that she believes encodes a transcriptional activator protein. She is trying to determine where this protein binds. Which of the following describes a technique she could use to answer her question? a. Chromatin immunoprecipitation to determine the DNA sequences to which the putative transcriptional activator protein binds b. RNA interference to determine the RNA sequences to which the putative transcriptional activator protein binds c. Crosslinking to determine the interactions the putative transcriptional activator protein has with other proteins d. Chromatin immunoprecipitation to determine the proteins to which the putative transcriptional activator protein binds e. RNA interference combined with crosslinking to determine all DNA and RNA sequences to which the putative transcriptional activator protein binds 26. A fly (Drosophila) with an XY genotype has a mutation in its sex-lethal (Sxl) gene that renders its protein product nonfunctional. Which of the following describes the sex of this fly? a. male b. female c. intersex d. male, but it will be sterile e. female, but it will be sterile

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 17_7e 27. A eukaryotic DNA sequence that activates transcription at distant promoters is called a(n): a. insulator. b. silencer. c. mediator. d. enhancer. e. repressor.

28. A mutation in the gene for the yeast regulatory protein GAL4 causes yeast to grow poorly on galactose. What is the function of GAL4? a. It is a substrate that binds and activates a transcriptional activator. b. It is a product that binds and activates a transcriptional repressor. c. It is a transcriptional activator for the galactose-digesting enzyme gene. d. It is a transcriptional repressor that prevents expression of yeast galactose-digesting enzymes. e. It is an enzyme that metabolizes galactose.

29. Flowering in Arabidopsis thaliania is suppressed by the FLC gene to prevent flowering at the wrong time of the year. At the right time of the year, the gene FLD stimulates flowering by repressing the activity of FLC. What type of genetic pathway is demonstrated in this scenario? a. a positive regulatory pathway b. a negative regulatory pathway c. a signal transduction pathway d. a biosynthetic pathway 30. Which of the following statements about response elements is INCORRECT? a. A single eukaryotic gene is regulated by only one unique response element. b. Response elements are composed of specific consensus sequences that are different from the sequences of other response elements. c. Different genes can possess a common regulatory element upstream of their start site. d. Multiple response elements allow the same gene to be activated by different stimuli. e. Response elements allow complex biochemical responses in eukaryotic cells.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 17_7e 31. Consider the regulation of galactose metabolism through GAL4. Which of the following would result from a mutation that allowed GAL3 to bind to GAL80 in the absence of galactose? a. Transcription of the genes involved in galactose metabolism would occur both in the presence and in the absence of galactose. b. GAL80 would be able to bind to GAL4, and transcription of the genes involved in galactose metabolism would be repressed. c. GAL4 would no longer be able to bind to the DNA; thus, transcription of the genes involved in galactose metabolism would occur. d. GAL80 would no longer be able to stimulate transcription of the genes involved in galactose metabolism. e. There would be no change in the regulation of galactose metabolism because GAL3 normally binds to GAL80 to cause a conformation change in GAL80.

32. Given the following figure, what would be the effect of a mutation that occurred in the insulator that prevented the binding of the insulator-binding protein?

a. Both Enhancer 1 and Enhancer II would be able to stimulate the transcription of Genes A and B. b. Enhancer I would no longer be able to stimulate the transcription of Gene B. c. Enhancer I will be able to hypermethylate Gene B. d. As long as the insulator sequence is not deleted, there will be no change in the transcription regulation of Gene A or B. e. Enhancer II will no longer be able to stimulate the transcription of Gene A. 33. Which of the following statements about regulation of eukaryotic gene expression is INCORRECT? a. The presence of a nuclear membrane separating transcription and translation in eukaryotes led to the evolution of additional mechanisms of gene regulation. b. In eukaryotes, most structural genes are found within operons. c. In eukaryotes, chromatin structure has a greater influence on the regulation of gene expression than in prokaryotes. d. There are a greater diversity of mechanisms for regulation of gene expression in eukaryotes than prokaryotes. e. Posttranslational regulation of histones is unique to eukaryotes.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 17_7e 34. Insulators can block the effects of enhancers only when they lie: a. between an enhancer and a promoter. b. upstream of a promoter. c. adjacent to a promoter. d. within the structural genes. e. within a consensus sequence.

35. Although operons are not common in eukaryotes, eukaryotic genes may be activated by the same stimulus.

Which of the following DNA regulatory sequences makes this coordinated gene expression possible? a. core promoter b. enhancer element c. response element d. boundary element e. silencer element 36. Which of the following events is known to make DNA more sensitive to digestion by DNase I? a. DNA methylation b. deacetylation of histone c. acetylation of histone d. formation of heterochromatin 37. The overall mechanism for degradation of a eukaryotic mRNA is generally initiated by: a. cleavage of the 5′ end. b. random cleavages throughout the mRNA strand. c. shortening of the poly(A) tail. d. recruitment of the chromosome-remodeling complex. e. removal of the 5′ cap.

38. Which of the following processes is also known as RNA silencing or posttranscriptional gene silencing? a. protein degradation b. transcriptional stalling c. RNA splicing d. transcriptional repression e. RNA interference

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 17_7e 39. In which part of the mRNA does degradation generally begin? a. at the 5′ end with the removal of the poly(A) tail b. at the 5′ end with the removal of the methyl cap c. at the 3′ end with the removal of the poly(A) tail d. at the 3′ end with the removal of the methyl cap e. Removal from either end is equally likely.

40. Regulation of gene expression using siRNAs is found in: a. prokaryotes only. b. eukaryotes only. c. prokaryotes and eukaryotes.

41. After translation, eukaryotic proteins can be modified by: a. acetylation. b. the addition of phosphate groups. c. the removal of amino acids. d. the addition of methyl groups. e. acetylation, addition of phosphate and methyl groups, or removal of amino acids.

42. Which type of control is illustrated by GAL4 in the control of genes for yeast galactose-metabolizing enzymes? a. negative inducible b. negative repressible c. positive inducible d. positive repressible 43. What type of cellular activity is known to occur within P bodies? a. active transcription b. active translation c. RNA degradation d. transcriptional stalling e. DNA hypermethylation 44. Alternative splicing is known to be important in the regulation of the: a. production of heat-shock elements. b. sexual development in Drosophila melanogaster. c. lac operon in E. coli. d. metallothionein gene. e. None of the answers is correct.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 17_7e 45. What are four things that influence the stability of eukaryotic mRNAs?

46. Let us assume that your goal is to determine whether or not the expression of your favorite gene (YFG) is regulated by miRNA. Also assume that the genome sequence is known for the organism from which YFG was isolated. Describe the logical experiment steps to achieve your goal.

47. Over the past decade, a significant finding in biology has been the identification of miRNAs and siRNAs and their role in regulating the development of many multicellular organisms. Briefly describe the four different ways these small RNAs influence gene expression.

48. Eukaryotic genes can be introduced into bacteria by recombinant DNA techniques. If the introduced gene encodes a protein that is also found in bacteria—for example, a universally used glycolysis enzyme—then expression of the eukaryotic gene may produce a protein that functions in the bacterial cell.

The mouse gene for a glycolysis enzyme is introduced into an E. coli cell that has a mutant gene for the bacterial version of the same enzyme. Even though the mouse enzyme should function in the bacterial cell and restore the cell's ability to perform glycolysis, it does not.

Provide two possible reasons why this experiment does not work and propose a solution to overcome one of the problems you suggest.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 17_7e 49. What are two distinct functions that transcriptional activator proteins perform in order to regulate gene transcription?

50. What would be the advantage of regulating gene expression at many levels rather than simply regulating at one level (such as at the start of transcription)?

51. In both prokaryotes and eukaryotes, groups of genes can be regulated simultaneously (coordinately expressed). However, each group accomplishes this task differently. Explain how coordinate expression differs in prokaryotes compared with eukaryotes.

52. In the nematode roundworm Caenorhabditis elegans, the LIN-14 protein controls the timing of certain cell divisions during development. LIN-14 protein levels are normally high in early development but decrease in the later stages. In a lin-4 mutant, the level of LIN-14 protein stays high throughout development, changing the pattern of cell divisions in the animal and producing defects in the shape of the animal. The lin-4 gene encodes a microRNA that binds to a sequence in the 3′ UTR of the lin-14 mRNA. a. How does the lin-4 microRNA likely regulate LIN-14 protein levels? Explain why the lin-4 mutant has high levels of LIN-14 throughout development. b. Mutations in the 3′ UTR of lin-14 have been identified that alter the sequence to which lin-4 normally binds. What effect would these mutations be expected to have on the expression of LIN-14 protein in the animal?

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 17_7e 53. We learned in Chapter 17 that GAL3 protein binds to GAL80 and frees up GAL4 to bind to UAS and activate transcription of genes involved in galactose metabolism. Assuming that you did not have prior knowledge of GAL3 protein but knew about GAL4 and GAL80, what experiments would you perform to identify the GAL3 gene and additional genes involved in the regulation of GAL4?

54. Histone methylation can have many different effects on gene expression. In some cases, histone methylation is associated with activation of transcription, whereas in other cases it can trigger the formation of heterochromatin and a decrease in transcription. If histone methylation has been detected in the region of gene X in yeast, describe an experiment that could distinguish whether the methylation is important to activate or repress transcription of gene X.

55. List at least two ways that regulation of genes for yeast galactose-metabolizing enzymes is different from regulation of E. coli lactose-utilizing enzymes.

56. Attenuation occurs in prokaryotes but has not been observed as a gene-expression mechanism in eukaryotes.

Why not?

57. What is the connection between DNA methylation, histone deacetylation, and gene regulation in eukaryotes?

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 17_7e 58. List five levels at which control of gene activity can take place in eukaryotes.

59. Explain how the poly(A)-binding protein that binds to the poly(A) tails located in the 3′end of an mRNA can play a key role in an mRNA degradation pathway that proceeds from the 5′ end of an mRNA in a 5′ → 3′ direction.

60. Describe the unusual posttranscriptional control of Drosophila sex determination. How does the cascade of events differ between males and females?

61. What is RNA crosstalk and how does it regulate gene expression?

62. What is the current hypothesis for how insulators create "neighborhoods" of gene regulation? Comment on their influence on enhancer activity.

63. Which gene-control mechanisms from the previous question are not used in prokaryotes, and why?

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 17_7e 64. What are three ways in which gene regulation is accomplished by modifying the structure of chromatin?

65. Define RNA silencing (or interference). Explain how siRNAs arise and how they potentially affect gene expression. How are siRNAs different from the antisense RNA mechanism?

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 17_7e 66. DNA sequences that might act as enhancers (regulatory elements) can be attached to the gene for green fluorescent protein (GFP), and the combined DNA sequence can be reintroduced into an organism. If the DNA sequence attached to GFP actually can act as an enhancer, glowing green pigment will be observed in particular regions of the organism. Gene A is expressed in developing nerve cells, muscle cells, and intestinal cells in the fruit fly embryo. The region of DNA around gene A is depicted below. The pieces of DNA indicated with lines 1–5 were attached to GFP and tested for the ability to activate gene expression in the fly embryo. In piece 5, the "X" indicates a deletion of a single base pair.

a. Based on the pattern of GFP expression when attached to either piece 1 or piece 2, propose a model for how gene A is activated in the different cells of the embryo. b. Propose an explanation for the difference in GFP expression caused by piece 3 and piece 4. c. What has the deletion "X" in piece 5 likely affected to cause the change of GFP expression compared to piece 2? How does this result refine the model proposed in (a)?

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 17_7e 67. The SUC2 gene in yeast encodes an enzyme to convert the sugar, sucrose, into glucose and fructose, which is necessary for yeast to use sucrose as a source of food. In the presence of glucose, SUC2 expression is switched off. But in the absence of glucose, SUC2 expression increases 100-fold. The expression of SUC2 in wild-type yeast and two mutant yeast strains is shown below. The mutations occur in two genes other than SUC2: SNF1 and SSN6.

Genotype Wild type snf1 ssn6

SUC2 Expression + glucose – glucose 1 100 <1 <1 100 100

a. Is the SNF1 gene normally important for activation or repression of SUC2 expression? b. Is the SSN6 gene normally important for activation or repression of SUC2 expression? c. If SNF1 and SSN6 work in the same pathway to regulate SUC2 expression, the order in which the genes acts can be one of two possibilities (where → indicates activation and — indicates inhibition):

Design an experiment to distinguish between these two possible orders. What would you expect the outcome of your experiment to be in each of the two possible cases?

68. Chimpanzee and human genome sequences are 96% identical. and yet the two species have extensive phenotypic differences. In what types of sequences are those 4% of genomic differences concentrated?

Explain how changes in such sequences can bring about large phenotypic differences.

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 17_7e Answer Key 1. a 2. d 3. a 4. a 5. b 6. e 7. d 8. e 9. d 10. e 11. c 12. c 13. b 14. d 15. a 16. c 17. b 18. d 19. a 20. d 21. d 22. a 23. b 24. b 25. a 26. a Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 17_7e 27. d 28. c 29. b 30. a 31. a 32. a 33. b 34. a 35. c 36. c 37. c 38. e 39. b 40. b 41. e 42. c 43. c 44. b 45.

(1) (2) (3) (4)

the 5′ cap the coding sequence the 3′ UTR the poly(A) tail

46. Isolate the total RNA and separate it on a gel to fractionate out the small RNAs (length: ~22–24 nucleotides).

Sequence the small RNAs. Because the miRNAs bind to target mRNA by virtue of either perfect or near-perfect sequence complementarity, the sequence of YFG could be BLASTed against the isolated miRNA sequences to identify if any one shows complementarity to YFG. One you have identified the miRNA, you could introduce it into the cell and observe the effect on the expression of your gene, by measuring mRNA or protein levels by Northern or Western blot.

47. Small interfering RNAs and microRNAs regulate gene expression through at least four distinct mechanisms: (1) cleavage of mRNA, (2) inhibition of translation, (3) transcriptional silencing, or (4) degradation of mRNA. Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 17_7e 48. (1) The prokaryotic RNA polymerase does not recognize the eukaryotic promoter, so there is no transcription.

Solution: A eukaryotic gene that will be introduced into bacteria for expression must be engineered to have a prokaryotic promoter.

(2) There is no Shine–Dalgarno sequence in the eukaryotic gene promoter, so even if an mRNA is produced, the prokaryotic ribosome will not efficiently recognize the translational start site.

Solution: If a prokaryotic promoter is used to express this gene, it should have the Shine–Dalgarno sequence within it.

(3) Prokaryotic genes do not have introns, so the cells do not have splicing enzymes and cannot remove introns from RNAs. If the mouse gene has introns, they will not be removed after it is transcribed in E. coli. If the RNA is translated, then the additional intron sequences in the unspliced RNA will result in an altered protein with extra amino acids or, if a stop codon just happens to be in the intron sequence, an incomplete protein.

Solution: Expression of the functional protein in bacteria requires engineering a version of the gene that does not have introns.

49. (1) They can bind DNA at a specific sequence.

(2) They can influence transcription rates by interacting with other transcriptional components.

50. It is wasteful to produce mRNAs or proteins that are not needed. Although it is relatively time-consuming for the cell to initiate transcription, translation can be executed rapidly if mRNAs are available. Thus, regulation at levels between transcription and translation may be wasteful of mRNAs but not of proteins, and such regulation poises the cell to produce a protein rapidly when it is needed—for example, see the discussion of T lymphocyte activation. Being able to regulate gene expression at several levels allows great flexibility in producing appropriate levels of proteins as quickly as necessary while stopping unnecessary production of mRNAs or proteins.

51. Prokaryotes have operons, in which structural genes share a single promoter and therefore are expressed together.

Eukaryotes have a variety of methods using response elements. A single response-element regulatory sequence may occur at several different genes, meaning that a single stimulus can increase transcription in all. However, a single gene may also be activated in multiple ways; each eukaryotic gene may be activated by a variety of stimuli, whereas the set of genes in a prokaryotic operon are all regulated by the same stimuli.

52. a. MicroRNAs typically bind to their target mRNA molecules and inhibit the translation of the mRNA into protein.

Since lin-4 microRNA can bind to lin-14 mRNA, it likely acts to prevent translation. We would expect that lin-4 microRNA is only itself expressed later in development, which is why LIN-14 protein levels are high at first but decrease during later stages of development, once lin-4 microRNA is present. In the lin-4 mutant, there is no microRNA to bind to the lin-14 mRNA, so translation occurs throughout development. b. If the target sequence of the lin-4 microRNA is altered, lin-4 will be unable to bind the lin-14 mRNA. Therefore, translation will occur, unimpeded by lin-4, resulting in high levels of LIN-14 throughout development.

This outcome is similar to that observed in lin-4 mutant animals. Thus, either a mutation in lin-4 or a specific mutation in lin-14 that affects the 3′ UTR can both have the same effect on the animal—an abnormally high level of LIN-14 protein, which causes abnormal cell division in late development.

53. A mutagenesis screen could be performed in yeast; you look for mutants that have the following characteristics: (a) galactose metabolism genes are not activated even in the presence of galactose, and (b) there are no mutations in the GAL4 and GAL80 genes. If you had hypothesized the existence of GAL3, then you might also require that, in the mutants, GAL4 and GAL80 constitutively show protein–protein interactions even in the presence of galactose.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 17_7e 54. To determine the role of methylation in regulating transcription of gene X, the simplest experiment is to remove the methylation and determine the consequent effect on gene X expression. It is possible to remove methylation in a number of ways—some pharmaceutical compounds can inhibit methylation enzymes; mutations in the genes that encode methylation enzymes will lower methylation throughout the genome. Regardless of the way in which methylation levels are reduced, the effect on gene X transcription can be measured. If gene X mRNA levels increase when methylation is removed, the histone methylation must normally inhibit gene X transcription. If gene X mRNA levels decrease when methylation is removed, histone methylation must normally activate gene X transcription.

55. (1) Yeast genes are not in an operon; E. coli genes are in an operon.

(2) The yeast GAL genes are mainly controlled by a transcriptional activator; E. coli genes are mainly controlled by a transcriptional repressor.

(3) The yeast GAL genes are under positive inducible control; the E. coli genes are under negative inducible control.

(4) Yeast is eukaryotic, so regulation of any genes, including galactose-utilizing enzymes, could include changes in chromatin structure, posttranscriptional modification of RNA, and RNA stability.

56. Attenuation requires that ribosomes begin translating an mRNA while it is still being transcribed. This may occur in prokaryotes because transcription and translation occur in the same cell compartment, the cytoplasm. In eukaryotes transcription occurs in the nucleus and translation occurs in the cytoplasm.

57. DNA methylation and histone deacetylation are associated with more tightly packaged chromatin, which represses transcription. Methylation of DNA recruits histone deacetylases to the modified region. Acetylation neutralizes positive charges of histones, reducing interaction with DNA, so deacetylation strengthens association of histones with DNA. 58.

(1) (2) (3) (4) (5) (6)

packaging of DNA into chromatin (chromatin structure) and modifications to DNA regulation of transcription processing of RNA transcripts regulation of RNA stability regulation of translation control of the activity of proteins after translation

59. The 5′ cap and the 3′ poly(A) tail of eukaryotic mRNA physically interact with each other, most likely through the bending around of the poly(A) tail so that the PABPs make contact with the 5′ cap. When the poly(A) tail has been shortened below a critical limit, PABP no longer binds to the poly(A) tail and therefore cannot protect the 5′ cap.

After the 5′ cap is removed, nucleases then degrade the mRNA by removing nucleotides from the 5′ to the 3′ end.

60. In response to the X:A ratio of 1:1, females transcribe Sxl (sex-lethal). The Sxl protein allows splicing of RNA from the transformer (tra) gene, so Tra protein is produced. Tra protein (with Tra-2 protein) regulates splicing of RNA from the double-sex (dsx) gene. Protein produced from spliced dsx RNA causes development of female characteristics.

In males, no Sxl RNA is produced, so tra RNA is not spliced and Tra protein is not produced.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 17_7e 61. RNA crosstalk is the interaction and competition between miRNAs, mRNAs, and other RNA molecules.

Competition for binding of miRNAs can lead to regulation of gene expression as can competition for protein binding of RNA molecules that regulate RNA splicing, stability, and translation.

62. Insulators (or boundary elements) may be involved in the formation of topologically associated domains (TADs) of chromatin. Each TAD is spatially separated from another TAD. A given TAD will allow an enhancer to interact with a promoter with the same TAD, but not a different TAD. Thus, an insulator can block the activity of an enhancers in a position-dependent manner.

63. (1) Chromatin structure: Although prokaryotes have proteins associated with their chromosomes, they don't have histones and do not package DNA in the same way that eukaryotes do.

(2) Processing of RNA transcripts: Prokaryotes do not splice introns or add a 5′ cap or poly(A) tail to their mRNA. (3) Also possible: Regulation of RNA stability, because transcripts in prokaryotes are generally more short-lived than in eukaryotes.

64. (1) modification of histone proteins (2) chromatin remodeling (3) DNA methylation 65. RNA silencing, or RNA interference, is a mechanism that inhibits gene expression through the action of small RNA molecules.

How siRNAs arise and affect gene expression: RNA silencing occurs when double-stranded RNA molecules are cleaved and processed to produce small single-stranded interfering RNAs (siRNAs). These siRNAs direct an enzyme complex to complementary sequences in mRNA molecules. The protein–RNA complex cleaves the mRNA in an area bound by the siRNA. Following the initial cleavage, the mRNA is further degraded. The cleavage and subsequent degradation of the mRNA make it unavailable for translation. The siRNAs may also stimulate DNA methylation at DNA sequences complementary to the siRNAs. DNA methylation in the nucleus stimulated by siRNAs also affects transcription.

Antisense RNA also binds to mRNAs, but it physically interferes with translation.

66. a. The DNA region surrounding gene A appears to have at least two enhancer sequences that can regulate expression of gene A. The region downstream (3′, piece 1) of gene A has an enhancer that activates gene A in the developing muscle cells. Enhancers can be located at a distance from the start of transcription and can function independently of orientation. Therefore, it is not uncommon for enhancer sequences to be 3′ of the genes they regulate. Piece 2 contains enhancer sequences that activate gene expression in the nerve cells and the intestinal cells. With the information from just piece 2, it is not possible to tell whether there is one enhancer that works in both tissues or whether there are two separate enhancers that work in each independently. b. The enhancer(s) activating expression in the nerve and intestinal cells are located in piece 4, whereas piece 3 does not appear to contain active enhancer sequences. c. Because GFP expression remains in intestinal cells but is lost in nerve cells, the deletion probably disrupts an enhancer that activates expression in nerve cells. More specifically, the deletion likely changes the binding site for an activator protein, preventing the activator from binding and switching on gene expression. This result also demonstrates that piece 2 probably contains two independent enhancer regions—one for expression in the nerve cells and one for expression in intestinal cells. The deletion has affected one without affecting the other.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 17_7e 67. a. SNF1 is involved in activating SUC2 expression. Because SUC2 is not active when SNF1 is mutated, SNF1 must normally be required to activate SUC2 expression. b. SSN6 is involved in repressing SUC2 expression. Because SUC2 is active even when glucose is present in ssn6 mutant yeast, SSN6 must normally be required to repress SUC2 expression. c. To test the two possible orders the simplest experiment is to generate a yeast strain that is mutant for both genes, an snf1 ssn6 double mutant, and observe the expression of SUC2 in the presence and absence of glucose.

If (1) is the correct order, SUC2 will be switched on (100 units in the table above) in the presence and absence of glucose. In this case, the mutant version of snf1 is unable to repress SSN6; however, because ssn6 is also mutated and cannot repress SUC2, SUC2expression will always be high. If (2) is the correct order, SUC2 will never be switched on (<1 unit in the table above) in the presence and absence of glucose. In this scenario, mutant ssn6 is unable to repress SNF1. But the nonfunctional snf1 is itself unable to switch on SUC2 expression, so SUC2 expression will always be low. Therefore, the phenotype of the snf1 ssn6 double mutant can distinguish between order (1) (SUC2 always on) and order (2) (SUC2 always off).

68. The changes are concentrated in regulatory sequences, which are sequences that affect the expression of other genes. A single regulatory sequence change can affect the expression of several to many other genes and thus affect many phenotypes.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 18_7e Indicate the answer choice that best completes the statement or answers the question.

1. The following nucleotide sequence is found in a short stretch of DNA: 5' TGCC 3' 3' ACGG 5' Suppose a depurination event occurred within the top strand. If this mutant strand were used as a template for replication, what is the MOST likely sequence of the newly synthesized strand? a. 3' AAGG 5' b. 3' TACC 5' c. 3' AGGG 5' d. 3' TTCC 5' e. 3' AGG 5' 2. Assume that a base-pair substitution mutation converts a DNA triplet (AAT) to another DNA triplet (AAA).

A second mutation now changes the AAA triplet to the GAA triplet. (UUA and CUU code for leucine and UUU codes for phenylalanine.) This second mutation is an example of a(n): a. transversion. b. intragenic suppressor. c. loss-of-function mutation. d. intergenic suppressor. e. frameshift.

3. Huntington's disease can strike at an earlier age and bring about a more rapid degeneration and death in successive generations within a family. This phenomenon can be explained by which mechanism? a. presence of a transposable element in the gene b. chronic exposure to mutagens in the environment c. successive expansion of a trinucleotide repeat in the coding sequence of the gene d. presence of an extra chromosome in the germ line e. absence of a gene product that is involved in DNA repair 4. What is the function of DNA glycosylases? a. to recognize and cleave phosphodiester bonds in DNA b. to recognize and remove modified bases from the sugar component of DNA c. to reattach the two parts of DNA that result from double-strand breaks d. to remove pyrimidine dimers from DNA of E. coli that result from exposure to UV light e. to prevent strand slippage during DNA replication

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 18_7e 5. Fragile-X syndrome is an example of a disease caused by what type of mutation? a. nonsense mutation b. frameshift mutation c. expanding nucleotide repeat d. loss-of-function mutation e. gain-of-function mutation 6. What is the consequence of a transversion mutation in duplex DNA? a. A purine is replaced by a pyrimidine, and a pyrimidine is replaced by a purine. b. A base pair is lost within the DNA of a gene, which causes a reading frameshift. c. A purine is replaced by another purine, and a pyrimidine is replaced by another pyrimidine. d. A base pair is added to the DNA within a gene, which causes a reading frameshift. e. The sequence of the DNA remains the same since the change involves proteins.

7. Suppose a research study shows that people who suffer from severe depression are homozygous for a mutation in the hypothetical DEP gene. Individuals without this form of depression have the following sequence at the beginning of the translated region of their DEP genes: 5′-ATG ACG TTT GAA ATT CAG TCT AGA-3′ (Met Thr Phe Glu Ile Gln Ser Arg). Affected individuals have the following sequence: 5′-ATG ACG TTT GAA ATT TAG TCT AGA-3′ (Met Thr Phe Glu Ile STOP). The mutation identified is most likely a: a. missense mutation. b. gain-of-function mutation. c. nonsense mutation d. frameshift mutation e. deletion 8. Ultraviolet light causes what type of DNA lesion? a. large deletions b. deaminated cytosines c. pyrimidine dimers d. mismatched bases e. depurinations 9. Upon transposing to a new site, transposable elements: a. add methyl groups to bases of the surrounding DNA. b. delete about 100 base pairs of DNA on each side of them. c. duplicate their transposase gene. d. express a gene that confers sensitivity to some common antibiotics. e. create a duplication of a target sequence on each side of them.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 18_7e 10. Which of the following is the MOST common transposable element in humans? a. copia b. Alu c. Ac d. Ty e. P 11. What do alkylating agents do? a. They cause pyrimidine dimers. b. They add methyl or ethyl groups to bases. c. They oxidize guanine. d. They deaminate cytosine. e. All of the answers are correct.

12. Which of the following statements about an animal bearing a somatic mutation is TRUE? a. Some, but not all, of the animal's offspring will also carry the mutation. b. All of the animal's offspring will carry the mutation. c. Both the animal and its offspring will show the mutant trait. d. The animal but not its offspring can be affected by the mutation. e. The gametes produced by the animal will all carry the mutation.

13. A new IS element is found in bacteria. Which of the following pairs of DNA sequences would MOST likely be found at each end of the IS element? (Only one of the two DNA strands is given.) a. 5´-GAGACTCTAC-3´ and 5´-GAGACTCTAC-3´ b. 5´-GAGACTCTAC-3´ and 5´-CATCTCAGAG-3´ c. 5´-GAGACTCTAC-3´ and 5´-CTCTGAGATG-3´ d. 5´-GAGACTCTAC-3´ and 5´-GTAGAGTCTC-3´ e. 5´-GAGACTCTAC-3´ and 5´-CAGACTCTAG-3´ 14. In the Ames test, what types of mutations are used to test for chemical mutagens? a. his– to his+ mutations b. pro– to pro+ mutations c. pro+ to pro– mutations d. his+ to his– mutations e. trp+ to trp– mutations

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 18_7e 15. A polypeptide has the following amino acid sequence: Met-Ala-Gln-Arg-Glu-Leu. This polypeptide was mutated to produce the following mutant sequence: Met-Ala-Gln-Thr-Gly-Thr-Ile-Thr. Which describes the MOST likely type of mutation that occurred? a. nonsense mutation b. missense mutation c. frameshift mutation d. in-frame mutation e. neutral mutation 16. A scientist discovers a mutant gene in which a nucleotide was deleted. Which of the following chemicals could potentially reverse this mutation? a. ethidium bromide b. hydroxylamine c. 5-bromouracil d. EMS e. nitrous acid 17. A single base substitution caused the amino acid sequence Met-His-Glu-Cys to be changed to Met-His. Which of the following describes the type of mutation that caused this change? a. transition mutation b. transversion mutation c. translesion mutation d. nucleotide deletion e. strand slippage 18. Assume that you have discovered a new chemical mutagen that modifies guanine so that it mispairs with adenine when adenine is in the template DNA strand during DNA replication. However, this mispairing is limited to when the modified guanine is being added to the newly replicating DNA strand. When the modified guanine is in the template DNA strand, it always pairs normally with cytosine being added to the growing newly synthesized strand. What type of mutation would you predict would be caused by the new chemical mutagen? a. A-to-G base substitutions b. A-to-C base substitutions c. A-to-T base substitutions d. A-to-G and A-to-C base substitutions e. G-to-T base substitutions

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 18_7e 19. Which of the following DNA repair systems does NOT involve the activity of a DNA polymerase? a. mismatch repair in humans b. nucleotide-excision repair in yeast c. photoreactivation in E. coli d. base-excision repair in E. coli e. All of choices provided involve the activity of a DNA polymerase.

20. Which of the following statements describes the possible parasitic nature of transposable elements? a. Transposable elements can increase in number within genomes without providing an advantage to the host. b. Transposable elements are collected within their genomes by host organisms so that the host will benefit but not the transposable elements. c. Transposable elements will provide an evolutionary advantage to host organisms by transposing as often as possible. d. Transposable elements will enhance their expression of transposase so that the hosts can evolve more quickly. e. Transposable elements will add methyl groups to their own DNA to reduce their own rate of transposition.

21. Which of the following mutagens is MOST likely to cause a frameshift mutation? a. base analog b. alkylating agent c. intercalating agent d. ionizing radiation e. UV light 22. Suppose that you identify a mutation in a gene caused by a single base substitution. Which of the following would be the BEST choice to use in an attempt to reverse this mutation? a. UV light b. ethidium bromide c. hydroxylamine d. acridine orange e. EMS

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 18_7e 23. Which of the following is characteristic of retrotransposons? a. They use transposase to transpose to new sites. b. They have inverted repeats at each of their ends. c. They transpose through an RNA intermediate. d. They make transposase. e. They are found only in prokaryotes.

24. Which of the following characterizes the mode of transposition of retrotransposons? a. It involves an RNA intermediate. b. It involves the initial synthesis of transposase. c. It involves the production of a protein repressor. d. It only occurs in nondividing host genomes. e. It requires inverted repeats at each end of the retrotransposon.

25. Which of the following kinds of mutations is MOST likely to be a null loss-of-function mutation? a. transition b. transversion c. frameshift d. missense e. induced 26. An example of a genetic disorder in humans that results from a loss-of-function mutation is: a. cystic fibrosis. b. achondroplasia. c. Huntington's disease. d. myotonic dystrophy. e. No correct answer is provided.

27. Consider two theoretical transposable elements in yeast, A and B. Each contains an intron, and each is transposed to a new location in the yeast genome. Suppose you then examine the transposons for the presence of the intron. In the new locations, you find that A has no intron but B does. From these facts, what can you conclude about the mechanisms of transposition for the two transposable elements? a. B probably makes a transposase. b. A probably has inverted repeats at each end of the element. c. B probably uses RNA as an intermediate in the transposition event. d. B probably makes a reverse transcriptase. e. A probably doesn't create a duplication of the host genome target sequence.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 18_7e 28. A _____ mutation changes a codon that specifies an amino acid into one that terminates translation. a. missense b. nonsense c. silent d. neutral e. reverse 29. The type of mutation that reverses the effects of a frameshift mutation without changing the frameshift and occurs with the same gene as the frameshift is called a(n): a. intergenic suppressor mutation. b. nonsense mutation. c. missense mutation. d. intragenic suppressor mutation. e. silent mutation.

30. Achondroplasia is a form of dwarfism that is inherited in humans as an autosomal dominant disorder. A survey in a small country showed that, within a two-year period, there were 12 children with normal parents born with this disorder out of a total of 420,000 births. What is the mutation rate in mutations/locus/generation? a. 1.4 × 10–5 b. 2.8 × 10–6 c. 2.5 × 10–5 d. 2.8 × 10–5 e. 7.4 × 10–6 31. Which of the following types of mutations does NOT lead to a change in the amino acid sequence of the gene product? a. missense mutation b. nonsense mutation c. neutral mutation d. silent mutation e. loss-of-function mutation 32. Bacterial insertion sequences encode which of the following gene products? a. reverse transcriptase b. DNA polymerase c. transposase d. repressor e. insertase Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 18_7e 33. The mutation shown in the diagram below can BEST be described as a _____ mutation.

a. missense b. nonsense c. silent d. neutral e. reverse 34. _____ mutations produce new activities and are usually dominant. a. Induced b. Spontaneous c. Forward d. Gain-of-function e. Lethal 35. A polypeptide has the following amino acid sequence: Met-Ala-Gln-Arg-Glu-Leu. This polypeptide was mutated to produce the following mutant sequence: Met-Ala. Which describes the MOST likely type of mutation that occurred? a. nonsense mutation b. missense mutation c. frameshift mutation d. in-frame mutation e. neutral mutation 36. Which of the following is a form of direct DNA repair? a. base-excision repair b. nucleotide-excision repair c. homologous recombination d. mismatch repair e. photoreactivation

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 18_7e 37. Which of the following statements about somatic mutations is FALSE? a. Some may give rise to cancers in humans and other animals. b. They may be inherited by daughter cells after cell division. c. They may result in inactive gene products of the mutated genes. d. They may result from both frameshift and base-pair substitution mutations. e. They may be inherited in the offspring of mutated individuals.

38. Composite transposons in bacteria are characterized by which of the following? a. They are composites of Ac and Ds elements. b. Each has the ability to transpose either by replicative transposition or nonreplicative transposition. c. They contain all deleted transposase genes. d. They contain more than one gene. e. No correct answer is provided.

39. Which of the following statements about chromosome 9 open reading frame 72 (C9orf72) and its association with amyotrophic lateral sclerosis (ALS) is FALSE? a. Expansion of the repeat GGGGCC in C9orf72 is observed in patients with ALS. b. mRNAs are transcribed from both strands of DNA of C9orf72 containing GGGGCC repeats. c. Proteins are translated from mRNAs GGGGCC repeats in the absence of a start codon. d. Proteins are translated from mRNAs with GGGGCC repeats in all three frames. e. Proteins translated from mRNAs with expanded GGGGCC repeats are primarily toxic to muscle cells.

40. Which of the following pairs of DNA repair systems will repair pyrimidine dimers in E. coli? a. mismatch repair and base-excision repair b. photoreactivation and mismatch repair c. nucleotide-excision repair and base-excision repair d. photoreactivation OR nucleotide-excision repair e. nonhomologous end joining and nucleotide-excision repair 41. Which of the following transposable elements are flanked by direct repeats of a short portion of the host genome? a. Tn10 b. L1 c. Activator (Ac) d. Alu e. All of the answers are correct.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 18_7e 42. Which of the following statements BEST describes the Activator (Ac) and Dissociation (Ds) transposable elements in maize (corn)? a. Ac elements cannot transpose unless a Ds element is present. b. Ac contains a functional transposase gene; Ds lacks a functional transposase gene. c. Ds contains a functional transposase gene; Ac lacks a functional transposase gene. d. Both Ac and Ds have functional transposase genes. e. Neither Ac nor Ds contains functional transposase genes.

43. Which of the following is required for an IS element in E. coli to be able to transpose? a. a gene for reverse transcriptase and long terminal repeats b. a gene for reverse transcriptase and inverted repeats c. a gene for transposase and inverted repeats d. a gene for transposase and long terminal repeats e. a gene for DNA polymerase and long terminal repeats 44. The disorder xeroderma pigmentosum is associated with a defect in what type of DNA repair system? a. mismatch repair b. base-excision repair c. photoreactivation d. nucleotide-excision repair e. homologous recombination 45. Insertion or removal of one or more nucleotide base pairs in DNA within a gene often results in a _____ mutation. a. transition b. frameshift c. reversion d. transversion e. suppressor 46. A mutation that changes a GC base pair to AT is a(n): a. transition. b. transversion. c. induced mutation. d. missense mutation. e. synonymous mutation.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 18_7e 47. Transposable elements are found: a. mainly in higher plants. b. mainly in animals, particularly in mammals. c. mainly in eukaryotes. d. mainly in prokaryotes. e. in practically all organisms.

48. Which of the following transposable elements have indirect repeats at each end? a. Ac in maize b. Ty in yeast c. copia in Drosophila d. Alu in humans e. All of the answers are correct.

49. Which of the following pairs of sequences would you expect to be found in the same transposable element? a. inverted repeats and a gene for transposase b. long terminal repeats and a gene for transposase c. inverted repeats and a gene for reverse transcriptase d. a gene for transposase and a gene for reverse transcriptase e. both long terminal repeats and a gene for transposase and inverted repeats and a gene for reverse transcriptase 50. A transposable element is found to use RNA as an intermediate in transposition. On the basis of this information, which of the following would you expect to be CORRECT? a. The transposable element also probably makes transposase. b. The transposable element may encode a reverse transcriptase. c. The transposable element is probably located in a bacterial genome. d. The transposable element probably contains inverted repeats at each end. e. The transposable element will not be able to transpose without a second copy also present in the genome.

51. Which of the following statements CORRECTLY describes nonsense mutations? a. They cause a nonfunctional amino acid to replace a functional amino acid. b. They change the nucleotide sequence of a gene but do not change the sequence of the resulting protein. c. They result in the insertion or deletion of a small number of nucleotides to the DNA. d. They convert a codon for a particular amino acid within a gene into a stop codon. e. They cannot revert to wild type.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 18_7e 52. Strains of E. coli that are defective in mismatch repair have very high levels of spontaneous mutagenesis.

Studies have shown that in wild-type strains A:C and G:T mispairings (as opposed to the normal A:T and G:C pairings) that occur during DNA replication are more likely than other mispairings (A:G, etc.) to be detected and repaired in these mismatch repair-proficient, wild-type strains. Which of the following types of base substitutions would you expect to be most common among the spectrum of spontaneous mutations created by the mutant mismatch repair-defective strains? a. A-to-C b. C-to-A c. G-to-C d. A-to-G e. A-to-T 53. Transposition can involve exchange of DNA sequences and recombination, which often leads to DNA: a. acetylation. b. rearrangements. c. condensation. d. repair. e. replication.

54. Practically all transposable elements that have been studied are associated with which of the following? a. indirect repeats at each end b. a gene for transposase c. a gene for reverse transcriptase d. a gene for RNA polymerase e. flanking direct repeats 55. A geneticist is studying a mutation in a population of turtles that causes their shells to become extremely brittle. She determines the mutation is caused by the loss of two nucleotides in the coding region of a gene.

Upon studying the mutant protein that is produced, she observes that it is 312 amino acids in length, as compared to the normal protein that is 588 amino acids in length. This mutant protein can no longer carry out its normal function of assisting in the stiffening of a turtle's shell. Which of the following could NOT describe this mutation? a. deletion b. loss-of-function mutation c. transversion d. frameshift mutation e. nonsense mutation

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 18_7e 56. It is estimated that transposable elements compose approximately what percentage of the human genome? a. less than 1% b. between 1% and 10% c. between 11% and 40% d. between 41% and 75% e. greater than 75% 57. A codon that specifies the amino acid Ile undergoes a single-base substitution to become the amino acid Met. Which of the following describes the type of mutation that must have occurred? a. a transition mutation at the first position of the codon b. a transversion mutation at the first position of the codon c. a transition mutation at the third position of the codon d. a transversion mutation at the third position of the codon e. a transition mutation or a transversion mutation at the third position of the codon 58. Helen has type I osteogenesis imperfecta (OI), a genetic skeletal disorder. Shown below is her DNA sequence for a portion of the coding region of the collagen type I gene, which contains the mutation responsible for her disorder. The corresponding wild-type sequence is shown also (only one DNA strand is shown in each case).

What type of mutation does Helen carry? a. missense b. nonsense c. silent d. deletion e. frameshift 59. A polypeptide has the following amino acid sequence: Met-Ala-Gln-Arg-Glu-Leu. This polypeptide was mutated to produce the following mutant sequence: Met-Ala-Gln-Gly-Glu-Leu. Which describes the MOST likely type of mutation that occurred? a. nonsense mutation b. missense mutation c. frameshift mutation d. in-frame mutation e. neutral mutation

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 18_7e 60. Assume that a mistake is made during DNA replication in a bacterium and a G is inserted into the newly synthesized DNA strand opposite a T in the template DNA strand. If this mistake is not repaired before the next round of DNA replication, what mutation will eventually result? a. A-to-G base substitution b. A-to-C base substitution c. A-to-T base substitution d. G-to-A base substitution e. C-to-A base substitution 61. Which of the following base changes in DNA is an example of a transition? a. A to C b. G to C c. C to A d. A to G e. A to T 62. Hybrid dysgenesis in Drosophila occurs: a. when a male and a female, each carrying a copia element, mate and produce offspring that have numerous mutations. b. in the offspring of a cross between a male that carries a copia element and a female that carries a P element. c. in the offspring of a cross between a male that carries a Ty element and a female that carries an Ac element. d. in the offspring of a cross between a male that carries a P element and a female that does not carry a P element. e. in the offspring of a cross between a male that carries an Alu element and a female that carries a Ty element.

63. Tumor suppressor proteins can assist in slowing down the cell cycle under appropriate conditions. In humans, the TP53 gene encodes a tumor suppressor called p53. Most mutations in the TP53 gene result in a mutant form of p53 that can no longer function to slow down the cell cycle, which can lead to a cell becoming cancerous. However, some mutant forms of p53 actually possess the ability to increase a cell's resistance to anticancer treatments. Which of the following BEST describes the latter type of mutation? a. loss-of-function mutation b. gain-of-function mutation c. reverse mutation d. suppressor mutation e. neutral mutation

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 18_7e 64. How do germ-line mutations differ from somatic mutations? a. Germ-line mutations involve small changes to DNA such as base-pair substitutions, while somatic mutations usually involve large deletions. b. Germ-line mutations occur during DNA replication, while somatic mutations do not. c. Germ-line mutations can be passed on to offspring, while somatic mutations cannot. d. Germ-line mutations are reversible, while somatic mutations are not. e. Germ-line mutations result in cancers, while somatic mutations do not.

65. Which of the following proteins does NOT play a role in nucleotide-excision repair? a. DNA polymerase b. DNA ligase c. DNA glycosylase d. DNA helicase e. single-strand-binding proteins Indicate one or more answer choices that best complete the statement or answer the question.

66. Which of the following enzyme activities play a role in nucleotide-excision repair? (Select all that apply.) a. DNA polymerase b. DNA ligase c. reverse transcriptase d. DNA helicase e. primase 67. Which of the following statements about silent mutations are FALSE? (Select all that apply.) a. Silent mutations arise due to the redundancy of the genetic code. b. Silent mutations never have phenotypic effects. c. Silent mutations do not change the identity of an amino acid. d. Silent mutations may alter binding sites for regulatory proteins. e. Silent mutations are typically gain-of-function mutations.

68. What is the difference between a mutagen and a carcinogen?

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 18_7e 69. (a) List the two major pathways in eukaryotes for repairing double-stranded breaks in DNA molecules. (b) Explain the need for each pathway.

70. A codon for the amino acid serine undergoes a transversion so that it now codes for threonine. A transversion at a different site in the same codon then suppresses the first mutation. Give the nucleotides in the original codon, the transition mutation, and the transversion suppressor. (Note: There are two possible answers.)

71. Explain how an individual with a suppressor mutation can be a double mutant but express a near normal phenotype.

72. To determine whether human exposure to radiation results in an increase in recessive mutations in the germ line, scientists can examine the sex ratio of children born to parents exposed to higher than normal radiation levels. Explain how and why the sex ratio might be affected by radiation exposure.

73. What are the differences between neutral mutations and silent mutations?

74. Explain how transposable elements might play a role in studying gene function.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 18_7e 75. Assume you are a geneticist and you're asked to examine the effects of a radiation leak that occurred at a facility in Iraq several years ago. How would you assess the level of mutations caused by this leak?

76. Explain how bacterial resistance to antibiotics can be efficiently transmitted by transposons.

77. You are working on an insulin-binding protein from fish. The beginning of the coding sequence of the gene is shown below. You find a mutant in the gene that cannot bind insulin (also shown below—the mutation is set in boldface type). Among a population of fish having the gene for the mutant protein, you find one that produces a variant of this protein that can now bind insulin again (DNA sequence also shown below). What kind of mutation is this new variant? (Use a genetic rather than a biochemical classification.)

Original sequence: atgttgtcctatgtgagttgcggcttgttg Mutant sequence: atg g tgtcctatgtgagttgcggcttgttg Variant sequence: atg g tgtcctatgtgagttgcggcttg g tg

78. Most transposable elements are flanked by direct DNA repeats. What is the significance of these direct repeats?

79. How do insertion sequences and composite transposons differ?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 18_7e 80. How can spontaneous mutations arise?

81. How do mutation rates differ among eukaryotic and prokaryotic organisms? What factors affect mutation rates?

82. Reversions, or reverse mutations (i.e., mutations that restore the wild-type phenotype initially lost by an earlier mutation), can be either of two types: back mutations or suppressor mutations. Explain how you could distinguish between these two different types of reversions using a test cross.

83. The following sequence represents the DNA template strand of a gene:

a. Give the sequence of the mRNA transcribed from this DNA strand and, using the genetic code provided in Figure 15.12, give the amino acid sequence of the protein it encodes. b. Give the amino acid sequence of the protein encoded by this sequence after the following mutations have occurred: • • • •

a transition at nucleotide 10 a transition at nucleotide 12 a transition at nucleotide 7 an insertion of a G immediately following nucleotide 10

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 18_7e 84. Why do disruptive DNA lesions, like deletions and insertions, sometimes not lead to frameshift mutations?

85. The mutagen EMS converts guanine (G) to O-6-ethylguanine (G*). O-6-ethylguanine (G*) forms base pairs with thymine (T) instead of cytosine (C). Suppose that exposure to EMS damages a DNA molecule as shown below:

a. Diagram the steps required for the incorporated G* to induce a stably inherited mutation. Your diagram should include all necessary rounds of replication. b. Characterize the mutation induced by EMS as a transition, transversion, or frameshift.

86. Transposable elements that transpose through an RNA intermediate are retrotransposons. There are two types of retrotransposons: those that have direct repeats at each end, often called long terminal repeats (LTRs), and those that do not have these repeats. Pick an example of each type of retrotransposon and give (1) its basic structure and (2) its possible evolutionary history.

87. Using diagrams, compare and contrast an insertion sequence with a composite transposon. Label the important features of both.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 18_7e 88. Why don't transposable elements that move through replicative transposition eventually take over the genome completely?

89. During mismatch repair, it is essential for the cell to be able to distinguish between the old template DNA strand and the newly synthesized DNA strand in order to make the appropriate correction of a mispairing created during DNA replication. How is this accomplished in E. coli?

90. How will the result of strand slippage on the newly synthesized strand differ from the result of strand slippage on the template strand?

91. In the following section of DNA, the arrows indicate transposase cleavage sites. Use diagrams to precisely describe and indicate how target site duplications are generated when transposable elements insert into DNA.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 18_7e 92. A diploid fungal cell is homozygous for a (TTG)5 trinucleotide repeat at a particular locus (see sequence below). During meiosis, an unequal crossover occurs at this locus between the second and third repeats on one homolog and between the third and fourth on the other. How many TTG repeats will each of the meiotic products have? (Assume that this species makes tetrads.)

ATGTTGTTGTTGTTGTTGTGA

93. Explain how UV light induces mutations in E. coli.

94. In a eukaryotic cell, two different transposable elements, each containing an intron, excise and insert in different locations within the genome. After transposition, the transposable elements are sequenced at their new sites. Explain the following sequencing results: a. Neither of the translocated transposable elements contains any intron sequences. b. One of the transposable elements contains the intron, but the other transposable element does not.

95. List all single-base substitutions that would change a codon for Leu to a nonsense codon (see Figure 15.12).

For each, indicate whether it would be a transition or transversion.

96. Why do insertions and deletions often have more drastic phenotypic effects than do base substitutions?

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 18_7e 97. What would be the result of a large deletion in the resolvase gene of a transposable element?

98. A company has invented a new low-calorie sugar substitute and wants to determine if the substitute might be carcinogenic, so researchers use it in the Ames test. The results show no increase in mutant bacterial colonies. They then perform feeding experiments in laboratory rats and find a significant increase in the incidence of cancer. Offer an explanation for why the Ames test did not accurately predict the carcinogenic potential of the sugar substitute and suggest a solution to the problem.

99. How do intergenic and intragenic suppressor mutations differ?

100. Do you believe that transposable elements (TEs) are best described as "parasitic" DNA sequences, simply using the host genome as a means to replicate and spread and providing no benefit for the host, or do you believe that they provide some adaptive advantage for the host genomes that house them? Explain. If you believe that TEs are advantageous to the host, give some specific examples to support this conclusion.

101. A DNA sequence encodes a protein with the amino acid sequence Met-Leu-Ser-Ile-Met-Ala. A mutation occurs in the DNA sequence so that now it encodes a protein with the amino acid sequence Met-Leu-Val. a. Propose an explanation for the type of mutation that produces the new amino acid sequence. b. Give an example of a second mutation that would produce the following amino acid sequence: Met-LeuVal-Ile-Met-Ala.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 18_7e 102. Explain how mutations can be both beneficial and harmful in biology.

103. Give the inverted repeat of the following sequences: a. 5′-ATCCGCT-3′ 3′-TAGGCGA-5′ b. 5′-AAATTT-3′ 3′-TTTAAA-5′ c. 5′-GGAATTCC-3′ 3′-CCTTAAGG-5′

104. What is the difference between a missense mutation and a nonsense mutation?

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 18_7e Answer Key 1. a 2. b 3. c 4. b 5. c 6. a 7. c 8. c 9. e 10. b 11. b 12. d 13. d 14. a 15. c 16. a 17. b 18. b 19. c 20. a 21. c 22. e 23. c 24. a 25. c 26. a Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 18_7e 27. a 28. b 29. d 30. a 31. d 32. c 33. a 34. d 35. a 36. e 37. e 38. d 39. e 40. d 41. e 42. b 43. c 44. d 45. b 46. a 47. e 48. a 49. a 50. b 51. d 52. d 53. b 54. e Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 18_7e 55. c 56. d 57. e 58. e 59. b 60. a 61. d 62. d 63. b 64. c 65. c 66. a, b, d 67. b, e 68. A mutagen is any agent that causes an increase in the mutation rate. A carcinogen is any chemical or physical agent that causes cancer. Some agents are both a mutagen and a carcinogen, and others are one but not the other.

69. (a) (b)

The two major pathways are homologous recombination and nonhomologous end joining.

Homologous recombination repairs a broken DNA molecule by using information in an identical or nearly identical molecule, usually a sister chromatid, when available. Homologous recombination generally puts the correct broken ends together precisely and restores the original molecule.

Nonhomologous end joining repairs double-stranded breaks without using a homologous template, usually when a sister chromatid is not available (as during G1 of the cell cycle). Nonhomologous end joining often joins broken pieces together that are not normally joined together, leading to deletions, insertions, and translocations.

Homologous recombination is more likely to repair the breaks correctly but does not operate during G1 because sister chromatids are not yet available. Nonhomologous end joining is able to operate during G1, but it is less likely to repair breaks correctly.

70. AGC (serine) → ACC (threonine) → UCC (serine) AGU (serine) → ACU (threonine) → UCU (serine) 71. A suppressor mutation is a mutation that occurs at a different site from an original mutation but acts to suppress the original mutation and restore at least some of the wild-type phenotype. Suppressor mutations partially cancel out the effects of a previous mutation.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 18_7e 72. The important mutations here are those in the germ line of the parents because these are the ones than can affect the sex ratio of their children. The parents may suffer somatic mutations also, which can affect their own health, but these are not expected to affect the sex ratio of their children.

Many of the mutations randomly induced by radiation will be recessive lethals. Some of these lethal mutations will be in genes located on the X chromosome. Even recessive lethal mutations will be expressed in males because the males are hemizygous for X-linked genes. Females will not express X-linked recessive lethal alleles unless they are homozygous, which is unlikely and rare. Therefore, an increase in the production of X-linked recessive lethal mutations due to radiation should result in a decrease in the ratio of males to females.

73. Neutral mutations are changes in DNA sequence that alter the amino acid coding sequence of a polypeptide but do not significantly change its biological function. Silent mutations, on the other hand, are changes in DNA sequence that produce synonymous codons specifying the same amino acid as the original, nonmutated sequence.

74. Because some transposable elements insert randomly throughout the genome and because the element has a known sequence, researchers can use the elements as a way to tag a gene in which that element has caused a disruption in that gene's function.

75. Compare the rates of cancer, gene mutations, chromosome abnormalities, and birth defects within a control (i.e., unexposed) population with these rates in the population in the area exposed to the radiation. Somatic mutations could be examined by studying cancer rates among survivors of the radiation leak. Germ-line mutations could be assessed by examining the levels of birth defects, chromosome abnormalities, and gene mutations in children born to people who had been exposed to radiation.

76. Composite transposons can carry antibiotic-resistance genes and can be transferred via conjugation, transduction (phage infection), and transformation. Transposable elements can also excise and insert into plasmid elements, which can be subsequently transferred by various means to different cells. Transposable elements can undergo homologous recombination with plasmids or other DNA sites within a cell containing the same transposable elements sequence and be transferred via conjugation, transduction, and so forth. Under the selective pressure antibiotic treatment, resistance will, of course, be highly advantageous to those bacterial cells containing resistance genes, and thus antibiotic resistance can spread very rapidly in bacterial populations.

77. The mutation in the new variant restores the original function of the gene by making a change in that gene; therefore, the mutation is either a revertant (reverse mutation) or it is an intragenic suppressor. Because the original mutation is still in the new variant as well as a new mutation, the new mutation must be an intragenic suppressor.

78. The direct repeats indicate that staggered cuts are made in the target DNA sequence when transposable genetic elements insert themselves. The direct repeats are a consequence of repairing the staggered cuts after insertion.

79. (1) Insertion sequences contain genes required for transposition (e.g., transposase) and reinsertion flanked by inverted terminal repeats. Insertion sequences only carry genes required for transposition and reinsertion and are, thus, relatively short sequences.

(2) Composite transposons are complex transposons that are generally any sequence of DNA flanked by two insertion sequences. Composite transposons can be up to thousands of base pairs long and carry various genes, such as antibiotic-resistance genes. One or both of the insertion sequences contain the transposase that enables the transposon to excise and reintegrate elsewhere within the genome.

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 18_7e 80. They can arise through natural consequences stemming from the chemical nature of DNA (e.g., tautomeric shifts, depurination, deamination, etc.). They can also arise from random mistakes that take place during normal processes, such as replication, crossing over, cell division, retrotransposon activity within genomes, wobble mispairing, and so forth.

81. Typical mutation rates for viral and bacterial genes are about 1 to 100 mutations per 10 billion cells (10–8 to 10–10).

The mutation rates for most eukaryotic genes are higher, about 1 to 10 mutations per million gametes (10–5 to 10–6). Mutation rates for eukaryotes are probably higher because the rates are calculated per gamete, and several cell divisions are involved in the production of a gamete, whereas mutation rates in prokaryotic cells and viruses are calculated per cell division. Mutation rates for different organisms vary considerably, and they probably reflect differing abilities to repair mutations, unequal exposures to mutagens, or biological differences in rates of spontaneously arising mutations. Mutation rates can be affected by several factors, including the following: Mutation rates can be affected by the frequency with which primary changes occur to DNA molecules.

Some mutations (e.g., mutations in genes that encode components of the replication machinery or DNArepair enzymes) can increase the overall rate of mutation of other genes.

Mutation rates can be affected by the ability to recognize mutations. For example, some mutations may appear to arise at a higher rate simply because they are easier to detect.

Mutation rates can be affected by the efficiency of DNA-repair systems (i.e., the probability that once a mutation occurs, it will be repaired).

Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 18_7e 82. If the reversion is a back mutation, then all the backcross progeny will be wild type; if the reversion is a suppressor mutation, then some backcross progeny will exhibit the mutant phenotype.

83. a. b.

RNA: 5′-AUG GCA CAG AGG GAA CUA UAA CUUGAU-3′ 5′-AUG GCA CAG GGG GAA CUA UAA CUU GAU-3′ → Met Ala Gln Gly Glu Leu STOP 5′-AUG GCA CAG AGA GAA CUA UAA CUU GAU-3′ → Met Ala Gln Arg Glu Leu STOP (no amino acid change) 5′-AUG GCA UAG AGG GAA CUA UAA CUU GAU-3′ → Met Ala STOP 5′-AUG GCA CAG ACG GGA ACU AUA ACU UGA U-3′ → Met Ala Gln Thr Gly Thr Ile Thr STOP

Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 18_7e 84. Because amino acid codons consist of three nucleotides, insertions and deletions that involve any multiple of three nucleotides will leave the reading frame intact. However, the addition or removal of one or more amino acids may still affect the phenotype. Insertions or deletions that do not alter the translation reading frame are called in-frame insertions and deletions.

85. a. At least two rounds of replication will be needed, as shown below.

b. The mutation is a CG to TA transition.

86. A prominent example of the first type of retrotransposon is the Ty element in yeast. It is about 6300 nucleotides in size and contains direct repeats, called deltas, of about 330 nucleotides. It also contains homologous genes to the gag and pol genes of retroviruses. The pol gene encodes a reverse transcriptase, which it can use for transposition.

Ty is clearly evolutionarily related to retroviruses, which also contain direct repeats (or LTRs) and gag and pol genes. In addition, most retroviruses contain an env gene (Chapter 8), which Ty doesn't have. Whether Ty evolved from a retrovirus or vice versa or whether they both evolved from some other entity is not known.

The copia element of Drosophila could also be used as an example of this class.

An example of the second type of retrotransposon would be the Alu element in primates, including humans. Alu is about 300 nucleotides in size, although many copies of Alu are smaller than this. There are neither direct repeats nor pol or gag genes. Therefore, Alu elements must utilize some other source of reverse transcriptase in the cell to facilitate transposition. There are over a million copies of the Alu element in humans, scattered throughout the genome as a prominent SINE (short interspersed element).

The Alu sequence seems to be related to the 7SL RNA gene, which has an internal promoter, and Alu has been referred to as a "pseudogene." Presumably, transcription of the Alu sequence leads to an RNA that can sometimes be converted to a DNA sequence with the ability to integrate throughout the genome. Presently, it appears that few Alu copies in humans are active in transposition, but some are, as evidenced by genetic disorders caused by an inappropriate integration of an Alu copy.

The LINE L1 could also be used as an example of this type of retrotransposon.

L1 is quite different from Alu in that it contains a gene for reverse transcriptase and is not a pseudogene.

Copyright Macmillan Learning. Powered by Cognero.

Page 30

Name:

Class:

Date:

Chap 18_7e

87.

88. Organisms have evolved mechanisms to regulate transposition, either by controlling the amount of transposase produced or by direct inhibition of the transposition event. The transposable elements themselves have also evolved mechanisms to limit their own level of transposition. The long-term well-being of the transposon is often dependent upon the long-term survival of its host. For example, the promoters of some transposase genes are often weak with low levels of expression.

89. In E. coli adenosines at GATC sites are normally methylated. This short sequence is a palindrome so both DNA strands will contain it. Immediately after DNA replication, the A at these sites in the new strand sequence will not be methylated, and there will be a short time period until it does get methylated. During this time, there will be a difference in methylation levels between the old template strand and the new strand. It is during this time that the machinery of the mismatch-repair system is able to distinguish between the two strands and make the appropriate correction.

90. Slippage on the newly synthesized strand will result in an insertion, whereas slippage on the template strand will result in a deletion.

Copyright Macmillan Learning. Powered by Cognero.

Page 31

Name:

Class:

Date:

Chap 18_7e

91.

92. Two products (nonrecombinant) will have five TTG repeats at the locus, one will have four (a deletion), and one will have six (an insertion).

93. DNA bases absorb UV light, resulting in the formation of covalent bonds between adjacent pyrimidines on the same strand of DNA. These pyrimidine dimers distort the DNA molecules and block replication by the most DNA polymerases. If the dimers are not repaired (e.g., by photolyase), the cell can replicate across the distorted region by using special replication machinery that is error-prone. Therefore, mutations result from error-prone replication in response to damage from UV light rather than directly from the damage itself.

94. a. The transposition process itself must result in loss of the introns. It is therefore likely that, in this case, the transposition process involves transcription and the elements moved via RNA intermediates (e.g., retrotransposons). b. Retention of the intron in one transposable element suggests that this particular transposable element moved through a DNA intermediate (i.e., not transcription dependent) and therefore didn't require an RNA intermediate.

This would likely be a class II transposable element and would make transposase. The other transposable element is likely a retrotransposon.

95. Nonsense codons are stop codons.

Leucine codons → UUA, UUG, CUU, CUC, CUA, CUG Stop codons → UGA, UAA, UAG Potential changes: Leucine codon Stop codon Mutation type UUA UAA transversion UUA UGA transversion UUG UAG transversion

Copyright Macmillan Learning. Powered by Cognero.

Page 32

Name:

Class:

Date:

Chap 18_7e 96. Insertions and deletions can change reading frames and thus alter all the amino acid coding information distal to the mutation. Base substitutions, on the other hand, will change only a single nucleotide, and therefore they are much less likely to alter significantly the function of the resulting polypeptide—unless, of course, the base substitution created a stop codon that resulted in a severely truncated polypeptide. Also, a base substitution at a wobble position (one chance in three, if random) will have no effect on amino acid coding information.

Note that mutations such as insertions, deletions, base substitutions, and so forth can also occur in DNA regulatory elements (promoters, enhancers, etc.), which could disrupt gene function.

97. Resolvase is an enzyme carried by some transposable elements that use a replicative transposition process. A transposable element with a deletion in its resolvase gene would not be able to resolve the cointegrate molecule formed during the transposition process into two separate transposable elements. Instead, a single cointegrate molecule containing two copies of the transposable element would be formed.

98. The Ames test can give results only for the substance applied to the bacterial test plates and only for the types of mutations detected by the Ames strains. Some mutations such as deletions cannot be detected with the Ames test, and some of these types of mutations may be carcinogenic in animals. In addition to DNA mutations, epigenetic changes can be carcinogenic, and chemicals that produce these changes would not be detected with the Ames test.

99. Intergenic suppressor mutations are suppressor mutations that occur on a different gene from the one having the original mutation and hence involve two different gene products. Sometimes the second gene is a tRNA gene that now makes a tRNA that recognizes the codon mutated in the original mutation. In contrast, intragenic suppressors occur within the same gene as the mutation they suppress, so only a single gene product is involved, usually a protein with two changes in it. The first mutation reduced or eliminated the activity of the protein, while the second mutation restored at least part of the activity.

Copyright Macmillan Learning. Powered by Cognero.

Page 33

Name:

Class:

Date:

Chap 18_7e 100. There is no correct answer to this question, and there is little agreement even among experts in this field.

As far as being parasitic and possibly detrimental to the host, TEs do create random mutations and chromosomal rearrangements. Most of these events would be expected to be either neutral or even detrimental to the host, and very few would be advantageous. Host genomes have evolved mechanisms to silence or at least reduce the activity of TEs such as DNA methylation and other epigenetic mechanisms. This suggests that the activities of TEs can be harmful unless controlled by the host.

TEs may be beneficial to the host in several ways. Sometimes the creation of genetic variability could be useful to the host genome. In a rapidly changing environment or even a somewhat static environment, a rare beneficial mutation may be so valuable that the species would be willing to trade the many deleterious mutations for a very few beneficial ones. In these cases, any mechanism that increases genetic variation may be valuable and this would include TEs. In addition, TEs themselves may evolve into a useful component of the host genome. The gene that makes the enzyme that replicates telomeres (telomerase, Chapter 12) may have come from a TE. In one plant a TE gene product has become a regulatory protein of plant genes, and in vertebrates part of the immune system seems to have come from a TE. As often happens in these types of controversies, it may turn out both points of view are somewhat valid. TEs may be somewhat parasitic and capable of causing harm if left unchecked. Hosts have evolved mechanisms to reduce the transposition activity of TEs but have not eliminated them completely from their genomes. Possibly, most genomes are in some kind of equilibrium with their resident TEs with a minimum of harm being done by them and an occasional useful adaptation coming from their presence.

101. a. A deletion of the first nucleotide (A) in the original serine codon, which changes the reading frame: AUG-CUC-AGU-AUA-AUG-GCC → AUG-CUC-GUA-UAA (stop codon) b. Insertion of an A residue just in front of the stop codon to change the reading frame: AUG-CUC-GUA-UAA (stop codon)-UGG-CC → AUG-CUC-GUA-AUA-AUG-GCC 102. Genetic variation arises through mutations. Mutations that undergo natural selection are key driving forces in evolution and essential for organisms to adapt to the changing environment. However, mutations may be detrimental to an individual organism's health or fitness.

Copyright Macmillan Learning. Powered by Cognero.

Page 34

Name:

Class:

Date:

Chap 18_7e 103. a. 5′-AGCGGAT-3′ 3′-TCGCCTA-5′ (inverted; nonpalindrome) b. 5′-AAATTT-3′ 3′-TTTAAA-5′ (inverted; palindrome) c. 5′-GGAATTCC-3′ 3′-CCTTAAGG-5′ (inverted; palindrome) Note that sequences (b) and (c) are identical when read 5′ → 3′ (top strand) or 3′ → 5′ (bottom strand).

Sequences like this are called palindrome sequences and are characteristic of restriction enzyme recognition/cleavage sites.

104. Missense mutations alter the coding sequence so that one amino acid is substituted for another. Nonsense mutations change a codon originally specifying an amino acid into a translation-termination codon.

Copyright Macmillan Learning. Powered by Cognero.

Page 35

Name:

Class:

Date:

Chap 19_7e Indicate the answer choice that best completes the statement or answers the question.

1. You have discovered a gene that enables organisms to accumulate gold in their tissues by concentrating trace amounts found in normal soil. You want to transfer this gene into a plant. Order the steps below that would accomplish this goal.

1. Infect the plant with the Rhizobium radiobacter strain.

2. Digest the gold gene and a Ti plasmid with appropriate restriction enzymes.

3. Insert the gold gene into the Ti plasmid.

4. Amplify the gold gene with PCR. 5. Transfer the recombinant Ti plasmid into Rhizobium radiobacter.

6. Use a selectable marker to identify plant cells that have integrated the recombinant plasmid into their genome. a. 4, 2, 3, 5, 1, 6 b. 4, 5, 2, 3, 1, 6 c. 4, 2, 5, 1, 6, 3 d. 4, 3, 2, 5, 1, 6 2. Which of the following statements is NOT correct? a. Coding sequences for gene products can be isolated from cDNA libraries. b. Antibodies are used for Northern blot analysis. c. The number of STR copies is variable throughout human populations. d. PCR amplification generates large numbers of linear DNA fragments. e. RNA molecules can be used as hybridization probes in Southern blot analysis.

3. Which of the following represents an appropriate cloning vector for cloning a gene into a bacterial cell? a. YAC b. Ti plasmid c. plasmid d. Rhizobium radiobacter e. lacZ 4. A scientist edits a mammalian genome by inserting a GFP reporter sequence downstream of a gene of interest.

Which of the following methods is she likely to have used to make this change? a. site-directed mutagenesis b. oligonucleotide-directed mutagenesis c. CRISPR-Cas9 genome editing followed by nonhomologous end joining (NHEJ) d. CRISPR-Cas9 genome editing followed by homologous recombination (HR) with a donor template e. mutagenesis with radiation

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 19_7e A fragment of DNA is cloned into a plasmid with a sequencing primer-binding site. After dideoxy sequencing, the gel pattern shown in this diagram is obtained.

5. On the strand that acted as the template in the sequencing reaction, what base of the cloned fragment was closest to the primer? a. G b. A c. C d. T 6. Antibodies are to Western blots as _____ is/are to Southern blots. a. RNA probes b. proteins c. DNA probes d. amino acids e. DNA or RNA probes 7. Which of the following would be MOST appropriate for cloning a gene that is 300 kb in size? a. plasmid b. cosmid c. phage lambda d. BAC e. yeast phage 8. Which of the following DNA sequences are contained within a cDNA library? a. exons b. introns c. promoters d. enhancers

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 19_7e 9. Which of the following statements CORRECTLY describes DNA ligase? a. DNA ligase forms hydrogen bonds between nucleotide bases. b. DNA ligase can seal nicks between amino acids. c. DNA ligase recognizes and cuts at specific sequences. d. DNA ligase is necessary for creating recombinant plasmids. e. DNA ligase is a requirement of a sequencing reaction.

10. Ampicillin resistance is to ampR as _____ is to lacZ. a. Bt toxin b. G418 c. cleavage of X-gal d. penicillin resistance e. gancyclovir

11. The recognition site for BamHI is

. Which of the following restriction sites when digested

would create a cohesive end that could be ligated to a BamHI digested DNA fragment? a. EcoRII

b. PvuII

c. EcoRI

d. BglII

e. CofI 12. Which of the following statements is NOT correct regarding type II restriction enzymes? a. They can create blunt ends. b. They make double-stranded cuts in DNA. c. They recognize specific sequences and make cuts further away from the recognition sequence. d. They are named based on their bacterial origin.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 19_7e 13. One technique for finding a gene of interest involves first generating a genetic map to find the general location of the gene and then identifying the specific location of the gene. What is this technique called? a. next-generation sequencing b. DNA fingerprinting c. positional cloning d. in silico gene discovery e. site-directed mutagenesis 14. In the introduction to this chapter, the use of the CRISPR-Cas9 system to make a deletion of exon 23 containing a premature stop codon in mdx mice was described. The resulting mice had partially restored dystrophin protein and muscle function. Which of the following CRISRP-Cas9 mediated changes might lead to a better restoration of dystrophin protein and muscle function? a. a deletion of exon 24 in addition to exon 23 b. a deletion of intron 23 c. a deletion of 30 bases within exon 23 d. a deletion of 25 bases within exon 21 e. an insertion of a corrected copy of exon 23 after the mutant exon 23 15. Which of the following is NOT a potential benefit of using transgenic plants? a. They can reduce the use of harmful chemical pesticides and thus provide an ecological benefit. b. They can generate restriction enzyme sites on a foreign gene of interest to be cloned. c. They often increase yields, providing more food per acre and reducing the amount of land needed for agricultural use. d. They can allow crops to be grown on land previously unavailable for productive agricultural use. e. They can be used to express large quantities of specific biological products more cheaply and quickly than by expression in animal systems.

16. Which of the following is NOT used in an in situ hybridization experiment for visualizing DNA? a. chromosome spread b. probe c. denaturation solution d. antibody e. microscope

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 19_7e 17. What is the function of dideoxynucleotides in Sanger DNA sequencing? a. They act as primers for DNA polymerase. b. They act as primers for reverse transcriptase. c. They cut the sequenced DNA at specific sites. d. They allow only the specific sequencing of the RNAs of a genome. e. They stop synthesis at a specific site, so the base at that site can be determined.

18. Which of the following statements is NOT true regarding the basic components required for a bacterial cloning vector? a. Selectable markers provide a means for preferentially allowing growth of only those bacterial cells that have been transformed with the cloning vector. b. Unique restriction enzyme sites allow for larger pieces of foreign DNA to be inserted into the bacterial cloning vector. c. Unique restriction enzyme sites provide a means for inserting the foreign DNA into the cloning vector at a specific known sequence site. d. A bacterial origin of replication ensures that the plasmid is replicated while present within the bacterial cell. e. Selectable markers provide a means for selecting cells that have been transformed with a recombinant plasmid.

19. A scientist is studying a normal tissue sample and a cancerous tissue sample. What method might she use to determine whether the transcription of gene X is upregulated in the cancerous tissue sample? a. carrying out a Southern blot using the cancerous tissue sample only b. carrying out a Southern blot using both samples c. carrying out a Northern blot using the cancerous tissue sample only d. carrying out a Northern blot using both samples e. carrying out a Western blot using the cancerous tissue sample only 20. Suppose you have previously located and cloned a human gene on chromosome 2. You now believe there is an important disease-related gene nearby on the same chromosome. What technique might you employ to locate the disease-related gene? a. chromosome walking b. DNA fingerprinting c. site-directed mutagenesis d. RNAi e. CRISPR-Cas genome editing

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 19_7e 21. What is the purpose of Taq polymerase in a PCR reaction? a. DNA denaturation b. primer annealing c. DNA synthesis d. heating of the reaction e. heating of the reaction and DNA denaturation A scientist carries out a ligation reaction designed to insert a foreign piece of DNA into a plasmid that contains the front end of the lacZ gene within the multiple cloning site (MCS). He carries out three transformations in parallel with lacZ– bacteria. The three transformations (i, ii, and iii) contain the following: i. sterile water i . a sample of the original plasmid iii. the ligation reaction He then plates the transformations on medium containing X-gal and ampicillin.

22. What results would indicate the scientist should proceed to culture colonies from (iii) that might contain the desired clone? a. Many blue colonies on all three plates. b. White colonies on plates (i) and (ii) and predominantly blue colonies on plate (iii). c. No colonies on plate (i), white colonies on plate (ii), and predominantly blue colonies on plate (iii). d. No colonies on plate (i), blue colonies on plate (ii), and predominantly white colonies on plate (iii). e. No colonies on plate (i), blue colonies on plate (ii), and predominantly blue colonies on plate (iii).

23. Which of the following is usually NOT a component used to generate cDNA? a. reverse transcriptase b. RNase c. ligase d. oligo(dT) chains e. restriction enzymes

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 19_7e 24. Which of the following statements is INCORRECT regarding the differences between a genomic library and a cDNA library? a. Genomic libraries contain fewer restriction enzyme sites, whereas cDNA libraries contain many more. b. A genomic library is prepared from total genomic DNA, whereas a cDNA library is prepared from mRNA. c. Genomic libraries contain much more sequence information and are much larger than cDNA libraries. d. Genomic libraries contain coding and noncoding (regulatory, intron, etc.) sequences, whereas cDNA libraries contain only coding sequences along with their associated 5′ and 3′ untranslated regions. e. cDNA libraries are generated with the use of reverse transcriptase, whereas genomic libraries are not.

You identify an RFLP in mice by digesting genomic DNA with HindIII enzyme and radiolabeling a piece of probe DNA. Southern blot analysis shows that the probe detects a 2-kb fragment in one strain of mice and a 4-kb fragment in another strain of mice. The two mice strains are crossed to produce an F1 generation. Two F1 siblings are mated to produce a dozen F2 progeny. You isolate genomic DNA from several F1 and F2 individuals, digest with HindIII, and perform a Southern blot with the same probe.

25. How many bands would be seen in the Southern blot of an F2 individual? a. one b. two c. three d. four e. one or two 26. Which of the following can be used for genetic engineering in plants? a. Ti plasmid b. Rhizobium radiobacter c. selectable markers d. Ti plasmids and Rhizobium radiobacter only e. Ti plasmids, Rhizobium radiobacter, and selectable markers 27. The lungfish Protopterus aethiopicus has a genome 38 times larger than that of humans. Most of the DNA in this species is noncoding repetitive DNA. What type of library would allow you to compare just the genes in lungfish to the genes in humans? a. cDNA library b. PCR library c. genomic library d. knockout library e. transgenic library

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 19_7e 28. All of the following are requirements of a bacterial cloning vector EXCEPT: a. origin of replication. b. unique restriction enzyme sites. c. Ti plasmid. d. selectable markers.

29. The difference between PCR and real-time PCR is that real-time PCR: a. can measure the amount of DNA amplified as the reaction proceeds, while standard PCR cannot. b. can amplify DNA a billion-fold within just a few hours, while standard PCR cannot. c. can determine the DNA sequence, while standard PCR cannot. d. uses DNA polymerase, while standard PCR does not. e. requires primers, while standard PCR does not.

30. A scientist mutates a gene in yeast and then looks to see what effect introducing this mutation has on the phenotype in yeast. This is an example of: a. DNA fingerprinting. b. next-generation sequencing. c. transgenic research. d. reverse genetics. e. positional cloning.

A student carries out PCR using the following steps: Step 1: 94°C for 1 minute Step 2: 60°C for 30 seconds Step 3: 72°C for 30 seconds 31. Which of the following lists the CORRECT terms for these three steps? a. denaturation of the double-stranded template, extension of the new DNA molecules, primer annealing b. denaturation of the double-stranded template, primer annealing, extension of the new DNA molecules c. denaturation of the double-stranded template, extension of the new DNA molecules, hybridization of the template d. degradation of the template, primer annealing, extension of the new DNA molecules e. hybridization of the single-stranded templates, primer annealing, extension of the new DNA molecules

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 19_7e A fragment of DNA is cloned into a plasmid with a sequencing primer-binding site. After dideoxy sequencing, the gel pattern shown in this diagram is obtained.

32. What was the sequence of the DNA strand that acted as the template in the sequencing reaction? a. 5' GCTAGCA 3' b. 5' ACGATCG 3' c. 5' TGCTAGC 3' d. 5' CGATCGT 3' You identify an RFLP in mice by digesting genomic DNA with HindIII enzyme and radiolabeling a piece of probe DNA. Southern blot analysis shows that the probe detects a 2-kb fragment in one strain of mice and a 4-kb fragment in another strain of mice. The two mice strains are crossed to produce an F1 generation. Two F1 siblings are mated to produce a dozen F2 progeny. You isolate genomic DNA from several F1 and F2 individuals, digest with HindIII, and perform a Southern blot with the same probe.

33. How many bands would be seen in the Southern blot of an F1 individual? a. one b. two c. three d. four e. one or two 34. All of the following are required to create a cDNA library EXCEPT: a. RNA polymerase. b. DNA polymerase. c. reverse transcriptase. d. dNTPs. e. mRNAs.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 19_7e 35. A gene does not contain the necessary restriction enzyme sites for cloning into a plasmid vector. What is a possible option? a. Increase the amount of gene used. b. Add linkers to generate new restriction enzyme sites. c. Use a cosmid as a cloning vector. d. Use dideoxy sequencing to obtain the sequence of the gene. e. Cut the DNA with a blunt end cutter.

36. Which of the following is a set of molecular techniques for locating, isolating, altering, and studying DNA segments? a. in situ hybridization b. gel electrophoresis c. molecular cloning d. Southern blotting e. recombinant DNA technology 37. The following are steps required to develop a knockout mouse. Place them in the CORRECT order.

1. Culture the cells on media with G418 and gancyclovir to select those that have been transformed by homologous recombination.

2. Implant the embryos in pseudopregnant mice.

3. Interbreed resultant chimeric progeny to produce homozygous knockout mice.

4. Disrupt the target gene with a neo+ marker, and link the tk+ marker to the target gene.

5. Transfer the disrupted gene (along with the tk+ marker) into cultured embryonic mouse cells.

6. Inject positive cells into early-stage mice embryos. a. 1, 4, 5, 6, 2, 3 b. 4, 1, 6, 5, 3, 2 c. 4, 5, 1, 6, 2, 3 d. 1, 4, 5, 2, 6, 3 e. 1, 6, 4, 5, 2, 3 38. Gel electrophoresis can be used to separate DNA on the basis of: a. size. b. electrical charge. c. nucleotide content. d. the probe used. e. both size and electrical charge.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 19_7e 39. Consider a tobacco plant cell that was able to express a toxin that was lethal to insects and was resistant to the antibiotic kanamycin. Which of the following genes would this cell contain? a. Bt b. kanamycin resistance gene product c. lacZ d. Bt and a kanamycin resistance gene product only e. Bt, kanamycin resistance gene product and lacZ 40. Which of the following statements does NOT describe a challenge of working at the molecular level? a. Cells contain thousands of genes. b. Individual genes cannot be seen. c. It is not possible to isolate DNA in a stable form. d. A genome can consist of billions of base pairs. e. No physical features mark the beginning or end of a gene.

41. You are interested in a particular segment of rhinoceros DNA and would like to clone it into a cloning plasmid.

You have the following restriction map of the region that includes the DNA of interest and the plasmid (E = EcoRI, H = HindIII, X = XbaI, S = SphI, N = NotI).

Which restriction enzymes would you choose to clone the DNA of interest into the cloning vector? a. E and H b. S c. X and N d. S and N

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 19_7e 42. How might RNAi be used to treat diseases in humans? a. siRNAs could be generated that would silence elevated levels of a harmful gene. b. An RNAi library could be generated to assay harmful gene expression levels. c. RNAi probes could bind to good genes and thus increase their expression. d. RNAi could be used to amplify large amounts of therapeutic proteins. e. RNAi cannot be used in humans, as it is a process found only in plants.

43. During gel electrophoresis, large DNA fragments will _____ small DNA fragments. a. migrate more rapidly than b. migrate at the same speed as c. migrate more slowly than d. cause degradation of e. separate into 44. Southern blotting is a technique used to transfer _____ to a solid medium. a. DNA b. RNA c. protein d. DNA and RNA e. DNA, RNA, or protein 45. Which of the following BEST describes knockout mice? a. They have a gene of interest that has been fully disabled. b. They have lower expression levels of a gene of interest. c. They have higher expression levels of a gene of interest. d. They have a point mutation in the gene of interest. e. They have a gene removed, which results in lowered fertility.

46. You try to generate a knockout mouse for a gene of interest, using the protocol described in this chapter. You do obtain progeny heterozygous for the knockout, as assessed by Southern blot. When you cross these heterozygous mice, you obtain viable progeny—some are heterozygous for the knockout, but none are found that are homozygous for the knockout. Which of following is the MOST likely explanation for this result? a. The target gene was not disrupted successfully. b. Genes in addition to the target gene were disrupted. c. The heterozygous mice in your cross did not mate properly. d. The target gene is haploinsufficient; that is, a mutant phenotype is observed when only one copy of the gene is disrupted. e. The target gene is essential for viability.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 19_7e 47. Which of the following is NOT a bacterial cloning vector? a. plasmid b. bacteriophage λ c. Rhizobium radiobacter d. bacterial artificial chromosome e. cosmid 48. Which of the following remains a problem of gene therapy? a. To date, there has not been a case of gene therapy curing a disease. b. We have yet to develop a vector for delivering the gene. c. Patients mount immune responses to the transferred gene and vectors. d. We do not have the ability to use somatic gene therapy at this time. e. We can only alter germ-line cells at this time.

Indicate one or more answer choices that best complete the statement or answer the question.

A student carries out PCR using the following steps: Step 1: 94°C for 1 minute Step 2: 60°C for 30 seconds Step 3: 72°C for 30 seconds 49. After subjecting his PCR reaction to gel electrophoresis, the student sees no PCR product on the gel. What error(s) might he have made? (Select all that apply.) a. He carried out step 1 at too low a temperature. b. He carried out step 1 at too high a temperature. c. He carried out step 2 at too low a temperature. d. He carried out step 2 at too high a temperature. e. He carried out step 3 for too short a time.

50. Which of the following is a current concern of genetic testing? (Select all that apply.) a. testing for a disease for which there is no cure or treatment b. the confidentiality of the tests c. the accuracy of the tests d. false negative results e. All of these options.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 19_7e A student carries out PCR using the following steps: Step 1: 94°C for 1 minute Step 2: 60°C for 30 seconds Step 3: 72°C for 30 seconds 51. After electrophoresing his PCR reaction, the student sees the desired PCR product on the gel as well as several smaller bands. What error(s) might he have made? (Select all that apply.) a. He carried out step 2 at too low a temperature b. He carried out step 2 at too high a temperature. c. He designed primers with repetitive sequences. d. He contaminated the template DNA sample. e. He carried out step 3 for too short a time.

A scientist carries out a ligation reaction designed to insert a foreign piece of DNA into a plasmid that contains the front end of the lacZ gene within the multiple cloning site (MCS). He carries out three transformations in parallel with lacZ– bacteria. The three transformations (i, ii, and iii) contain the following: i. sterile water i . a sample of the original plasmid iii. the ligation reaction He then plates the transformations on medium containing X-gal and ampicillin.

52. The scientist observed colonies on plate (i). Which of the following might be reasons(s) for this? (Select all that apply.) a. The plasmid he used has a kanamycin resistance marker instead of an ampicillin resistance marker. b. He transformed cells that were not actually ampicillin sensitive. c. Too high a concentration of ampicillin was added to the plates. d. Too low a concentration of ampicillin was added to the plates. e. The water sample was contaminated with the original plasmid.

53. A scientist wishes to screen a DNA library for a gene of interest which has not yet been isolated. Which of the following are strategies she could use? (Select all that apply.) a. She could create a probe using a similar gene from another organism. b. She could randomly synthesize probes and hope one will recognize the gene of interest. c. She could use the entire chromosome on which the gene is found as a probe. d. If the gene of interest produces a protein and its amino acid sequence is known, she could create a pool of synthetic probes corresponding to all possible nucleotide sequences that would express that protein or peptide. e. If gene of interest produces a protein, she could use antibodies that recognize the protein of interest if the DNA library was an expression library.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 19_7e 54. A gene in the fruit fly D. melanogaster is found to affect neuronal function when mutated. A homologue of this gene is identified in mice. Which of the following methods would it make most sense for a scientist to use to study the function of the gene in mice? (Select all that apply.) a. Forward genetics b. Reverse genetics c. Mutagenesis with radiation d. CRISPR-Cas9 genome editing targeting the gene of interest followed by nonhomologous end joining (NHEJ) e. CRISPR-Cas9 genome editing targeting the gene of interest followed by homologous recombination (HR) with a donor template 55. Which of the following statements are TRUE? (Select all that apply.) a. Reverse genetics requires that a gene sequence be known. b. Reverse genetics can only be used to study genotype and not phenotype. c. Forward genetics begins with phenotypic analysis followed by genotypic analysis. d. Forward genetics is no longer used in the 21st century to examine gene function. e. Forward genetics is also called classical genetics and was the primary approach for genetic analysis prior to the genomics era.

56. A scientist edits a mammalian gene by changing a single AT base pair to a GC base pair. Which of the following methods might she have used to make this change? (Select all that apply.) a. site-directed mutagenesis b. oligonucleotide-directed mutagenesis c. CRISPR-Cas9 genome editing followed by nonhomologous end joining (NHEJ) d. CRISPR-Cas9 genome editing followed by homologous recombination (HR) with a donor template e. mutagenesis with radiation 57. A scientist attempts to use the CRISPR-Cas9 system to edit a single gene within a cell. She finds that, in addition to editing the desired gene, she has edited several other genes as well. What are reasonable options for her to try next in order to target just the gene of interest? (Select all that apply.) a. Add less Cas9 enzyme. b. Create a longer sgRNA. c. Try a different sgRNA that will still pair within the gene of interest. d. Use separate crRNA and tracrRNA molecules. e. Use a higher-fidelity version of Cas9.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 19_7e Figure A below shows a restriction map of a rare prokaryotic gene with its direction of transcription indicated by the arrow. Figure B shows the unique restriction sites contained within a plasmid-cloning vector. The blackened region in Figure A represents the amino acid coding sequence of a protein that can be used in humans as a vaccine. The striped region in Figure B is a highly active, constitutive (unregulated) prokaryotic promoter region. Letters indicate the cleavage sites for different restriction enzymes. Known DNA sequences are indicated by short thick lines.

58. Explain how you would isolate and then insert the coding region (Figure A) under the control of the indicated promoter in the cloning vector (Figure B) to produce large amounts of the protein in bacterial cells. Assume that the cloning vector carries the gene for tetracycline (an antibiotic) resistance.

59. A cDNA encoding a protein that is specific for a particular strain (strain Q) of bacteria has been cloned. How would you determine whether an infection contains this particular bacterial strain?

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 19_7e 60. You are handling a paternity lawsuit brought against five potential fathers by a woman. You isolated DNA from the mother, the child, and all the potential fathers. After using PCR to amplify specific polymorphic loci from each individual, you run a gel of the amplified products and stain with ethidium bromide to visualize the DNA fingerprints (shown below). Mo = mother; Ch = child; M1–M5 = potential fathers. B1–B10 indicate marker bands present in the child. Do these results suggest that any of the men could be the child's biological father?

Explain your answer.

61. The haploid human genome contains about 3 × 109 nucleotides. On average, how many DNA fragments would be produced if this DNA was digested with restriction enzyme PstI (a 6-base cutter)? RsaI (a 4-base cutter)? How often would an 8-base cutter cleave?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 19_7e 62. A pedigree and Southern blot results in humans are shown in the following figure. Filled-in figures represent individuals expressing a dominant trait (hypothetical) for blue tongue. What can you say about the potential linkage relationship between the allele responsible for the trait? Shaded regions within the homologs represent sequences that hybridize to the probe used for the Southern analysis. Arrows indicate cleavage sites used for the Southern analysis.

63. List two applications of recombinant DNA technology.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 19_7e 64. You have two probes that detect RFLPs that are closely linked. You have been asked to help settle a paternity case. The two possible fathers have the following RFLP pedigrees. The child and mother are also probed with these two RFLP probes. (The data below the pedigrees are two Southern blots.)

Which male is more likely to be the father? Explain your reasoning.

65. The CRISPR-Cas9 system was the original CRISPR system used for genome editing. Describe at least two modifications of the CRISPR-Cas system and how they are improving or expanding the applications of the system.

66. What are Northern analyses used for? Describe the steps involved in performing a Northern analysis, and describe how levels of gene expression are determined.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 19_7e 67. The full-length (i.e., containing the entire protein coding region) cDNA for a specific eukaryotic gene in humans is 1500 nucleotides long. You screen a pig genomic library with this cDNA and isolate two genomic clones of different lengths. Both clones are sequenced and found to be 1900 and 2100 nucleotides long from start codon to stop codon. How would you explain the presence of two genomic clones in pigs and the discrepancies in their lengths compared to the cDNA probe?

Figure A below shows a restriction map of a rare prokaryotic gene with its direction of transcription indicated by the arrow. Figure B shows the unique restriction sites contained within a plasmid-cloning vector. The blackened region in Figure A represents the amino acid coding sequence of a protein that can be used in humans as a vaccine. The striped region in Figure B is a highly active, constitutive (unregulated) prokaryotic promoter region. Letters indicate the cleavage sites for different restriction enzymes. Known DNA sequences are indicated by short thick lines.

68. After trying to isolate and then insert the coding region (Figure A) under the control of the indicated promoter in the cloning vector (Figure B), you found that all the transformed bacterial cells contained either one of two smaller portions of the coding region for the gene, and some of the fragments were inserted backward (with regard to reading frame) into the cloning vector. How would you explain these observations?

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 19_7e 69. Cronin et al. (Science, 2009, 325: 340–343) used RNA interference to study the immune response of the fruit fly, Drosophila melanogaster, to the bacterial pathogen Serratia marcescens. They identified several members of the JAK-STAT cell–cell signaling pathway in their analysis. a) The PIAS gene is a negative regulator of JAK-STAT signaling (i.e., it inhibits the signaling pathway). RNAi to block PIAS activity causes flies to die significantly earlier than control flies when exposed to bacteria. Does JAK-STAT signaling appear to protect the flies from the bacteria? Explain. b) The upd gene encodes a ligand that activates JAK-STAT signaling. RNAi that blocks upd function causes flies to survive longer than control flies when exposed to bacteria. Explain how this is or is not consistent with your answer from part (a).

70. Cloned eukaryotic genes are not always able to be expressed in bacterial cells. What are some possible reasons for this?

71. List four seminal techniques that have transformed recombinant DNA technology and furthered the field of biotechnology.

72. You are examining a 1000-bp DNA fragment as part of a forensic analysis. If the limit of detection of this molecule is 9 × 108 molecules, what is the minimum number of PCR cycles you would have to run to detect a PCR product generated from a single molecule of the template?

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 19_7e Answer Key 1. a 2. b 3. c 4. d 5. b 6. e 7. d 8. a 9. d 10. c 11. d 12. c 13. c 14. c 15. b 16. d 17. e 18. b 19. d 20. a 21. c 22. d 23. e 24. a 25. e 26. e Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 19_7e 27. a 28. c 29. a 30. d 31. b 32. a 33. b 34. a 35. b 36. e 37. c 38. e 39. d 40. c 41. c 42. a 43. c 44. a 45. a 46. e 47. c 48. c 49. d, e 50. e 51. a, c, d 52. b, d, e 53. a, d, e 54. b, d, e Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 19_7e 55. a, c, e 56. a, b 57. c, e 58. If the prokaryotic gene is already cloned into a vector, a large population of the bacteria containing the vector should be grown, the plasmid DNA should be isolated, and enzymes C and D should be used to cut the plasmid. You should then isolate and purify fragment C–D after gel electrophoresis.

If the prokaryotic gene is not cloned into a vector, then the only way to make copies of it to clone into the expression vector is to use PCR. In order to do this, make primers that anneal to the known sequences indicated in Figure A that flank the coding region, and amplify the region between them using PCR. Isolate the PCR product, purify it, and digest it with restriction enzymes C and D. Next, digest the cloning vector in Figure B with enzymes C and D, mix with gene fragment C–D, from one of the two previous steps, and ligate together using DNA ligase. Transform the ligation products into tetracycline-sensitive bacterial cells and plate out on media containing tetracycline.

Incubate the plates overnight. Bacterial colonies growing on these plates the next day should contain the cloning vector and fusion product (i.e., coding regions fused to promoter). It would be wise at this point to sequence the fusion product contained in one or two of the isolated bacterial colonies to confirm the integrity of the promoter– coding region sequence. After sequencing, to ensure that no nucleotide sequence mistakes were introduced during the cloning procedures or PCR, scale up the growth cultures to produce sufficient protein in the bacterial cells for subsequent isolation and purification steps.

59. Use the cloned DNA as a probe and perform a Southern analysis of bacteria in the infection. For example, isolate DNA from bacteria in the infection, digest it with a restriction enzyme, gel-fractionate the fragments, blot the fragments onto a nylon membrane, and probe the digested bacterial DNA with the cloned cDNA under stringent conditions. Because the cDNA is a specific indicator for strain Q, if a positive signal is shown with the cDNA probe, then the infection contains at least some strain Q bacterial cells.

If the DNA sequence of the cDNA were known, PCR primers could be designed to amplify the sequence from the infection site tissue or isolated bacteria. In order for this to work reliably, the PCR primers would have to be shown not to amplify any other DNA sequences except this cDNA. Note, however, that even if a positive signal for strain Q is observed in the Southern analysis or the PCR, it doesn't necessarily mean that strain Q is the cause of the infection: It shows only that strain Q bacteria are (or more specifically, that this sequence is) present in the infection.

60. Yes, the fourth man (M4) could be the father, but none of the other men tested could be. The alleles present in the child will be a combination of alleles present in the genomes of the mother and the father. Because of this, all bands that are present in the child, but absent from the mother, must have been contributed by the biological father.

Marker bands 1, 3, 4, 5, 7, and 10 present in the child are missing in the mother, but all of them are present in M4.

None of the other suspected fathers contains all six of these marker bands (or alleles). M1 has only B5 and B7; M2 has B1, B3, B4, B5, and B7, but not B10; M3 has only B4, B5, and B10; and M5 has B1, B3, and B7.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 19_7e 61. (1) A 6-base cutter will cleave random-sequence DNA on average once every 46 = 4096 nucleotides. Therefore, on average there would be 3 × 109/4096 = 732,422 cleavage sites in this much DNA and thus 732,422 different fragments.

(2) A 4-base cutter would cut, on average, once every 44 = 256 nucleotides. Therefore, you'd expect 3 × 109/256 = 11,718,750 different fragments.

(3) An 8-base cutter would cut, on average, once every 48 = 65,536 nucleotides.

62. The pedigree and Southern blot results shown above indicate that the mutant allele responsible for the dominant trait is very likely to be linked to homolog B. This is an important result because determining which chromosome is carrying a mutant allele for a trait is often an important first step in beginning the process of isolating the gene, mapping it on a chromosome, and so forth. Because the 1.0-kb RFLP on homolog B is linked to the mutant allele, it can also be used as a reliable marker for the disease and can thus be used for diagnostic testing.

63. 1) pharmaceutical drugs (e.g., human insulin, human growth hormone) 2) bacteria-designed special processes (e.g., leaching minerals from ore) 3) agricultural products (e.g., Bt corn) 4) genetic testing 5) gene therapy 6) gene mapping 7) DNA fingerprinting as a tool in criminal investigations

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 19_7e 64. Probe #1 identifies three alternative RFLP patterns at a single locus: a (two small bands—apparent in grandfather #1), A (one large band—apparent in the child), and A' (one very small band and one mid-sized band—apparent in the mother). Probe #2 identifies three alternative RFLP patterns at a second locus: b (one mid-sized band— apparent in grandfather #1), B (one large band—apparent in the child), and B' (two small bands—apparent in the mother).

Dad #1's pedigree Grandfather is homozygous for both loci: ab/ab Dad #1 is heterozygous for both loci. He must have inherited one chromosome, ab, from the grandfather. Thus, the other chromosome, AB, must have come from the grandmother. We can conclude that Dad #1 has the genotype AB/ab.

Dad #2's pedigree Dad #2 is also heterozygous for both loci. Grandfather #2's genotype is Ab/ab (heterozygous for the A and a alleles, homozygous for the b allele) and Grandmother #2's genotype is aB/ab (homozygous for the a allele, and heterozygous for the B and b alleles).

Since Dad #2 is also heterozygous at both loci, he must be Ab/aB. The mother is homozygous at both loci but has the third allele of each: A'B'/A'B'. The child's genotype is AA'BB'.

He must have received A'B' together (in cis) from his mother, and therefore he must have received AB together from his father.

Dad #1 is the more probable father because he carries A and B alleles together in cis on the same chromosome, so he can donate them both on that same chromosome to the child.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 19_7e 65. Answers may vary. Examples that may be included are: Using modified Cas9 so that it makes only single-stranded (ss) cuts in the DNA (vs. double-stranded cuts). Repair through homology directed repair (HDR) is more likely with ss cuts and will lead to more precise genome edits.

Cpf1, a Cas family member, can be used to make staggered cuts and create complementary sticky ends. This makes repair more accurate and allows insertion of DNA sequences with overhangs that are complementary to the sticky ends at the cuts.

If a Cas9 recognized PAM is not located at the desired position in the genome, Cas proteins from species other than S. pyogenes (source of Cas9) can be used to bind to different PAMs. Also, variants of Cas9 have been engineered to have less stringent and shorter sequence requirements for binding.

CRISPR-Cas systems have been modified to include Cas enzymes that are incapable of cleaving DNA. These cleavage-inactivated Cas enzymes do allow for target DNA binding. Various functional domains/proteins can be tethered to cleavage-inactivated Cas enzymes to bring about various downstream effects. For instance, the CRISPRi system tethers a KRAB domain (chromatin remodeling domain) to a cleavage-inactivated Cas enzyme.

This leads to the modification of chromatin at the target site causing transcriptional inhibition.

CRISPR has been modified to allow delivery of an error-prone DNA polymerase to target sites where it can introduce mutations allowing for evolution to be studied experimentally.

The CRISPR system has been modified for viral detection for diagnostic purposes.

The CRISPR system has been manipulated through the CAMERA system to create a DNA alteration record of various events (changes in gene expression, developmental changes, etc.)

66. Northern analyses are used to examine gene expression at the mRNA level. Steps are the following: (a) Isolate and purify RNA from cells. (b) Fractionate RNA using gel electrophoresis. (c) Blot the RNA to a solid support membrane (usually nylon). (d) Probe the membrane (containing blotted RNA) with a specific labeled sequence. (e) Wash the membrane to remove unbound probe. (f) Expose the membrane to film and develop an autoradiogram.

Note that the higher the intensity of the signal, the greater the relative gene expression level.

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 19_7e 67. The pig genome must contain two separate genes encoding related isoforms for the polypeptide, each of which has sufficient sequence similarity to hybridize effectively with the human cDNA probe. The human and pig genes are therefore very likely to be related evolutionarily (i.e., descended from a common ancestral gene sequence). The coding sequence for this particular gene in pigs may include more amino acids than the putative homologous (i.e., evolutionarily related) gene in humans. The extra sequence length of the genomic clone in pigs (versus the human cDNA probe) is also likely to be caused by the presence of intron sequences.

If we can assume that the pig genomic clones contain all the coding sequences for the polypeptide product, and if the human and pig gene products are similar in size, then most of the extra pig genomic DNA is most likely intron sequences. Note, however, that in this case we are assuming that we isolated complete (i.e., containing all polypeptide coding sequences) pig genomic clones during library screening. This assumption is not always valid because truncated clones (both cDNA and genomic) containing only partial coding domains for genes can also hybridize with probe sequences if enough sequence similarity is present to allow hybridization. Also note that experimental conditions (i.e., hybridization and washing temperatures, etc.) can also play a very significant role in determining success or failure of library screening experiments.

68. There must be a previously unknown cleavage site for restriction enzymes C and D contained within the coding region of the gene. For example:

69. a) JAK-STAT signaling seems not to be protective in the immune response to bacteria. When PIAS is inhibited by RNAi, JAK-STAT signaling will increase (since removal of a negative regulator will cause an increase in activity of the pathway). Since the functional result of this increase in signaling is an earlier death, JAK-STAT signaling must not be protective and instead can actually accelerate the death of the animal. b) If the activating ligand of JAK-STAT signaling is blocked by RNAi, then the activity of the pathway should be reduced. When the signaling activity is reduced, we observe an increase in survival of the animals. This result is consistent with the observations from part (a), where increasing the signal caused a quicker death; here, reducing the signal caused a slower death. Therefore, these two results together demonstrate that JAK-STAT signaling is a negative influence on survival of the flies in response to infection with Serratia.

Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 19_7e 70. (1) Expression of the protein in bacteria may be toxic or detrimental to the bacterial cell. The protein may also be degraded by the bacterial cell or aggregate into an insoluble mass (called an inclusion body) that would be difficult or impossible to dissolve.

(2) If the eukaryotic gene is a genomic sequence, the bacterial cell may not be able to recognize the eukaryotic promoter sequence. And even if a transcript were produced, if the eukaryotic gene contained introns, the bacterial cell could not process them.

(3) If the gene is transcribed in the bacteria cell, it may not be translated or may be translated inefficiently, because the eukaryotic mRNA would lack a Shine–Dalgarno ribosome binding site.

(4) If the eukaryotic protein is translated, it still may require specific posttranslational processing, carbohydrate residue addition, and so forth—which the bacterial cell cannot perform—for it to fold correctly, become active, or protect itself from degradation.

71. Answers may vary. Four techniques are explicitly stated in Section 19.1, but there are others that may be mentioned. Techniques that can be included are: DNA cloning (use of restriction enzymes and other methods) DNA sequencing PCR Transformation and transfection techniques Genome engineering (CRISPR-Cas, ZFNs TALENs, etc.)

Genomics (microarrays, WGS, single-cell genomics, etc.)

Bioinformatics 72. PCR will amplify the DNA fragment exponentially. If you start with a single copy, you will have two copies after one cycle, four copies after two cycles, and so forth. The number of copies can be generalized by the formula: Nt = N0 * 2t where Nt is the number of copies after t cycles and N0 is the number of starting copies. If our minimum detection limit is 9 × 108, we can set that as our Nt, set our N0 as 1 (number of starting copies), and solve for t.

9 × 108 = 1 * 2t t = log2 (9 × 108) = 29.7 Therefore, it will take a minimum of 30 cycles to be able to detect a single molecule of the starting template using PCR amplification.

Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 20_7e Indicate the answer choice that best completes the statement or answers the question.

1. The expression of miRNAs was compared in normal cells and cancer cells using a microarray. miRNAs from cancer cells were labeled with green fluorescent tags, and miRNAs from normal cells were labeled with red fluorescent tags. miRNA #1 was overexpressed in cancer cells, miRNA #2 was not expressed in either cell type, miRNA #3 was underexpressed in cancer cells, and miRNA #4 was equally expressed in both cell types. Which of the following microarrays illustrate these findings?

a.

b.

c.

d.

e.

2. Which of the following is TRUE of noncoding DNA? a. It is not essential for life in most organisms. b. It can contain sequences to which proteins can bind and influence gene expression. c. It can contain genes that code for rRNA and tRNA. d. It contains a high percentage of introns. e. These regions have a low gene density.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 20_7e 3. In mass spectrometry, a molecule is ionized and its migration rate in an electrical field is determined. What information does the migration rate tell about the molecule? a. It indicates mass because small molecules migrate more rapidly than larger molecules. b. It indicates electrical charge because negatively charged molecules migrate more rapidly than positively charged molecules. c. It indicates the solubility of the molecule because insoluble molecules form aggregates and migrate much more slowly than soluble molecules. d. It indicates the relative abundance of the molecule in the sample. e. It indicates the complexity of the proteome because greater proteome diversity takes a longer time to process.

4. What information CANNOT be learned from the structure of a protein? a. the location of the active site of an enzyme b. possible interaction sites with other molecules c. targets for potential drugs d. a source of insight into the function of an unknown protein e. the number of introns in the gene that encodes the protein 5. Which of the following is involved in the RNA sequencing process? a. conversion of the RNA to complementary DNA (cDNA) sequences b. direct sequencing of RNA molecules c. extraction of DNA from a cell d. generation of RNA from cDNA sequences e. isolation of RNA from a cell followed by fragmentation of RNA

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 20_7e The following microarray data are shown for four human genes. Two samples were hybridized to the microarray in this experiment. One sample was cDNA generated from mRNA of a tissue culture cell line that was originally derived from a B-cell leukemia. The other sample was cDNA generated from normal, mature neural tissue.

Vertical stripes mean hybridization to cancerous B-cell cDNA only.

Horizontal stripes mean hybridization to normal neural cell cDNA only.

Completely filled in means hybridization to both cDNA types.

Blank means hybridization to neither cell's cDNA.

6. One spot on the microarray contains sequences unique to a gene for an RNA polymerase subunit. One spot contains sequences unique to the skeletal muscle myosin gene. Which spot would be expected to contain the myosin gene sequences? a. spot 1 b. spot 2 c. spot 3 d. spot 4 e. none of the spots

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 20_7e A researcher is interested in studying Drosophila genes that are turned on in response to UV light exposure.

She uses a microarray to carry out her experiment and is particularly interested in eight genes shown in the microarray below.

Vertical stripes mean hybridization to UV light–exposed cell cDNA only.

Horizontal stripes mean hybridization to nonexposed cell cDNA only (control).

Completely filled in means hybridization to both cDNA types.

Blank means hybridization to neither cell's cDNA.

7. Which genes appear to be downregulated in the UV light–exposed group? a. 1, 3, and 5 b. 2 and 7 c. 4, 6, and 8 d. 6 only e. 4 and 8 only 8. Two genes that evolved from the same common ancestral gene but are now found as homologs in different organisms are called: a. orthologs. b. paralogs. c. heterologs. d. pseudologs. e. lincolnlogs.

9. The set of all proteins encoded by the genome is called the: a. genome. b. transcriptome. c. metabolome. d. proteome. e. glycome.

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 20_7e 10. Which of the following is NOT an example of the importance of gene duplication in genome evolution? a. The human olfactory multigene family consists of about 1000 genes used in our sense of smell. b. Ribosomal protein S4 is found as a single-copy gene in all three domains of life. c. After a segmental duplication, the original copy of the gene can continue its function while the new copy undergoes mutation, leading to new functions. d. The globin gene family in humans consists of 13 genes that encode globin-like molecules, most of which produce proteins that carry oxygen. e. These are all examples of how gene duplication plays a role in genome evolution.

11. The study of the expression, localization, and interactions of the complete set of proteins found in a given cell is called: a. genomics. b. bioinformatics. c. metagenomics. d. proteomics. e. transcriptomics.

12. Which of the following statements is TRUE of genetic maps? a. They are derived from direct analysis of DNA. b. They are based on frequencies of recombination between loci. c. They have a high level of detail rivaling that of physical maps. d. They always accurately correspond to physical distances between genes. e. They are more accurate than physical maps.

13. Which of the following information CANNOT be learned from RNA sequencing? a. the types and number of RNA molecules produced by transcription b. the presence of alternatively processed RNA molecules c. information about differential expression of the two alleles in a diploid individual d. different RNA molecules generated by bidirectional or overlapping transcription of DNA sequences e. the active sites of the enzyme the RNA encodes 14. A section of a genome is cut with three enzymes: A, B, and C. Cutting with A and B yields a 10-kb fragment.

Cutting with B and C yields a 2-kb fragment. What is the expected result from a digest with A and C, if the C site lies in between the A and B sites? a. a 12-kb fragment b. an 8-kb fragment c. an 8-kb and a 2-kb fragment d. a 10-kb and a 2-kb fragment e. a 10-kb, an 8-kb, and a 2-kb fragment Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 20_7e 15. A _____ is a group of evolutionarily related genes that arose through repeated evolution of an ancestral gene. a. microarray b. multigene family c. proteome d. haplotype e. contig Species Saccharomyces cerevisiae (yeast) Drosophila melanogaster (fruit fly) Mus musculus (mouse) Pan troglodytes (chimpanzee) Homo sapiens (human)

Genome Size (millions of base pairs) 12 170 2,627 2,733 3,223

Number of Predicted Genes 6,144 13,525 26,762 22,524 ~20,000

16. Based on the table above, which species has the LARGEST amount of noncoding DNA within its genome? a. yeast b. fruit fly c. mouse d. human e. It cannot be determined from the information provided in the table.

17. A human gene with a disease phenotype is going to be mapped by positional cloning. Which would be the MOST useful for this task? a. information about bacterial orthologs of the gene b. an EST database of the human genome c. microarray data of tissues in which the gene is expressed d. data about the inheritance of SNP markers in families with the disease e. whole-genome-shotgun clones of the human genome 18. Imagine a prokaryote has a genome of 1 million base pairs. How many genes would you predict it contains? a. approximately 10 b. approximately 100 c. approximately 1000 d. approximately 5000 e. approximately 10,000.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 20_7e 19. Two-dimensional polyacrylamide gel electrophoresis (2D-PAGE) is a technique wherein proteins are separated in one dimension by a chemical property and in a second dimension by a different property. By what property are proteins separated in the second dimension? a. electrical charge b. molecular mass c. posttranslational modification d. isotope e. hydrophobic index 20. Short regions of chromosome 1 were sequenced in a population of Drosophila. These sequences from three different flies are shown below. Identify the SNP haplotype for Fly #3.

Fly #1: 5' ATGGCACGAAGCTAAGAATA 3' Fly #2: 5' ATGGCTCGACGCTCATAATA 3' Fly #3: 5' ATGGCGCGATGCTAAGAATC 3' a. 5' GTAGC 3' b. 5' CGATG 3' c. 5' GTAGA 3' d. 5' NNAGA 3' e. 5' CATCG 3'

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 20_7e The following microarray data are shown for four human genes. Two samples were hybridized to the microarray in this experiment. One sample was cDNA generated from mRNA of a tissue culture cell line that was originally derived from a B-cell leukemia. The other sample was cDNA generated from normal, mature neural tissue.

Vertical stripes mean hybridization to cancerous B-cell cDNA only.

Horizontal stripes mean hybridization to normal neural cell cDNA only.

Completely filled in means hybridization to both cDNA types.

Blank means hybridization to neither cell's cDNA.

21. One spot on the microarray contains sequences unique to a gene for an RNA polymerase subunit. One spot contains sequences unique to the skeletal muscle myosin gene. Which spot would be expected to contain the RNA polymerase gene sequences? a. spot 1 b. spot 2 c. spot 3 d. spot 4 e. none of the spots 22. In the genetic map of the human genome, one map unit is approximately 850,000 bp. For the genome of the eukaryotic yeast Saccharomyces cerevisiae, one map unit is approximately 3000 bp. Why is a map unit so different in these two different types of organisms? a. A map unit is the amount of measured recombination between two linked points in a genome, and humans have more SNPs than Saccharomyces cerevisiae that can serve as genetic markers. b. An increased number of noncoding introns in humans suppresses recombination and leads to larger physical distances for each map unit. c. Saccharomyces cerevisiae has fewer predicted genes than humans, which means that less recombination is required in yeast cells. d. The amount of homologous recombination per DNA length must be lower in humans than in Saccharomyces cerevisiae. e. Humans have more chromosomes and, therefore, have more linkage groups than Saccharomyces cerevisiae.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 20_7e 23. Two-dimensional polyacrylamide gel electrophoresis (2D-PAGE) is a technique wherein proteins are separated in one dimension by a chemical property and in a second dimension by a different property. By what property are proteins separated in the first dimension? a. electrical charge b. molecular mass c. posttranslational modification d. isotope e. hydrophobic index 24. Which of the following statements about eukaryotic genomes is TRUE? a. A substantial part of the genomes of most multicellular organisms consists of protein coding sequences. b. People differ not only at millions of individual SNPs but also in the number of copies of many larger segments of the genome. c. In humans, the number of proteins is approximately equal to the total number of genes. d. There is a directly proportional relationship between the size of a eukaryotic organism and the number of genes it has. e. The number of genes among multicellular eukaryotes is related to phenotypic complexity.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 20_7e The following microarray data are shown for four human genes. Two samples were hybridized to the microarray in this experiment. One sample was cDNA generated from mRNA of a tissue culture cell line that was originally derived from a B-cell leukemia. The other sample was cDNA generated from normal, mature neural tissue.

Vertical stripes mean hybridization to cancerous B-cell cDNA only.

Horizontal stripes mean hybridization to normal neural cell cDNA only.

Completely filled in means hybridization to both cDNA types.

Blank means hybridization to neither cell's cDNA.

25. The gene sequences contained in spot 2 are present only in the B-cell cancer sample. What is a possible function for this gene, based on the microarray data? a. This gene may be involved in neurotransmitter release at the synapse since B-cells are known to be involved in cell-to-cell signaling. b. This gene may be involved in activating cell growth, so it is switched on in the cancer cell line that is growing rapidly. c. This gene is involved in the immune function of the B-cell since that is the cell type from which this cancer line was initially derived. d. This gene may be involved in neurotransmitter release at the synapse or in activating cell growth. e. This gene may be involved in activating cell growth or in the immune function of the B-cell.

26. For a physical map of a chromosome, distances are measured in units of: a. map units. b. RFLPs. c. centiMorgans. d. base pairs. e. contigs.

27. Loci that are far apart: a. have a higher recombination rate than loci that are close together. b. exhibit a recombination frequency of greater than 50%. c. always segregate together in meiosis. d. influence the phenotype to a lesser degree than loci that are close together. e. have less attraction than close loci.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 20_7e 28. Crossing over is often reduced around centromeric regions of chromosomes. If you were trying to construct a genetic map of two linked marker loci in this region, what result might you obtain? a. The recombination frequencies would be low, and you would deduce that the markers were very close to one another. b. The recombination frequencies would be low, and you would deduce that the markers were very far from one another. c. The recombination frequencies would be high, and you would deduce that the markers were very close to one another. d. The recombination frequencies would be high, and you would deduce that the markers were very far from one another. e. The recombination frequencies would be too low to detect, and you would not be able to estimate the distance between genetic markers.

29. You've cloned a 2-kb fragment of DNA into a bacterial cloning vector and want to construct a restriction map of the insert. You amplify the 2-kb insert using PCR, purify it, and subject it to differential digestion with the enzymes EcoRI and HindIII, gel-fractionate the digests, and visualize the restriction patterns by staining the gels with ethidium bromide to generate the following results. Using these data, indicate the order of restriction sites in the DNA insert.

a. EcoRI, HindIII, EcoRI, HindIII, EcoRI b. HindIII, EcoRI, HindIII, EcoRI, HindIII c. HindIII, EcoRI, HindIII, HindIII, EcoRI d. HindIII, EcoRI, EcoRI, HindIII, EcoRI e. EcoRI, EcoRI, HindIII, EcoRI, HindIII 30. The Earth BioGenome Project (EBP) has the goal of sequencing all eukaryotic species on earth by 2028.

How will these sequence data help scientists and society in general? a. clarify evolutionary relationships between organisms b. discover new species c. clarify how human activity is affecting species d. aid in the design of biotechnological products e. all of the above

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 20_7e 31. Linkage disequilibrium is the nonrandom association among SNPs within a haplotype. Over time, would you expect linkage disequilibrium among markers to be maintained, to increase, or to decrease? a. decrease, because random mutation will generate new SNPs that will give rise to new haplotypes b. decrease, because crossing over will break up the physical linkage among markers c. increase, because unequal crossing over between SNPs will decrease the physical distance between them d. increase, because positive selection will cause the haplotype to become more common e. be maintained, because there is no selective pressure to change the frequency of haplotypes in the population Species Saccharomyces cerevisiae (yeast) Drosophila melanogaster (fruit fly) Mus musculus (mouse) Pan troglodytes (chimpanzee) Homo sapiens (human)

Genome Size (millions of base pairs) 12 170 2,627 2,733 3,223

Number of Predicted Genes 6,144 13,525 26,762 22,524 ~20,000

32. Based on the table above, which species has the HIGHEST gene density? a. yeast b. fruit fly c. mouse d. chimpanzee e. human 33. Which of the following statements about prokaryotic genomes is FALSE? a. Bacteria can obtain new genetic information through horizontal gene transfer. b. All prokaryotes have a genome that consists of a single circular chromosome. c. Prokaryotes with the smallest genomes tend to be in species that occupy restricted habitats, such as bacteria that live inside other organisms. d. In many sequenced genomes, the function of a significant percentage of genes is currently unknown. e. The number of genes between species can vary from a few hundred to a few thousand.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 20_7e 34. Crossing over is often reduced around centromeric regions of chromosomes. If you were trying to construct a genetic map of two linked marker loci in this region, how would the genetic map correspond to the physical map? a. The genetic map will underestimate the distance between markers because the low recombination rate will suggest that the markers are very close to one another when they could be physically far apart. b. The genetic map will underestimate the distance between markers because the markers will segregate independently in meiosis. c. The genetic map will overestimate the distance between markers because the low recombination rate will make it appear that they are far from one another when they could be physically close to one another. d. The genetic map will overestimate the distance between markers because rates of crossing over are not the same at all places on all chromosomes. e. The genetic and physical maps will precisely correspond since recombination rates at the centromere are lower and more predictable.

35. If a restriction enzyme cuts a circular DNA into three fragments, how many restriction sites are there in the DNA? a. two b. three c. four d. six e. five 36. A linear DNA fragment is cut with a restriction enzyme to yield two fragments. How many restriction sites are there for the enzyme in this fragment? a. one b. two c. three d. four e. It cannot be determined with this information.

37. How much of the human genome codes for proteins? a. less than 2% b. approximately 24% c. approximately 50% d. approximately 74% e. more than 82%

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 20_7e 38. A set of overlapping DNA fragments that form a contiguous stretch of DNA is called a: a. chromosome. b. sequence. c. map. d. contig. e. clone.

39. The genome of the rice plant was completely sequenced in 2002. What features of the rice genome make it useful for sequencing efforts in other grass species, such as wheat and corn? a. These other grass species have genomes that are considerably larger than that of rice. b. The grasses have a common evolutionary ancestor. c. A smaller, more easily sequenced genome can serve as a model for the mapping and isolation of genes in the larger genomes. d. Many genes are present in the same order in related grass species' genomes (the genomes are collinear). e. All of the answers are correct.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 20_7e 40. Given the 2D-PAGE gel below, which one of the following conclusions is true? [Figure after G. Gibson and S. Muse. 2004. A Primer of Genome Science, 2e. Sinauer Associates, Inc. p. 274, Fig. 5.4.]

a. Cathepsin B has a smaller molecular mass than actin and approximately the same charge as actin. b. Proteosome B chain has a higher charge and a larger molecular mass than actin. c. ER60 has a smaller molecular mass than cathepsin heavy chain and approximately the same charge as cathepsin heavy chain. d. Calreticulin has a lower charge than glutathione-S-transferase and a smaller molecular mass than glutathione-S-transferase. e. ATP synthase D chain has a higher charge than transthyretin and approximately the same molecular mass as transthyretin.

41. A BLAST search is done to: a. find similar gene or protein sequences. b. find the chromosomal location of a sequence. c. predict the three-dimensional structure of a protein from its amino acid sequence. d. find restriction sites and SNPs in a sequence. e. determine the conditions under which a gene is expressed.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 20_7e 42. Below are the steps involved in the RNA-seq process. Place these steps in the CORRECT order.

1. Conversion of the RNA to cDNA 2. Fragmentation and preparation of the cDNAs for sequencing 3. Isolation of the RNA molecules of interest from cells 4. Assembly of the sequencing reads into RNA transcripts 5. Sequencing of the cDNA a. 3, 1, 2, 5, 4 b. 1, 2, 4, 5, 3 c. 3, 4, 1, 2, 5 d. 1, 5, 4, 3, 2 e. 1, 5, 2, 3, 4 43. Which of the following is NOT a reason that metagenomics is particularly useful for studying microbes? a. Many bacteria that cannot be cultured in the laboratory can now be studied. b. The composition of natural microbial communities can be reconstructed. c. The microbiome of individuals can be compared to understand the basis of disease. d. Previously unknown species of bacteria and new gene families have been discovered. e. The genomes of somatic cells in a human individual can vary significantly.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 20_7e A researcher is interested in studying Drosophila genes that are turned on in response to UV light exposure.

She uses a microarray to carry out her experiment and is particularly interested in eight genes shown in the microarray below.

Vertical stripes mean hybridization to UV light–exposed cell cDNA only.

Horizontal stripes mean hybridization to nonexposed cell cDNA only (control).

Completely filled in means hybridization to both cDNA types.

Blank means hybridization to neither cell's cDNA.

44. Which genes appear to be involved in the UV light stress response in Drosophila? a. 1, 3, and 5 b. 2 and 7 c. 4, 6, and 8 d. 6 only e. 4 and 8 only 45. Which technique would not be used to find a gene for a functional protein in a sequenced region of a genome? a. Scan the region for ORFs. b. Scan the region for start and stop codons. c. Scan an SNP database that contains sequences in the region. d. Scan the region for promoter sequences. e. Scan the region for intron splice sites.

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 20_7e 46. Which of the following is a disadvantage of map-based sequencing over shotgun sequencing? a. Map-based sequencing minimizes the amount of repeat sequencing in which the same region is sequenced several times. b. Map-based sequencing yields well-characterized physical maps as well as actual sequences. c. The relationship of overlapping fragments to sequences is known before sequencing begins. d. Map-based sequencing is more time-consuming because extensive mapping is required before sequencing starts. e. Restriction patterns can be used to identify overlapping clones.

Species Saccharomyces cerevisiae (yeast) Drosophila melanogaster (fruit fly) Mus musculus (mouse) Pan troglodytes (chimpanzee) Homo sapiens (human)

Genome Size (millions of base pairs) 12 170 2,627 2,733 3,223

Number of Predicted Genes 6,144 13,525 26,762 22,524 ~20,000

47. Based on the table above, which species has the LOWEST gene density? a. yeast b. fruit fly c. mouse d. chimpanzee e. human

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 20_7e 48. Which of the following genes are paralogs?

a. A1 and A2 b. A1 and B1 c. A1 and B2 d. A2 and B1 e. B2 and B2

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 20_7e A researcher is interested in studying Drosophila genes that are turned on in response to UV light exposure.

She uses a microarray to carry out her experiment and is particularly interested in eight genes shown in the microarray below.

Vertical stripes mean hybridization to UV light–exposed cell cDNA only.

Horizontal stripes mean hybridization to nonexposed cell cDNA only (control).

Completely filled in means hybridization to both cDNA types.

Blank means hybridization to neither cell's cDNA.

49. Which genes appear to be upregulated in the UV light–exposed group? a. 1, 3, and 5 b. 2 and 7 c. 4, 6, and 8 d. 6 only e. 4 and 8 only 50. The average gene in the human genome is approximately _____ base pairs in length. a. 700 b. 7000 c. 17,000 d. 27,000 e. 57,000

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 20_7e 51. Which of the following statements about reference genomes is FALSE? a. The "human genome" is not the universal genome sequence for a human being. b. An individual may have varying genome sequences in different somatic cells. c. The issues associated with reference genomes are more pronounced in humans than bacteria. d. The issues associated with reference genomes must be considered for experimental results based on such genomes. e. The term "pangenome" refers to the entire set of genes possessed by all members of a particular species.

52. You've cloned a 2-kb fragment of DNA into a bacterial cloning vector and want to construct a restriction map of the insert. You amplify the 2-kb insert using PCR, purify it, and subject it to differential digestion with the enzymes EcoRI and HindIII, gel-fractionate the digests, and visualize the restriction patterns by staining the gels with ethidium bromide to generate the following results. Using these data, which two loci would be closest on a genetic map if the frequency of crossing over is equal in this region?

a. The EcoRI sites that are 100 bp apart would show the smallest distance in cM. b. The HindIII sites that are 750 bp apart would show the smallest distance in cM. c. The HindIII site and the EcoRI site that are 200 bp apart would show the smallest distance in cM. d. The EcoRI site and the HindIII site that are 300 bp apart would show the smallest distance in cM. e. The EcoRI sites that are 150 bp apart would show the smallest distance in cM. 53. A new species of plant was discovered and its genome was sequenced. A BLAST search was performed on a small section of one of the chromosomes, and one gene in this region showed 98% similarity to a gene in Arabidopsis. Based on this information, which of the following statements is FALSE? a. The gene in the new species of plant and the gene in Arabidopsis are homologous. b. The gene in the new species of plant and the gene in Arabidopsis likely encode proteins with similar functions. c. The gene in the new species of plant and the gene in Arabidopsis are paralogs. d. The new species of plant and Arabidopsis are evolutionarily related. e. The new species of plant and Arabidopsis likely have a common ancestor.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 20_7e 54. In the following list of cloned fragments, which are the fragments needed to make the longest possible contig with the least amount of overlap?

a. A, B, F b. A, B, C, E c. B, C, D, F d. B, C, D, E, F e. A, B, C, D, E, F 55. Which class of genes is the result of an ancient gene duplication? a. orthologs b. paralogs c. heterologs d. pseudologs e. lincolnlogs 56. Prokaryotes with large genomes: a. are endosymbiotic. b. live in constant environments. c. require less time for cell division. d. live in complex habitats. e. have multiple linear chromosomes.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 20_7e 57. For a genetic map of a chromosome, distances are measured in units of: a. restriction sites. b. RFLPs. c. centiMorgans. d. base pairs. e. contigs.

Indicate one or more answer choices that best complete the statement or answer the question.

58. Choose the statements correctly describing the differences between map-based sequencing and whole genome shotgun sequencing. a. Map-based sequencing requires the assembly of contigs whereas whole genome shotgun sequencing does not. b. Map-based sequencing relies on sequence overlaps whereas whole genome shotgun sequencing does not. c. Currently, map-based sequencing is the primary method used for sequencing of genomes. d. Map-based sequencing will have lower sequence coverage on average than whole genome shotgun sequencing. e. Map-based sequencing and whole genome shotgun sequencing both require computing power to assemble genome sequences.

59. Explain why the greatest diversity of human SNPs is found among African populations.

60. Below is a genetic map for three loci. Which two of these loci would show the smallest physical distance in base pairs, if the frequency of crossing over is equal in this chromosome region? Explain your answer.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 20_7e 61. Compare the fields of structural, functional, and comparative genomics. What is the purpose of each?

62. How can a BLAST search reveal information about protein function?

63. The transcriptome of a genome contains more components than the proteome. Explain why this is true.

64. Compare orthologs and paralogs. Include one feature they have in common and one way in which they are different.

65. Briefly describe how RNA sequencing is carried out.

66. Explain why genetic and physical map distances may differ in relative distances between two genes on a chromosome.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 20_7e 67. Construct a genetic map for a region of a genome, given the following information: the observed recombination frequency between genes D and E is 10%. The observed recombination frequency between genes D and F is 13%. The observed recombination frequency between genes E and F is 23%. Draw the map given by this information.

68. You obtain six BACs (of known order, shown below) and six sequence-tagged sites (STSs, of unknown order) derived from an unfinished portion of the genome sequence of Monodelphis domestica, the gray short-tailed opossum.

Using PCR, you test each BAC for the presence (+) or absence (–) of each of the STSs. You obtain the following results:

Construct a physical map that shows the order of the STSs and their approximate locations within each BAC.

69. There is clear scientific evidence that many species may become extinct within a few decades. Describe at least two human activities that have contributed to the threat to biodiversity, and for each suggest an activity that may slow or reverse the dangerous trend.

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 20_7e 70. Describe how a reporter sequence could be used to reveal information about gene expression.

71. How has the study of microbiomics shed light on the understanding of obesity?

72. To test for gene expression patterns in different tissue types, mRNA is prepared from mouse cells that are precursors to skin cells and from mouse cells that are precursors to red blood cells. The mRNA from each cell type is converted to cDNA and hybridized to a microarray. Spots containing unique sequences from five genes of the microarray are shown in the following diagram. Color them in as specified.

Vertical stripes for hybridization to skin cell cDNA only Horizontal stripes for hybridization to red blood cell cDNA only Filled in completely for hybridization to both cDNA types Blank for hybridization to neither skin nor red blood cell cDNA

73. You have determined that a bacterial strain you are working with contains a single type of plasmid. You isolate the plasmid DNA and digest separate portions of it with each of two different restriction enzymes, BamHI and HpaI, and also perform a double digest using both enzymes. You then fractionate the enzyme digests on an agarose gel and stain the gel with ethidium bromide to visualize the restriction fragment patterns. Your results are shown below. Size markers (in nucleotide base pairs) are indicated at the left side of the gel. Using these data, construct a possible restriction map for the plasmid.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 20_7e 74. Explain why large-scale structural proteomics studies are challenging at present. Describe one method that might move this discipline forward.

75. The transcription factor Twist helps specify development of the mesoderm in the Drosophila embryo. You wish to identify other genes that Twist (as a transcription factor) switches on or off. You have wild-type embryos available, as well as embryos that are mutant for the Twist gene. Design an experiment using microarrays that could allow you to identify genes switched on or off by the Twist transcription factor.

76. What is metagenomics, and what important issues can it address about microbial communities?

77. In general, what kind of information can be obtained from homologous genes?

78. What role might segmental duplication play in evolution?

79. How has metagenomics of ocean samples contributed to our understanding of energy in biological systems?

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 20_7e 80. How are genetic maps different from physical maps?

81. Describe one major difference in the organization or content of prokaryotic and eukaryotic genomes.

82. How are SNPs used to search for genes causing disease?

Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 20_7e Answer Key 1. a 2. b 3. a 4. e 5. a 6. c 7. e 8. a 9. d 10. b 11. d 12. b 13. e 14. b 15. b 16. d 17. d 18. a 19. b 20. a 21. a 22. d 23. a 24. b 25. e 26. d Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 20_7e 27. a 28. a 29. a 30. e 31. b 32. a 33. b 34. a 35. b 36. a 37. a 38. d 39. e 40. a 41. a 42. a 43. e 44. c 45. c 46. d 47. e 48. a 49. d 50. d 51. c 52. c 53. c 54. c Copyright Macmillan Learning. Powered by Cognero.

Page 30

Name:

Class:

Date:

Chap 20_7e 55. b 56. d 57. c 58. a, d, e 59. Since neutral (or nearly neutral) mutations accumulate over time, the most ancient populations have the greatest genetic diversity. Evidence from many sources points to an African origin of the human species, which would be consistent with the great SNP diversity found among modern-day Africans.

60. A and B. If the frequency of crossing over is equal in a region, closer distances in base pairs means less chance of crossing over between two linked loci.

61. Structural genomics is concerned with the actual base sequence of a genome and the organization of genes and other sequences. The purpose is to determine the sequence of a genome and produce maps that show the physical relationship between functional parts like genes, gene regulation elements, and centromeres.

Functional genomics is concerned with the RNAs and proteins encoded by the sequences determined by the structural genomic studies. The purpose is to determine which parts of the genome encode RNAs and proteins as well as functions of each RNA or protein.

Comparative genomics takes the information from structural and functional genomic studies of different organisms and compares the sequences and functional parts of the genomes. The purpose is to find motifs that are common to different organisms and to find divergences between similar organisms that can clarify their differences.

62. The sequence of a protein-encoding gene whose function is unknown can be entered into a BLAST search. This search will compare the unknown sequence to known sequences in the database. If the BLAST search reveals that there is a gene with high sequence similarity in another species, it is highly probable that the unknown sequence has a similar protein function as the protein revealed in the BLAST search.

63. The transcriptome includes all mRNAs that will be transcribed into proteins that are represented in the proteome.

The transcriptome also includes RNAs that are not mRNAs and will not be translated, like tRNAs, rRNAs, and other transcripts that function as RNA molecules.

64. Common features: Both orthologs and paralogs are homologs.

Both orthologs and paralogs have similar sequences and encode RNAs or proteins of similar or the same function.

Differences: Orthologs are homologs in different organisms.

Paralogs are homologs in the same organism.

Copyright Macmillan Learning. Powered by Cognero.

Page 31

Name:

Class:

Date:

Chap 20_7e 65. First, the cells of interest are lysed and the RNA is extracted. Next, the type of RNA of interest is isolated. For example, if sequencing protein-encoding RNA is the goal, then the mRNA is separated from the other RNAs. This RNA is converted into cDNA using reverse transcriptase. The cDNA is then fragmented into overlapping sequences of approximately 200 bp in length and sequenced. The cDNA sequences are then assembled using the overlapping regions.

66. Genetic maps are based on recombination distances, and these are not consistent in all portions of the genome.

Recombination may be suppressed in one region of the chromosome or enhanced in another, which may skew the calculated distance. Because physical maps are based on actual DNA base-pair information, they provide a more precise distance measure.

67.

68. The order of the STSs is shown in the diagram below. The exact location of the STSs within the BACs cannot be determined from the data provided. For instance, we know that STS 6 is within the region of overlap between BACs A, B, and C, but we cannot be more specific than that. For STS 4, we can be slightly more specific because the region of overlap between BACs C, D, and E is smaller, and therefore the range of possible locations of STS 4 is smaller, as well.

69.

70. Genomic fragments are cloned into vectors. The coding region of the gene of interest is replaced with a reporter sequence, such as the gene that encodes GFP, but the gene's regulatory sequences are left intact. The vector is then inserted into an embryo, creating a transgenic organism. The organism is then observed for the presence of the GFP, which fluoresces under UV light. Because the reporter sequence (GFP) is under the regulatory controls of the gene of interest, the location of the fluorescence reveals the location that the gene of interest would normally be expressed.

71. One study examined the gut microflora of obese and lean people. Two groups of bacteria are common in the human gut: Bacteroidetes and Firmicutes. Researchers discovered that obese people have relatively more Firmicutes than do lean people and that the proportion of Firmicutes decreases in obese people who lose weight on a low-calorie diet. These same results were observed in obese and lean mice. In an elegant experiment, researchers transferred bacteria from obese to lean mice. The mice that received bacteria from obese mice extracted more calories from their food and stored more fat, suggesting that gut microflora might play some role in obesity.

Copyright Macmillan Learning. Powered by Cognero.

Page 32

Name:

Class:

Date:

Chap 20_7e

72.

73.

Copyright Macmillan Learning. Powered by Cognero.

Page 33

Name:

Class:

Date:

Chap 20_7e 74. Large-scale structural proteomics studies are difficult because X-ray crystallography and NMR, the methods used to determine high-resolution structures, require significant preparatory work and scientist intervention. Thus, determining the structure of a single protein can be time consuming, let alone the proteins in an entire proteome.

Answers may vary with respect to suggesting a method to move the discipline forward. Gaining an understanding of sequence/structure relationships and building computer algorithms to predict structures based on these relationships will be helpful. Additionally, it would be helpful if methods can be devised to quicken the pace of structure determination and/or allow for the determination of several structures at once. Currently, homology identification with protein sequences with known structure can help with structural proteomics.

75. Microarrays allow you to measure the expression levels of all the genes within an organism. Therefore, it would be possible to compare the expression level of each gene in wild-type animals with its expression level in Twist mutant animals. Since these animals lack functional Twist transcription factor, any observed changes in expression between the two groups should be due to the absence of Twist. Genes whose expression level increases in the mutant are likely to be negatively regulated (switched off) by Twist since, in its absence, their levels go up. Genes whose expression level decreases in the mutant are likely to be activated (switched on) by Twist. Unfortunately, we cannot tell from these microarray experiments if Twist directly switches on or off all of the genes identified—it is still possible that a gene Twist regulates (gene Y) can then in turn regulate its own target gene (gene Z), so that the loss of Twist causes a change in gene Y expression, leading to a change in gene Z expression without Twist directly switching on or off gene Z. 76. Metagenomics is a field within genetics that sequences the genomes of an entire group of organisms that inhabit a common environment. This process can allow the study of microbes that are not able to be cultured in the laboratory and can assist in studying the community structure of microorganisms.

77. Homologous genes often have related functions, so the function of an unknown gene can be deduced if the function of its homologs is known.

78. After a segmental duplication arises, the original copy of the gene can continue its function while the new copy undergoes mutation. These changes may eventually lead to new functions.

79. Analyses of bacterial genomes found in ocean samples led to the discovery of proteorhodopsin proteins, which are light-driven proton pumps. Subsequent research demonstrated that these proteins are found in diverse microbial groups and are a major source of energy flux in the world's oceans.

80. Genetic maps are created using genetic recombination frequencies and are measured in map units or centiMorgans, while physical maps use techniques that determine the physical distance and are measured in base pairs. Physical maps are more accurate and have higher resolution than genetic maps.

Copyright Macmillan Learning. Powered by Cognero.

Page 34

Name:

Class:

Date:

Chap 20_7e 81. (1) Prokaryotic genomes generally have less DNA and fewer genes than eukaryotic genomes.

(2) Prokaryotic genomes have fewer repeated sequences and noncoding, intragenic sequences than eukaryotic genomes.

(3) Most prokaryotic genomes are contained in one chromosome; most eukaryotic genomes are contained on several chromosomes.

(4) The prokaryotic genome shows evidence of much horizontal gene transfer.

(5) The eukaryotic genome shows evidence of much gene duplication.

(6) In general, eukaryotic genomes contain many introns, repeated sequences, and transposable elements.

82. When an SNP is physically close to a disease-causing locus, it will tend to be inherited along with the diseasecausing allele. People with the disease will tend to have different SNPs from those of healthy people. A comparison of SNP haplotypes in people with a disease and in healthy people can reveal the presence of genes that affect the disease; because the disease gene and the SNP are closely linked, the location of the disease-causing gene can be determined from the location of associated SNPs.

Copyright Macmillan Learning. Powered by Cognero.

Page 35

Name:

Class:

Date:

Chap 21_7e Indicate the answer choice that best completes the statement or answers the question.

1. A group of scientists is studying memory in mice. They feed sodium butyrate, a histone deacetylase inhibitor, to a group of older mice and then subject them to memory exercises. How do you predict the memory of treated mice might compare with the memory of untreated mice? a. Histone acetylation increases with age, so sodium butyrate may improve memory in older mice. b. Histone acetylation increases with age, so sodium butyrate may impair memory in older mice. c. Histone acetylation decreases with age, so sodium butyrate may improve memory in older mice. d. Histone acetylation decreases with age, so sodium butyrate may impair memory in older mice. e. Histone acetylation is not associated with memory in mice.

2. Which of the following molecular processes leading to epigenetic changes exclusively represses gene expression? a. DNA methylation b. histone acetylation c. histone methylation d. RNA molecules e. All of the above.

3. Which of the following is NOT affected by genomic imprinting? a. Angelman syndrome b. Prader-Willi syndrome c. Beckwith-Wiedemann syndrome d. Igf2 alleles in mice e. Down syndrome 4. Which of the following is NOT true of imprinted loci? a. Some imprinted alleles have biased expression, rather than being all on or off. b. Imprinting occurs in clusters of 3 to 12 genes. c. Imprinted loci contain protein-coding and noncoding RNA genes. d. Imprinted loci occur on sex chromosomes but not on autosomes. e. Some alleles are imprinted only in certain tissues or at certain times of development.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 21_7e 5. Which of the following is NOT true of bisulfite sequencing? a. Methylated cytosines are unaffected by sodium bisulfite. b. DNA is treated with sodium bisulfite, which chemically converts unmethylated cytosines to uracil. c. Unmethylated cytosines are detected as thymine during sequencing. d. All thymines revealed in DNA sequencing of bisulfite treated DNA were cytosines in the original, untreated DNA. e. The sequence of genomic DNA treated with sodium bisulfite is compared to the sequence of untreated genomic DNA.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 21_7e You are studying epigenetic regulation of the Bdnf gene in mice and its association with memory. You train mouse B to avoid an adverse stimulus and then isolate genomic DNA from its brain. Mouse A is untrained and used as a control. You take the brain DNA samples and split each of them into two parts. One part is treated with sodium bisulfite and the other part is left untreated. You then perform PCR for the Bdnf locus and digest the PCR product with the restriction enzyme AccII that cuts DNA at the sequence CGCG. The digested DNA is then separated by agarose gel electrophoresis to obtain the following DNA fragments: Mouse A (Bisulfite DNA) 21.2 kb 7.4 kb 5.8 kb 5.7 kb 4.9 kb 3.5 kb

Mouse A (Untreated DNA) 19.3 kb 7.4 kb 5.1 kb 3.9 kb 3.5 kb 2.7 kb 2.2 kb 1.9 kb 1.8 kb 0.7 kb

Mouse B (Bisulfite DNA) 19.3 kb 7.4 kb 5.7 kb 5.1 kb 4.9 kb 3.5 kb 1.9 kb 0.7 kb

Mouse B (Untreated DNA) 19.3 kb 7.4 kb 5.1 kb 3.9 kb 3.5 kb 2.7 kb 2.2 kb 1.9 kb 1.8 kb 0.7 kb

6. How many of the AccII sites were methylated at the Bdnf locus in the trained mouse? a. 1 b. 2 c. 3 d. 4 e. 5

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 21_7e 7. Which of the following stimulates the transcription of Xist on the inactive X chromosome? a. Xist b. Tsix c. Jpx d. Xite e. Xist is not transcribed on the inactive X chromosome.

8. How does histone acetylation affect chromatin structure? a. neutralizes positive histone charges and results in more tightly packed chromatin b. neutralizes negative histone charges and results in more tightly packed chromatin c. increases positive histone charges and results in more tightly packed chromatin d. neutralizes positive histone charges and results in more loosely packed chromatin e. neutralizes negative histone charges and results in more loosely packed chromatin 9. Consider a gene in mice that is normally maternally imprinted. If a deletion occurred in the gene, rendering it nonfunctional, what would be the expected phenotypic result? a. If the mutation occurred in the maternal copy of the gene, the mouse would be wild-type, but if the mutation occurred in the paternal copy of the gene, the mouse would be mutant. b. If the mutation occurred in the maternal copy of the gene, the mouse would be mutant, but if the mutation occurred in the paternal copy of the gene, the mouse would be wild type. c. If the gene behaves in a dominant manner, the mouse would be mutant, but if the gene behaves in a recessive manner, the mouse would be wild type. d. Because the gene is maternally imprinted, all female mice would be mutant, but all male mice would be wild type. e. Because the gene is maternally imprinted, all female mice would be wild type, but all male mice would be mutant.

10. Which of the following inhibits the expression of Xist on the active X chromosome? a. Xist b. Tsix c. Jpx d. Xite e. X-inactivation center

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 21_7e 11. The promoter of gene A has histones that are acetylated on lysine 27 of histone H3, and the promoter of gene B has histones that are methylated on lysine 4 of histone H3. Which gene is being actively transcribed? a. gene A, because histone acetylation is associated with gene expression b. gene A, because histone acetylation is associated with gene silencing c. gene B, because histone methylation is associated with gene expression d. gene B, because histone acetylation is associated with gene silencing e. Both genes are actively transcribed because both of those marks are associated with gene expression.

12. What effect would the deamination of 5-methylcytosine in a promoter have on the expression of that gene? a. The gene will have increased expression because this chemical reaction will result in unmethylated cytosine. b. The gene will have increased expression because this chemical reaction will result in thymine. c. The gene will have increased expression because this chemical reaction will result in uracil. d. The gene will have decreased expression because this chemical reaction will result in unmethylated cytosine. e. The gene will have decreased expression because this chemical reaction will result in uracil.

13. Which of the following is an example of "crosstalk" in epigenetics? a. DNA methylation leads to histone deacetylation. b. DNA methylation leads to histone acetylation. c. H2B ubiquitination leads to H3K79 methylation. d. DNA methylation leading to either histone acetylation or deacetylation. e. DNA methylation leads to histone acetylation and H2B ubiquitination leads to H3K79 methylation.

14. Which statement BEST described how imprinting varies over the course of development? a. Imprinted alleles are permanently silenced from embryo development through adulthood. b. Imprinting is only active during embryo development, but expression returns to normal during adulthood. c. Imprinting is only active during gametogenesis, where alleles are selectively silenced before being transmitted to offspring. d. Imprinting randomly flickers on and off during development. e. Imprinted alleles differ by tissue type and time of development.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 21_7e 15. In a female human cell, what would be the effect on X inactivation if a mutation occurred in the Xist gene making it nonfunctional? a. One of the two X chromosomes would become inactivated. b. Both of the X chromosomes would become inactivated. c. Neither of the X chromosomes would become inactivated. d. Only the maternal X chromosome would become inactivated. e. Only the paternal X chromosome would become inactivated.

16. How does the DNA methylation pattern compare between embryonic stem cells, somatic cells, and induced pluripotent stem cells? a. Embryonic stem cells show the highest amount of DNA methylation, while somatic cells show the lowest. b. Embryonic stem cells show the highest amount of DNA methylation, while induced pluripotent stem cells show the lowest. c. Induced pluripotent stem cells have less DNA methylation than differentiated cells but more than embryonic stem cells. d. Induced pluripotent stem cells show the highest amount of DNA methylation, while somatic cells show the lowest. e. Somatic cells show the highest amount of DNA methylation, while induced pluripotent stem cells show the lowest.

17. Which of the following is NOT a defining feature of paramutation? a. The expression pattern is transmitted to future generations. b. The altered allele is able to convert other alleles. c. No DNA sequence differences exist between alleles. d. Mutation of coding sequences generates new alleles. e. These are all defining features of paramutation.

18. What is the molecular basis of the epigenetic effect of royal jelly? a. inhibits Dnmt3, causes less DNA methylation b. inhibits Dnmt3, causes more DNA methylation c. inhibits Hdac3, causes less histone acetylation d. inhibits Dnmt3, causes less histone methylation e. inhibits Dnmt3, causes more histone methylation

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 21_7e 19. Which of the following statements is NOT true of DNA methylation? a. It is associated with transcription repression. b. It occurs on cytosines adjacent to guanines. c. Cytosines on both strands are usually methylated. d. It is carried out by DNA methyltransferases. e. Methylation is permanent and cannot be reversed.

20. How do epigenetic marks compare in monozygotic twins? a. They are similar early in life but are increasingly dissimilar with age. b. They are different at birth, but a similar environment during childhood causes them to become more similar with time. c. They stay the same throughout their lives because monozygotic twins are genetically identical. d. DNA methylation patterns are similar, but histone acetylation patterns are very different.

21. A beekeeper notices that her bee population is declining rapidly. Upon further study, she notices that there are no queen bees present in her population. She suspects this is due to exposure to an environmental agent.

Which of the following could be a possible mechanism for the environmental agent? a. The environmental agent inhibits Dnmt3. b. The environmental agent causes royal jelly to be produced in abnormally high amounts. c. The environmental agent mimics a Dnmt3 siRNA. d. The environmental agent is causing extensive chromosome rearrangements within the genome of the queen bees. e. The environmental agent is a methyltransferase enzyme that acts on genes that encode queen bee characteristics.

22. A DNA fragment with the following base sequence has some cytosine bases that are methylated (indicated by C* ) and others that are unmethylated. To determine the location of methylated and unmethylated cytosines, researchers sequenced this fragment both with and without treatment with sodium bisulfite. Give the sequence of bases that will be read with and without bisulfite treatment. a. With treatment: —AAAGGGCTTTTT— Without treatment: —AAAGGGCCCTTT— b. With treatment: —AAAGGGCCCTTT— Without treatment: —AAAGGGCTTTTT— c. With treatment: —AAAGGGTCCTTT— Without treatment: —AAAGGGCTTTTT— d. With treatment: —TTTCCCGGGAAA— Without treatment: —AAAGGGCTTTTT— e. With treatment: —AAAGGGGGGTTT— Without treatment: —AAAGGGCCCTTT—

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 21_7e You are studying epigenetic regulation of the Bdnf gene in mice and its association with memory. You train mouse B to avoid an adverse stimulus and then isolate genomic DNA from its brain. Mouse A is untrained and used as a control. You take the brain DNA samples and split each of them into two parts. One part is treated with sodium bisulfite and the other part is left untreated. You then perform PCR for the Bdnf locus and digest the PCR product with the restriction enzyme AccII that cuts DNA at the sequence CGCG. The digested DNA is then separated by agarose gel electrophoresis to obtain the following DNA fragments: Mouse A (Bisulfite DNA) 21.2 kb 7.4 kb 5.8 kb 5.7 kb 4.9 kb 3.5 kb

Mouse A (Untreated DNA) 19.3 kb 7.4 kb 5.1 kb 3.9 kb 3.5 kb 2.7 kb 2.2 kb 1.9 kb 1.8 kb 0.7 kb

Mouse B (Bisulfite DNA) 19.3 kb 7.4 kb 5.7 kb 5.1 kb 4.9 kb 3.5 kb 1.9 kb 0.7 kb

Mouse B (Untreated DNA) 19.3 kb 7.4 kb 5.1 kb 3.9 kb 3.5 kb 2.7 kb 2.2 kb 1.9 kb 1.8 kb 0.7 kb

23. What can you conclude about the epigenetic effect of learning on the Bdnf locus in mice? a. DNA methylation decreases, causing a decrease in Bdnf expression. b. DNA methylation increases, causing a decrease in Bdnf expression. c. DNA methylation decreases, causing an increase in Bdnf expression. d. DNA methylation increases, causing an increase in Bdnf expression. e. DNA methylation does not alter Bdnf expression.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 21_7e 24. How many AccII sites are found at the Bdnf locus? a. 4 b. 6 c. 9 d. 10 e. 16 25. Which of the following is NOT true of honeybees? a. Queen bees and worker bees are both female. b. Worker bees are smaller than queen bees. c. Queen bees develop by being fed royal jelly. d. Worker bees produce eggs to maintain the hive population. e. Queen bees and worker bees are genetically identical.

26. What histone modifications are associated with DNA repair? a. H3K27 acetylation b. H3K79 methylation c. H2B ubiquitination d. H3K27 acetylation and H2B ubiquitination e. H3K79 methylation and H2B ubiquitination 27. What effect does histone acetylation have on transcription? a. decreases expression b. decreases translation but not transcription c. no change in expression d. increases expression e. either increases or decreases expression 28. You are scientist who studies the epigenetic effects of maternal behavior in rats. Which of the following experimental manipulations will lead to an INCREASED stress response in young rats whose mothers lick and groom them? a. injection of glucocorticoid agonist (that would stimulate the glucocorticoid receptor) into young rat brains b. injection of histone deacetylase inhibitors into young rat brains c. injection of DNA demethylase inhibitors into young rat brains d. feeding glucocorticoid receptor inhibitors during early postnatal development

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 21_7e 29. What epigenetic marks occur during X inactivation? a. H3K27 acetylation b. H3K79 methylation c. H2B ubiquitination d. H3K27 methylation e. H3S10 phosphorylation 30. When does maternal X inactivation occur in mice? a. at fertilization b. eight-cell stage of embryo c. blastocyst maturation d. at birth e. within 10 days after birth 31. How would you expect DNA methylation to alter gene expression? a. greatly increase expression b. moderately increase expression c. no change in expression d. measurably decrease expression e. either increase or decrease expression 32. How is the epigenetic trait induced by vinclozolin transmitted to offspring? a. DNA methylation is increased in sperm of treated males, and this methylation pattern is inherited by offspring. b. Treated males are more successful at mating, so more offspring inherit the trait. c. Females exposed to the drug produce fewer eggs, and the resulting offspring exhibit this same phenotype. d. Treated females have increased DNA methylation in neuronal cells, which alters their mating behaviors.

33. How does genomic imprinting affect gene expression? a. increases expression of imprinted alleles b. lowers expression of imprinted alleles c. completely silences expression of imprinted alleles d. increase of decrease expression of imprinted alleles, depending on the situation e. Genomic imprinting can lower the expression of imprinted alleles or completely silence them.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 21_7e 34. What type of enzyme would MOST likely promote increased gene expression? a. DNA methyltransferase b. DNA demethylase c. DNA polymerase d. DNA ligase e. DNAse 35. Consider a hypothetical gene in mice called the can gene. The mutant form of this gene causes skin cancer and behaves in a dominant manner. A mouse with skin cancer was studied to determine the location of the mutation in the can gene. However, when scientists sequenced both maternal and paternal copies of the gene, they could not identify any mutation in the gene. What is a possible explanation for the mouse's skin cancer? a. One of the copies of the can gene could be hypermethylated and thus transcriptionally inactive in the mouse. b. One of the copies of the can gene could have an inversion that renders it nonfunctional. c. Both copies of the can gene must have the exact same mutation and, therefore, the sequencing reaction could not detect the location of the mutation. d. One of the copies of the can gene could be deacetylated and thus transcriptionally inactive in the mouse. e. The only possible explanation is that it is a different gene at a different locus that is causing the skin cancer in the mouse.

36. Which of the following stimulates the transcription of Tsix on the active X chromosome? a. Xist b. Tsix c. Jpx d. Xite e. Tsix is not transcribed on the active X chromosome.

37. What do you expect would develop if you injected Dnmt3 siRNA into honeybee larvae? a. a male drone b. a female worker c. a female queen d. a male drone or a female worker e. a female worker or a queen

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 21_7e 38. What type of RNA is Xist? a. lncRNA b. mRNA c. rRNA d. siRNA e. tRNA 39. Kit+Kit+ mice were crossed with Kit+Kitt mice. The embryos resulting from this cross were then injected with miRNAs specific to Kit mRNA. What phenotype will these embryos develop to have? a. All mice will have the wild-type phenotype. b. All mice will have the mutant phenotype. c. The Kit+Kit+ mice will have the wild-type phenotype, and the Kit+Kitt mice will have the mutant phenotype. d. The Kit+Kit+ mice will have more mutant phenotypes than wild-type phenotypes, and the Kit+Kitt mice will have the mutant phenotype. e. The Kit+Kit+ mice will have the wild-type phenotype, and the Kit+Kitt mice will have a 1:1 ratio of the wild-type to mutant phenotype.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 21_7e You are studying epigenetic regulation of the Bdnf gene in mice and its association with memory. You train mouse B to avoid an adverse stimulus and then isolate genomic DNA from its brain. Mouse A is untrained and used as a control. You take the brain DNA samples and split each of them into two parts. One part is treated with sodium bisulfite and the other part is left untreated. You then perform PCR for the Bdnf locus and digest the PCR product with the restriction enzyme AccII that cuts DNA at the sequence CGCG. The digested DNA is then separated by agarose gel electrophoresis to obtain the following DNA fragments: Mouse A (Bisulfite DNA) 21.2 kb 7.4 kb 5.8 kb 5.7 kb 4.9 kb 3.5 kb

Mouse A (Untreated DNA) 19.3 kb 7.4 kb 5.1 kb 3.9 kb 3.5 kb 2.7 kb 2.2 kb 1.9 kb 1.8 kb 0.7 kb

Mouse B (Bisulfite DNA) 19.3 kb 7.4 kb 5.7 kb 5.1 kb 4.9 kb 3.5 kb 1.9 kb 0.7 kb

Mouse B (Untreated DNA) 19.3 kb 7.4 kb 5.1 kb 3.9 kb 3.5 kb 2.7 kb 2.2 kb 1.9 kb 1.8 kb 0.7 kb

40. How many of the AccII sites were methylated at the Bdnf locus in the untrained mouse? a. 1 b. 2 c. 3 d. 4 e. 5

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 21_7e 41. Which of the following stimulates the transcription of Xist on the active X chromosome? a. Xist b. Tsix c. Jpx d. Xite e. Xist is not transcribed on the active X chromosome.

42. What protein factors are recruited by Xist that causes epigenetic silencing of the X chromosome? a. RNA polymerases b. DNA polymerases c. polycomb group proteins d. histone acetyltransferases e. DNA demethylases 43. Which of the following is NOT an example of epigenetics? a. independent assortment of alleles b. genomic imprinting c. miRNAs d. paramutation e. X inactivation 44. You are a researcher trying to understand the epigenetic mechanisms of queen bee development. Which of the following experimental manipulations would probably NOT lead to the development of a queen bee? a. feeding royal jelly to a female honeybee larva b. feeding decitabine, a DNA methyltransferase inhibitor, to a female larva c. injecting Dnmt3 siRNA into a female honeybee larva d. activation of a DNA cytosine deaminase enzyme in a female larva e. feeding royal jelly to an adult worker honeybee 45. How long is Xist RNA? a. 21 bp b. 500 bp c. 1500 bp d. 17,000 bp e. 230,000 bp

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 21_7e 46. Which of the following inhibits the expression of Xist on the inactive X chromosome? a. Xist b. Tsix c. Jpx d. Xite e. Xist expression is not inhibited on the inactive X chromosome.

47. What characteristic of DNA allows methylation patterns to be maintained through replication and cell division? a. semiconservative replication b. recombination and repair c. the absence of uracil d. topoisomerases e. deoxyribonucleotide synthesis Indicate one or more answer choices that best complete the statement or answer the question.

48. When does paternal X inactivation occur in mice? (Select all that apply.) a. at fertilization b. eight-cell stage of embryo c. blastocyst maturation d. at birth e. within 10 days after birth 49. How do enzymes and proteins that are unable to bind directly to DNA introduce epigenetic marks on the DNA? a. They may interact with transcription factors bound to DNA. b. They may bind to pre-existing chromatin marks. c. They may interact with DNA Polymerase during semiconservative replication. d. They may interact the methyl groups on CpG islands. e. They may interact with noncoding RNA molecules.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 21_7e 50. In corn, red kernels (PP) are typically dominant to white kernels (pp). A red plant was crossed with a white plant to produce a Pp heterozygote with red kernels. However, in one ear of corn, a partially white kernel appeared unexpectedly. This kernel was grown into an adult plant, and for several generations of crosses, the offspring of this original kernel all had a mix of partially red and partially white kernels. However, when the seeds were grown in the presence of a demethylating drug, all of the resulting plants had red kernels. Explain why the mixed-color kernel appeared, why the offspring of this kernel also had the mixed-color phenotype, and why the demethylating drug had its effect.

51. Compare the effects of the environment on the epigenetics of glucocorticoid receptor expression in mice and humans.

52. Explain how the b1 locus in corn exhibits paramutation.

53. List three types of molecular mechanisms that alter chromatin structure and underlie many epigenetic phenotypes.

54. Explain how bisulfite sequencing helps researchers detect methylated cytosines.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 21_7e 55. Explain how studies in mice suggest that diet can cause epigenetic changes that are transmitted through generations.

A study of adults who developed prenatally during Dutch Hunger Winter, a period of severe caloric restriction, demonstrates caloric restriction during prenatal development lead to epigenetic changes that increased in the incidence of adult obesity.

56. What was the biochemical basis of the epigenetic change observed?

57. Describe three important features that define paramutation.

58. Explain how DNA methylation is passed on to daughter cells.

A study of adults who developed prenatally during Dutch Hunger Winter, a period of severe caloric restriction, demonstrates caloric restriction during prenatal development lead to epigenetic changes that increased in the incidence of adult obesity.

59. Was the observed level of change (i.e. increase or decrease) consistent amongst the loci examined? If not, suggest why.

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 21_7e 60. What is the "genetic conflict hypothesis," and how does it relate to genomic imprinting?

61. How does DNA methylation contribute to the repression of transcription?

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 21_7e Answer Key 1. c 2. a 3. e 4. d 5. d 6. b 7. c 8. d 9. a 10. b 11. e 12. b 13. d 14. e 15. c 16. c 17. d 18. a 19. e 20. a 21. e 22. a 23. c 24. c 25. d 26. e Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 21_7e 27. d 28. a 29. d 30. c 31. d 32. a 33. e 34. b 35. a 36. d 37. c 38. a 39. d 40. d 41. e 42. c 43. a 44. e 45. d 46. e 47. a 48. b, c 49. a, b, e

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 21_7e 50. This type of highly variable, mottled, mixed phenotype is typical of an epigenetic phenomenon of semistable methylation. In this case, when the P gene is methylated, its expression is reduced, causing the white portions of kernels. Methylation states are heritable, so the white phenotype can be passed along. However, methylation is not stable, like a change in DNA sequence is, and methylation can be gained or lost in subsequent generations. With that instability in mind, the P gene appears to switch between active and inactive states, causing the mottled or mixed appearance. More methylation may correlate with less P expression and less pigment. Lower methylation may correlate with more P expression and more pigment. Growing the corn on the demethylating drug would prevent methylation of the P gene, so it would always be active, resulting in a reappearance of the completely redpigmented kernels.

51. A mother rat licks and grooms her offspring while nursing them. Offspring exposed to more licking and grooming develop a different pattern of DNA methylation compared with offspring exposed to less licking and grooming.

These differences in DNA methylation affect the acetylation of histone proteins that persist into adulthood and alter the expression of the glucocorticoid receptor gene, which plays a role in hormonal responses to stress.

Numerous studies have demonstrated that stress during childhood and adolescence produce a number of adverse effects that persist into adult life. Those who experienced childhood abuse had a greater degree of methylation of the glucocorticoid receptor gene, a gene involved in the stress response, than those who had not experienced abuse.

52. Both the B-I and B' alleles have tandem repeats, but the chromatin structure of the two alleles differs: The B-I allele has more open chromatin. The tandem repeats are required for high expression of the B-I allele and pigment production, but only when the chromatin surrounding the repeats is in an open configuration. The more closed configuration of the B' allele may prevent the repeats from interacting with the promoter of b1 and stimulating transcription. The tandem repeats that are required for paramutation encode 25-nucleotide-long siRNAs. Some siRNAs are known to modify chromatin structure by directing DNA methylation to specific DNA sequences.

53. (1) changes in patterns of DNA methylation (2) chemical modifications of histone proteins (3) RNA molecules that affect chromatin structure and gene expression 54. Genomic DNA is first treated with sodium bisulfite, which chemically converts unmethylated cytosine to uracil.

Uracils are then detected as thymine during sequencing. However, 5-methylcytosine is not chemically altered by treatment with bisulfite and is detected as cytosine during sequencing. By sequencing genomic DNA with and without bisulfite treatment, researchers are able to determine the locations of all copies of 5-methylcytosine in the DNA.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 21_7e 55. In one study, researchers fed male mice either a normal (control) diet or a diet low in protein. They then bred mice in both groups to control females fed a normal diet. The males were then separated from the females and never had any contact with their offspring. The offspring were raised and their lipid and cholesterol levels examined. The offspring of males fed a low-protein diet exhibited increased expression of genes involved in lipid and cholesterol metabolism and a corresponding decrease in levels of cholesterol, compared to the offspring of males fed a normal diet. They also observed numerous differences in DNA methylation in the offspring of the two types of fathers.

In another study, researchers fed male rats a high-fat diet and they gained weight. They then bred these males to females that had been fed a normal diet. The offspring were also fed a normal diet. The daughters of the male rats on the high-fat diet had normal weight, but as adults they developed a diabetes-like condition of impaired glucose tolerance and insulin secretion.

56. Methylation.

57. (1) The newly established expression pattern of the converted allele is transmitted to future generations, even though the allele that brought about the alteration is no longer present with it.

(2) The altered allele is now able to convert other alleles to the new phenotype.

(3) There are no associated DNA sequence differences in the altered alleles.

58. Methylation of CpG sequences means that two methylated cytosine bases sit diagonally across from each other on opposing strands. Before replication, cytosine bases on both strands are methylated. Immediately after semiconservative replication, the cytosine base on the template strand will be methylated, but the cytosine base on the newly replicated strand will be unmethylated. Special methyltransferase enzymes recognize the hemimethylated state of CpG dinucleotides and add methyl groups to the unmethylated cytosine bases, creating two new DNA molecules that are fully methylated. In this way, the methylation pattern of DNA is maintained across cell division.

59. No, the change was not consistent. Some loci showed an increase in methylation while some showed a decrease in adults born from caloric-restricted mothers.

As discussed in Chapter 17, methylation is associated with transcriptional repression. Loci that showed an increase in methylation might encode gene products that promote higher metabolic rates. Loci that showed a decrease in methylation might encode gene products that promote lower metabolic ranges.

60. The genetic conflict hypothesis suggests that there are conflicting evolutionary pressures on alleles of maternal versus paternal origin with respect to genes that affect early embryonic and fetal growth. For genes that affect offspring size and weight, alleles that maximize these traits will be favored by evolution in the father as these traits are associated with offspring survival. However, larger fetal size and weight can adversely affect the mother because of the risks involved in giving birth to larger babies, and because nutrient depletion in a mother with a large baby may affect her ability to give birth to other babies in the future. Thus, genomic imprinting will evolve in a mother to repress or silence expression of alleles that maximize offspring size and weight.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 21_7e 61. The methyl group of 5-methylcytosine sits within the major groove of the DNA, which is recognized by many DNA binding proteins. The presence of the methyl group in the major groove inhibits the binding of transcription factors and other proteins required for transcription to occur. 5-Methylcytosine also attracts certain proteins that directly repress transcription. In addition, DNA methylation attracts histone deacetylase enzymes, which remove acetyl groups from the tails of histone proteins, altering chromatin structure in a way that represses transcription.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 22_7e Indicate the answer choice that best completes the statement or answers the question.

1. Epigenetic changes in development are important in that the gene expression of cells that make up a particular organ or tissue type is often defined by epigenetic markers. Which of the following epigenetic regulations is known to cause more long-term gene silencing during the course of development? a. chromatin remodeling b. DNA methylation c. histone modifications d. DNA ubiquitination 2. The insect hormone ecdysone is known to trigger apoptosis during metamorphosis that turns larva into adult structures. Which of the following experiments would confirm the activation of apoptosis by ecdysone? a. Inhibition of rpg and hid gene via RNAi should result in excess cells that would normally be eliminated. b. Overproduction of rpg and hid genes via constitutive promoter should prevent the full metamorphosis. c. Inhibition of receptor binding to ecdysone should result in excess apoptosis. d. Overproduction of ecdysone receptor should result in inhibition of rpg and hid genes. e. Inactivation of grim should result in elimination of excess number of cells.

3. A lymphocyte pool plays a role in which process? a. self-antigen generation b. clonal selection c. antibody production d. genetic maternal effect e. major histocompatibility complex (MHC) recognition

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 22_7e The diagram below illustrates the normal expression of three different groups of genes (A, B, and C) within four different whorls of a developing flower. Normally, whorl 1 produces sepals, whorl 2 produces petals, whorl 3 produces stamens, and whorl 4 produces carpels. Furthermore, expression of class A genes inhibits class C genes, and expression of class C inhibits class A.

4. What kind of mutation would produce sepals in whorls 1 and 2 and carpels in whorls 3 and 4? a. defective in A b. defective in B c. defective in C d. defective in A and B e. defective in B and C 5. What would be the identity of the four whorls if gene B was ectopically expressed in whorls 1 and 2? a. sepals, petals, stamens, and carpels b. petals, petals, petals, petals c. petals, petals, stamens, stamens d. stamens, carpels, stamens, carpels e. sepals, petals, sepals, petals

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 22_7e 6. The proteins Dorsal and Bicoid both act as morphogens in Drosophila embryogenesis. However, the establishment of the concentration gradients of each protein differs. Which of the following statements CORRECTLY describes this difference? a. The dorsal gene is transcribed maternally while the bicoid gene is transcribed in the developing embryo. b. The dorsal gene is transcribed in the developing embryo whereas the bicoid gene is transcribed maternally. c. The concentration gradient of Dorsal is established through asymmetric subcellular localization whereas the concentration gradient of Bicoid is established through selective degradation. d. The concentration gradient of Dorsal is established through asymmetric subcellular localization whereas the concentration gradient of Bicoid is established through asymmetric localization of the bicoid mRNA and thus protein. e. The concentration gradient of Dorsal is established through differential rates of translation whereas as the concentration gradient of Bicoid is established through a transport system.

7. Which of the following pairing of genes and developmental processes is INCORRECT? a. bicoid and nanos—anterior posterior axis b. antennapedia complex—head and anterior thoracic segment c. cactus and toll—dorsal development d. hunchback—ventral development e. bithorax complex—posterior thoracic and abdominal segment 8. In wild-type flowering plants, carpel development is determined by what class of gene products? a. A gene products b. A + B gene products c. B + C gene products d. C gene products e. A + C gene products 9. Regulation of apoptosis involves initial activation of the enzymes called _____, which act as proteases that can trigger a cascade of cellular protein degradation, bringing about eventual DNA fragmentation. a. apaptase b. acetylase c. demethylase d. nucleases e. caspases

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 22_7e The diagram below illustrates the normal expression of three different groups of genes (A, B, and C) within four different whorls of a developing flower. Normally, whorl 1 produces sepals, whorl 2 produces petals, whorl 3 produces stamens, and whorl 4 produces carpels. Furthermore, expression of class A genes inhibits class C genes, and expression of class C inhibits class A.

10. What kind of mutation would produce sepals in whorls 1 and 4 and petals in whorls 2 and 3? a. defective in A b. defective in B c. defective in C d. defective in A and B e. defective in B and C 11. The primary function of differentiated B lymphocytes within the vertebrate immune system is to produce: a. cytotoxic chemicals. b. hormones. c. antibodies. d. ROS. e. complements.

12. The product of the bicoid gene is critical for the establishment of _____ structures in the Drosophila embryo. a. anterior b. posterior c. dorsal d. ventral e. segment

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 22_7e 13. Activin is a protein that normally exists in a gradient within certain tissues in developing embryos. Certain undifferentiated cells cultured without activin produce epidermal cells. However, activin induces these same cells to adopt other fates (e.g., blood, muscle, heart), depending on the concentration of activin. Based on these observations, activin acts as a(n): a. transcription factor. b. cell-surface receptor. c. morphogen. d. second messenger. e. activator.

14. The Bicoid protein in fruit flies specifies anterior development. In the absence of Bicoid protein in an egg, the embryo will develop two tails. The bcd– mutation is recessive and exhibits a maternal effect. Which of the following CORRECTLY describes the embryos produced from a cross between a bcd+/ bcd– male and a bcd– /bcd– female? a. All embryos will be normal. b. All embryos will have two tails. c. Three-fourths of the embryos will have two tails. d. One-half of the embryos will have two tails. e. One-fourth of the embryos will have two tails.

15. The development of the nematode worm C. elegans is highly reproducible. During development of the hermaphrodite, 1090 somatic cells are generated, but only 959 somatic cells are found in the mature adult.

What is the MOST likely explanation for the change in cell number given its reproducibility? a. inhibition of genes that stimulate cell division b. activation of genes that inhibit cell division c. apoptosis of cells during development d. necrosis of cells during development e. stochastic variations in cell division 16. Which of the following mechanism does NOT contribute to the diversity of antibody and T-cell receptors recognizing different antigens? a. junctional diversity b. somatic hypermutation c. somatic recombination d. pairing of different light and heavy chains e. pairing of different MHC loci

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 22_7e 17. Which of the following statements about apoptosis is FALSE? a. It involves DNA degradation. b. It involves cell membrane blebbing. c. It typically elicits an inflammatory response. d. It involves shrinkage of the nucleus. e. It involves shrinkage of the cytoplasm.

18. The eyeless gene in fruit flies, Small eye gene in mice, and Aniridia gene in humans have high homology in DNA sequence and their mutations result in defects in eye development. Which of the following statements is NOT consistent with this observation? a. The same genes in distantly related organisms can shape similar developmental pathways. b. The eyes of insects and mammals must have evolved independently during evolution. c. A common pathway underlies eye development in all three organisms. d. All three genes evolved from a common ancestral sequence.

19. A flower is made up of four concentric rings of modified leaves called whorls. Which of the following lists the individual whorls in CORRECT order going from the outside toward the inside of the flower? a. petals, sepals, carpels, and stamens b. sepals, petals, stamens, and carpels c. carpels, stamens, petals, and sepals d. stamens, carpels, sepals, and petals e. carpels, petals, sepals, and stamens 20. Which of the following statements about Hox genes is INCORRECT? a. All Hox genes found in organisms other than Drosophila have a conserved pattern of identical clustered arrangements. b. When a vertebrate Hox gene is moved to a new location within the Hox gene complex, the timing of its expression is altered. c. The Hox genes encode transcription factors that help to determine the identity of the body region. d. They are usually organized as clusters, and the order on the chromosome is important for their expression pattern. e. The recent studies suggest the role of miRNA in controlling the expression of some Hox genes.

21. Which of the following statements about the gene dorsal in D. melanogaster is FALSE? a. dorsal is transcribed maternally. b. When the developing embryo is a multinucleate syncytium, dorsal mRNA and Dorsal protein are uniformly distributed. c. When the developing embryo is a syncytial blastoderm, Dorsal protein shows a nonuniform distribution. d. Dorsal acts in the cytoplasm. e. Dorsal protein is a transcription factor that specifies ventral cell fates.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 22_7e In the nematode worm C. elegans, loss-of-function ced-3 mutations lead to more cells than normal due to a defect in apoptosis. Loss-of-function mutations in ced-9 cause embryonic lethality due to ectopic cell deaths.

The phenotype of ced-9; ced-3 double mutants is the same as ced-3 single mutants; that is, there are more cells than normal due to a defect in apoptosis. Genes ced-3 and ced-9 are known to act in the same genetic pathway.

22. Which of the following pathways is consistent with the genetic data? a. ced-3 → ced-9 b. ced-9 → ced -3 c. ced-3 → ced-9 d. ced-9 → ced-3 23. Which of the following statements accurately describes antibodies? a. Kappa and lambda are the two types of immunoglobulin heavy chains. b. Genes coding for antibody constant regions undergo somatic recombination to generate diversity. c. Antigens are always proteins that are exposed on the surface of foreign invaders. d. Antigen-binding sites are located at the tips of the two arms of the Y-shaped antibody. e. The variable region segments (V genes) are only subjected to somatic recombination to contribute to different antigen recognition as they come in direct contact with antigens.

24. Which specific group of genes specifies the identities of individual body segments? a. segmentation genes b. gap genes c. homeotic genes d. pair-rule genes e. segment-polarity genes

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 22_7e The diagram below illustrates the normal expression of three different groups of genes (A, B, and C) within four different whorls of a developing flower. Normally, whorl 1 produces sepals, whorl 2 produces petals, whorl 3 produces stamens, and whorl 4 produces carpels. Furthermore, expression of class A genes inhibits class C genes, and expression of class C inhibits class A.

25. What kind of mutation would produce sepals in all four whorls? a. defective in A b. defective in B c. defective in C d. defective in A and B e. defective in B and C 26. Which of the following groups of genes is directly regulated by egg-polarity genes? a. homeotic genes b. gap genes c. Hox genes d. pair-rule genes e. segment-polarity genes 27. Which of the following is the CORRECT order of gene groups expressed in normal Drosophila development?

1. Gap genes 2. Homeotic genes 3. Pair-rule genes 4. Segment-polarity genes 5. Egg-polarity genes a. 1, 2, 3, 4, 5 b. 4, 2, 5, 1, 3 c. 3, 1, 2, 5, 4 d. 5, 2, 4, 3, 1 e. 5, 1, 3, 4, 2 Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 22_7e 28. Which of the following statements accurately describes the organization of the immune system? a. Humoral immunity centers around the production of specialized T cells that can mount an attack on cells that express foreign MHC. b. Cellular immunity involves specialized lymphocytes called B cells, which mature in the bone marrow. c. The primary immune response results in generation of memory cells, which allows the body to mount a more effective attack upon repeated exposure to the same antigen. d. The theory of clonal selection supports the presence of a large pool of lymphocytes that get engaged in attacking one specific antigen. e. Autoimmune disease is specifically referring to the self-destructive behavior of T cells mounting an attack on other T cells, which leads to a compromised immunity.

29. What is apoptosis? a. autophagy b. a process that generates the anterior–posterior axis c. antibody production d. genetic maternal effect e. programmed cell death 30. Which of the following choices lists the CORRECT sequence of gene expression during Drosophila development?

1. Segmentation genes 2. Egg-polarity genes 3. Homeotic genes a. 1, 2, 3 b. 2, 3, 1 c. 2, 1, 3 d. 3, 2, 1 e. 3, 1, 2 31. Which of the following genes determines the identity of individual segments during development? a. bicoid b. nanos c. cactus d. disheveled e. dorsal

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 22_7e The diagram below illustrates the normal expression of three different groups of genes (A, B, and C) within four different whorls of a developing flower. Normally, whorl 1 produces sepals, whorl 2 produces petals, whorl 3 produces stamens, and whorl 4 produces carpels. Furthermore, expression of class A genes inhibits class C genes, and expression of class C inhibits class A.

32. What kind of mutation would produce carpels in whorls 1 and 4 and stamens in whorls 2 and 3? a. defective in A b. defective in B c. defective in C d. defective in A and B e. defective in B and C 33. The loss of a tadpole's tail during amphibian metamorphosis and the separation of the digits in human hands during embryonic development are caused by: a. necrosis. b. autophagy. c. cellular digestion. d. apoptosis. e. morphogenesis.

34. Which of the following statements about homeotic genes is INCORRECT? a. Homeotic genes were first identified in invertebrates, but their homologs were found in almost all organisms. b. All homeotic genes contain consensus sequences that result in a specific DNA-binding motif. c. In vertebrates, homeotic genes are organized into two distinct clusters. d. All homeotic genes encode transcription factors that direct further gene expression. e. Homeotic genes exhibit a relationship between their order on the chromosome and the timing of their expression.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 22_7e Indicate one or more answer choices that best complete the statement or answer the question.

35. Which of the following characteristics of vaccines are required in order for them to be effective in immunization? (Select all that apply.) a. They must contain the entire pathogen. b. They must present antigenic surfaces recognized by immunoglobulin molecules. c. They must stimulate lymphocyte activity. d. They must stimulate the production of large quantities of plasma cells. e. They must stimulate the production of memory cells.

In the late 1940s Edward Lewis discovered mutations in the Antennapedia gene that lead to legs developing on the head of a fly instead of antenna. These mutations are dominant and lead to ectopic expression of the Antennapedia gene.

36. Based on these data, how would you expect the Antennapedia gene to be expressed in the mutant? (Select all that apply.) a. in every segment b. in a more anterior segment c. in a more posterior segment d. earlier e. later In the nematode worm C. elegans, loss-of-function ced-3 mutations lead to more cells than normal due to a defect in apoptosis. Loss-of-function mutations in ced-9 cause embryonic lethality due to ectopic cell deaths.

The phenotype of ced-9; ced-3 double mutants is the same as ced-3 single mutants; that is, there are more cells than normal due to a defect in apoptosis. Genes ced-3 and ced-9 are known to act in the same genetic pathway.

37. Which of the following statements might be TRUE about the genes ced-3 and ced-9? (Select all that apply.) a. ced-3 encodes a caspase. b. ced-9 encodes a caspase. c. ced-9 encodes a functional homolog of the Drosophila gene reaper. d. ced-3 stimulates phagocytosis of a cell in which it is expressed. e. ced-9 activity leads to DNA degradation.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 22_7e In the late 1940s Edward Lewis discovered mutations in the Antennapedia gene that lead to legs developing on the head of a fly instead of antenna. These mutations are dominant and lead to ectopic expression of the Antennapedia gene.

38. Certain mutations in the Antennapedia gene can lead to antenna developing in a mesothoracic segment where legs should have developed. What characteristics would you expect from this class of mutations? (Select all that apply.) a. They are loss-of-function. b. They are recessive. c. They are gain-of-function. d. They are dominant. e. They result in ectopic Antennapedia expression.

39. Xenotransplantation involves transplanting tissues or organs from one species to another. Some medical researchers are exploring the possibility of transplanting pig organs into humans to address the worldwide human organ donor shortage. What strategies might help with the success of such transplantations? (Select all that apply.) a. Create transgenic pigs that lack a pig-specific sugar molecule that is absent in humans for use as organ donors. b. Inject human organ transplant recipients with a pig-specific sugar molecule that is absent in humans prior to transplantation. c. Treat pigs with immunosuppressive drugs prior to harvesting their organs for donation. d. Treat human organ transplant recipients with immunosuppressive drugs following xenotransplantation. e. Administer pig bone marrow to the human organ transplant recipient prior to transplantation.

40. Which of the following are postulates of the "evo-devo" field of biology? a. Ontogeny always capitulates phylogeny. b. Developmental pathways in distantly related organisms are often controlled by homologous genes. c. In different species, homologous genes that act in similar developmental pathways always lead to similar phenotypes. d. Evolutionary adaptions in development can be achieved through changes in expression of regulatory gene products. e. Activation or inhibition of gene expression in one species can mimic evolutionary adaptations of a distinct species.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 22_7e 41. What are the role and relationship of egg-polarity genes, segmentation genes, pair-rule genes, gap genes, and homeotic genes in Drosophila development?

42. List the five major classes of developmental genes in Drosophila in the order (earliest to latest) that they regulate development.

43. What are morphogens, and what do they do?

44. Briefly describe when and how Dolly the sheep was cloned.

45. Compare and contrast the terms cell determination (or cell fate determination) and differentiation.

46. What are Hox genes?

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 22_7e 47. What three sets of proteins are required for immune responses?

48. (a) In what sense does each person's immune system become adapted to that person's environment and to the set of antigens to which that person is exposed? (b) Explain in terms of the clonal selection theory how a person's immune system responds to and comes to be "adapted" to the person's environment and personal exposure to antigens.

49. What is apoptosis? Describe the important roles that apoptosis plays during development of an organism.

Provide specific evidence that apoptosis is important to normal development.

50. What is an autoimmune disease?

51. What is the function of segmentation genes and homeotic genes in Drosophila?

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 22_7e 52. The following diagram describes the genetic regulation of flower development in angiosperms. In each box, write the class of gene, and below write the flower structures determined by those gene(s).

53. Why do you think the mutation rate in immunological genes is unusually high?

54. How do lymphocytes violate the general principle that all cells in an organism contain the same set of genetic information?

55. The genes that determine flower development were divided into three classes by Meyerowitz and his colleagues: class A, class B, and class C. What would the flowers produced by a plant with a mutation in a class B gene look like?

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 22_7e 56. What are homeotic genes and what do they do?

57. When animals like Dolly the sheep are cloned using techniques that transfer a nucleus from a donor animal into the egg of a recipient animal, is the resultant clone genetically identical to the nuclear donor animal?

58. What have the cloning experiments in plants and animals confirmed with regard to developmental processes?

59. What is the role of the major histocompatibility complex (MHC) in tissue-graft rejection?

60. What is the difference between apoptosis and necrosis?

61. (a) Explain briefly the roles of Bicoid and Nanos proteins in determining the anterior–posterior axis of the fruitfly embryo. (b) Explain how the Bicoid and Nanos protein gradients are established in the fruit-fly embryo. (c) Bicoid mutant embryos lack head structures and instead have structures at the anterior end of the embryo that are characteristic of the posterior end. Propose an explanation of this phenotype.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 22_7e 62. During development, cells can respond to the presence of hormones and other signals with changes in gene expression that can bring about cell differentiation. In most cases, the differentiated state needs to be maintained through subsequent cell divisions and long after the inducing signal has disappeared. Describe a mechanism that can maintain the differentiated state after the inducing signal has disappeared.

63. List at least two processes involved in apoptosis that are controlled by caspases.

64. Describe the epigenetic changes that occur in cells during development and the role they play in differentiation of cell types within a multicellular organism.

65. Distinguish between the genetic mechanisms used to generate antibody diversity within a person's immune system and those used to generate MHC antigen diversity between different individuals.

66. What is meant by the phrase "ontogeny recapitulates phylogeny"? Is this an accurate statement?

67. What are some of the different signals that can initiate programmed cell death (apoptosis)?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 22_7e Answer Key 1. b 2. a 3. b 4. b 5. c 6. d 7. d 8. d 9. e 10. c 11. c 12. a 13. c 14. b 15. c 16. e 17. c 18. b 19. b 20. a 21. d 22. c 23. d 24. c 25. e 26. b Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 22_7e 27. e 28. c 29. e 30. c 31. d 32. a 33. d 34. c 35. b, c, e 36. b, d 37. a, d 38. a, b 39. a, d, e 40. b, d, e 41. The major axes and regions of the embryo are first established by egg-polarity genes. Next, distinct patterns within each of these large regions are determined by the action of segmentation genes (gap genes, pair-rule genes, and segment-polarity genes). The gap genes define large sections, the pair-rule genes define regional sections of the embryo and affect alternate segments, and the segment-polarity genes affect individual segments. Finally, the homeotic genes establish unique identities for each segment.

42. Egg-polarity genes → gap genes → pair-rule genes → segment-polarity genes → homeotic genes 43. Morphogens are proteins whose concentration gradients in a developing organism affect the developmental fate of the surrounding region.

44. Dolly the sheep was cloned in 1996 by researchers at the Roslin Institute of Scotland. Workers isolated the genetic material from a single differentiated udder cell of a 6-year-old adult, white-faced, Finn Dorset ewe, fused this genetic material with an enucleated egg cell, and stimulated the egg electrically to initiate development. After culturing the modified egg in the laboratory for a week, they implanted the embryo into a Scottish black-faced surrogate mother. Dolly, the first mammal cloned from a differentiated adult cell, was born on July 5, 1996. Since Dolly, many other sheep, mice, and calves have been cloned from differentiated adult cells.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 22_7e 45. Cell fate determination involves a commitment to a particular lineage of development. It involves a restriction of cell potency that is usually irreversible. Cell fate determination typically occurs in the absence of any obvious physical changes distinguishing determined cells from their undetermined or differently determined neighbors. In contrast, cell differentiation involves cells taking on specialized functional roles. Differentiation typically involves phenotypic changes that lead to recognizable cell types (e.g., muscle cells, nerve cells). Determination precedes differentiation and is required for the latter. Both processes involve changes in gene expression.

46. Hox genes are homeobox-containing genes that were originally found in Drosophila and are now known to be present in all animals that have been studied, including nematodes, beetles, sea urchins, frogs, birds, and mammals.

Hox genes have also been discovered in fungi and plants, indicating that they arose early in the evolution of eukaryotes.

47.

(1) Antibodies (2) T-cell receptors (3) Major histocompatibility antigens 48. (a) Each person's immune system normally is capable of making antibodies against virtually any antigen that might be encountered in the person's lifetime. So everyone normally starts with the same or similar complete immune response repertoire. However, each person's immune system normally only makes antibodies in significant quantities against the antigens to which the person is actually exposed. Also in many cases a person's immune system "remembers" the initial exposure and is able to mount a more effective defense against subsequent exposures, providing long-term immunity. Therefore, each person's immune system adapts itself uniquely to the set of antigens that it encounters. (b) The clonal selection theory states that, initially, there is a large pool of millions of different lymphocytes, each capable of responding to only one antigen so that millions of different antigens can be detected. When a lymphocyte encounters its specific antigen and binds to it, the lymphocyte is stimulated to proliferate rapidly, producing a large population of genetically identical descendant cells (a clone) that is able to produce a large quantity of the antibody that is specific to that particular antigen. Thus, the immune system normally only mounts an active defense against the antigens that it actually encounters. In many cases, some members of the selected clone, called memory cells, remain in circulation for a long time and provide lasting immunity to the pathogen that bears that particular antigen.

49. Apoptosis is genetically programmed cell death. It involves specific biochemical pathways that lead to degradation of DNA, breakdown of the nucleus and cytoplasm, and ingestion of the cell contents by macrophages. Apoptosis plays a critical role in development. As animals develop, excess cells are produced and must be culled by apoptosis later in development. In many cases, whole body structures must be removed, and this is accomplished at the cellular level by apoptosis. The importance of apoptosis is highlighted by abnormal phenotypes of mutants that are defective in apoptosis, including mutants in fruit flies that exhibit no apoptosis and die early in development with an excess of cells.

50. An autoimmune disease is a disease of the immune system whereby the body's ability to make a distinction between "self" and "nonself" is compromised so that an immune reaction is initiated against antigens present on the body's own cells. An example of an autoimmune disease is rheumatoid arthritis.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 22_7e 51. (1) After the major axes of development are established, the three types of segmentation genes act sequentially to determine the number and organization of the embryonic segments in the fruit fly Drosophila.

(2) Homeotic genes define the identity of individual segments in Drosophila. Homeotic genes contain a consensus sequence, called the homeobox, that encodes a DNA-binding domain. The products of homeotic genes are DNAbinding proteins that regulate the expression of other genes. Genes containing homeoboxes are found in many other organisms in addition to Drosophila.

52. (1) Class B genes (2) Class A genes (3) Class C genes (4) Sepals (5) Petals (6) Stamens (7) Carpels 53. For the immune system to serve its purpose, it continually strives to anticipate future threats by producing vast numbers of structurally diverse potential antigen-recognition sites. The immune system has in essence been “engineered” over evolutionary time to be a hypervariable system so that it can produce highly diverse populations of antigen-recognition sites for antigens that the organism hasn’t even been exposed to. For example, the somatic recombination process that joins the V, J, D, and C gene segments in developing B cells has evolved to be imprecise and to involve random deletions or additions of several nucleotides at the junctions of the recombining segments.

54. Antibody genes in lymphocytes are organized in segments, and germ-line DNA contains different versions (alleles) of each of these segments. During lymphocyte maturation, gene rearrangements occur (through somatic recombination within a single chromosome), generating many different combinations of the different segments (V, D, and J), which results in the immense variety of different antibodies capable of being produced in the lymphocyte pool. Other contributing factors to this diversity are (i) the random association of light and heavy chains in different combinations, (ii) the random addition and deletion of nucleotides at the junctions of the segments during lymphocyte maturation, and (iii) the high mutation rate in immunological genes.

55. Petals are determined by class A + class B genes. Stamens are determined by class B + class C genes. Without proper class B function, petals and stamens would not be produced, and the resulting flower would instead have only sepals (first and second whorls) and carpels (third and fourth whorls).

56. Homeotic genes play crucial roles in developmental processes in higher organisms by coding for regulatory proteins (e.g., transcription factors) that activate other genes that control segment-specific characteristics of body parts.

Homeotic genes control the fate and ultimate development of body segments. Mutations in homeotic genes cause body parts to appear in the wrong segments of the body.

57. No. Although the clone will share identity with the donor animal for genes encoded by the nuclear genome, it will in contrast inherit its cytoplasmic genes (i.e., mitochondrial genes) from the egg donor.

58. Cloning experiments have demonstrated that genetic material is not lost or permanently altered during development, and that development must involve selective, coordinated expression of genes.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 22_7e 59. When tissues are transferred from one species to another, or from individual to individual within a species, the transplanted tissues are usually rejected by the host animal. Early studies demonstrated that this tissue-graft rejection is caused by an immune response that occurs when antigens on the surface of the grafted tissue are detected and attacked by T cells in the host organism. The antigens that elicit graft rejection are referred to as histocompatibility antigens, and they are encoded by a cluster of genes called the major histocompatibility complex (MHC). The MHC genes are highly variable genes, with more than 100 different alleles for some MHC loci.

Because each individual possesses five or more MHC loci, and there are many possible alleles at each locus, no two individuals (with the exception of identical twins) produce the same set of histocompatibility antigens. The exceptional variation in histocompatibility antigens provides each individual with a unique identity for their own cells and allows the immune system to distinguish self from nonself. This variation in the MHC is also the cause of rejection in organ transplants.

60. (1) Apoptosis is a form of programmed cell death that is usually initiated by the cell itself. Apoptosis is essential to development and embryogenesis in animal and plant cells. In fact, most multicellular animals cannot complete development if the apoptosis process is blocked.

(2) Necrosis is the process of cells dying in an uncontrolled, unregulated manner after they are injured.

61. (a) Bicoid exists in a gradient in the egg and embryo such that its concentration is high at the anterior end and low at the posterior end. Nanos is distributed in a gradient in the opposite orientation. Bicoid stimulates anterior development by influencing expression of genes needed for anterior development. Nanos inhibits anterior development and stimulates development of posterior structures. All along the anterior–posterior axis, cells experience varying levels of Bicoid and Nanos and can differentiate appropriately based on this positional information. (b) Before fertilization, bicoid mRNA is synthesized in the egg and other cells of the ovary and accumulated at the anterior end of the egg, where it is tethered to the cytoskeleton. Similarly, nanos mRNA accumulates at the posterior end. After fertilization, the asymmetrically distributed mRNAs are translated to form the corresponding proteins. Because these proteins are degraded rapidly, the proteins form a gradient from high at the end where they are created to low at the opposite end. (c) The bicoid mutant lacks head structures because there is no Bicoid protein to stimulate development of the head. The presence of posterior structures in the anterior region of the mutant embryo implies that Bicoid not only stimulates anterior development but also inhibits posterior development. Therefore, posterior structures are able to develop in anterior regions in the absence of Bicoid protein.

62. One way that a particular gene expression state can be maintained in a cell and during cell divisions is through epigenetic changes. These include heritable changes in DNA and chromatin structure such as histone modification and DNA methylation. These modifications affect gene expression, and the patterns of modifications can be replicated during DNA replication and inherited during cell division. Therefore, once a gene is inactivated, for example in response to a signal, the inactive state can be maintained in the cell and passed on to daughter cells and later cell descendants.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 22_7e 63. (1) Caspase activity results in the cleavage of proteins that are essential to cell function, such as actin and other proteins that support the nuclear membrane and cytoskeleton. As a consequence, cells undergoing apoptosis lose their shape.

(2) Caspases also cleave a regulatory protein that normally functions to maintain an enzyme that degrades DNA (i.e., DNase) in an inactive form. Cleavage of this regulatory protein releases DNase from inhibition and thereby activates it, resulting in subsequent breakdown of cellular DNA, which eventually leads to cell death.

64. Cell differentiation is driven by regulation of gene expression. Genes whose products are needed for the structure and function of a particular cell are turned on, and those whose products are not needed are turned off. Epigenetic changes are heritable alterations to DNA and chromatin structure that can affect gene expression. Because these changes are heritable during cell division, gene expression states (on or off) established within a cell early in development can be passed on to daughter cells and other descendant cells. Epigenetic changes can include DNA methylation, modification of histone proteins, and chromatin remodeling. In early stages of development, genes that may be required later in development are often held in a transiently silent state by histone modifications. These modifications are generally flexible and easily reversed, allowing for later activation. Genes can be more permanently silenced by DNA methylation.

65. A huge diversity of antibodies is produced from a small number of genes through mechanisms that include somatic recombination, segment joining (junctional diversity), and somatic hypermutation. Somatic recombination produces many different light and heavy chains by bringing together various combinations of V segments to form the polypeptide-encoding genes. Segment joining is imprecise, and a few random nucleotides are frequently lost or gained at the junctions of the recombining segments, generating junctional diversity. Somatic hypermutation is a mechanism that causes an unusually high rate of mutations in antibody-encoding genes, generating even more diversity. Finally, a large number of combinations of different light and heavy chains are possible. Because of these mechanisms, a person's immune system is capable of making antibodies against virtually any antigen that the person might encounter.

The diversity of MHC antigens between individuals is also impressive, but the mechanism by which this diversity is generated is quite different. This diversity is based on the number of antigen-encoding loci (at least five) and the large number of alleles at each locus (as high as 100). Because of the number of loci and the large number of alleles, no two individuals (except identical twins) would be expected to have the same set of histocompatibility antigens.

66. This overused statement, which is not accurate, was originally used to describe the belief that organisms repeat their evolutionary history through their development. For example, a human embryo would thus pass through fish, amphibian, reptilian, and mammalian stages before developing human traits. However, current research based on the study of evolution through the analysis of development (often called the "evo-devo" approach) strongly suggests that the same genes (i.e., homologs) are often involved in mediating the developmental pathways in distantly related organisms.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 22_7e 67. (1) Pathogen infection by viruses and so forth can activate immune cells to secrete substances onto an infected cell, causing that cell to undergo apoptosis. This defensive measure would prevent spread of the pathogen in an organism.

(2) DNA damage can activate apoptosis to prevent replication of mutated sequences.

(3) Damage to mitochondria may cause the initiation of apoptosis.

(4) Accumulation of misfolded proteins in the endoplasmic reticulum may signal the cell to undergo apoptosis.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 23_7e Indicate the answer choice that best completes the statement or answers the question.

1. Which of the following are regulatory molecules whose normal function is to inhibit gene expression but which often have reduced activity in many tumor cells? a. histones b. miRNAs c. reverse transcriptases d. growth factors e. kinases 2. An abnormally low level of expression of a particular microRNA might facilitate the development of a cancer by which of the following methods? a. inducing the increased expression of a tumor-suppressor gene b. inducing the increased expression of DNA-repair genes c. promoting a decreased level of telomerase activity d. inhibiting the G1 to M transition in the cell cycle e. increasing the expression of an oncogene A new form of kidney cancer was identified in a family in which many of the family members were affected.

This family's pedigree is shown below.

* 3. Upon genetic analysis of the cancerous tissue, a mutated form of a gene, named the kid gene, was discovered in all affected individuals but not in the unaffected individuals. What is the MOST likely function of the normal, nonmutated kid gene? a. DNA-repair gene b. proto-oncogene c. tumor-suppressor gene d. telomerase gene e. acetyltransferase gene

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 23_7e 4. Genes that encode components of the cytoskeleton and extracellular matrix often contribute to which of the following processes? a. DNA repair b. mutation c. metastasis d. primary tumor formation e. signal transduction 5. Which of the following result(s) directly from metastasis? a. primary tumors b. secondary tumors c. tumor vascularization d. decreased DNA repair e. increased apoptosis 6. A drug is injected into a cell that inhibits the removal of phosphate groups from MPF. Which of the following would result? a. MPF would remain in its active state. b. The cell would not be able to transition into mitosis. c. Cyclin B levels would not be able to increase. d. The cell would not be able to progress through the G1/S checkpoint. e. Cyclin D could no longer be activated.

7. A cancer cell line was identified as having inappropriate expression of telomerase. Which of the following would be the BEST treatment that could make this cell line mortal? a. a drug that increases the levels of telomerase expression b. a drug that methylates the promoter of the telomerase gene c. a drug that repairs the mutation within the coding region of the telomerase d. a drug that allows the telomerase to be expressed but at lower levels e. a drug that inhibits a tumor-suppressor gene in the cell line 8. Which of the following is the major event associated with the retinoblastoma cancer? a. a translocation involving chromosomes 9 and 22 b. both copies of a tumor-suppressor gene being inactivated c. a translocation resulting in the enhanced expression of an oncogene d. a mutation resulting in an activated RAS oncogene e. inactivation of a major DNA repair system

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 23_7e 9. In Burkitt lymphoma, there is increased expression of the MYC gene. Which of the following statements BEST explains the reason for the increased expression of this gene? a. A chromosome deletion has removed a tumor-suppressor gene. b. A chromosome deletion has removed an oncogene. c. A chromosome duplication involves a segment with an oncogene. d. A translocation has brought the MYC gene next to a different regulatory region. e. The MYC gene has been amplified.

10. Which of the following groups of proteins is NOT commonly known to include oncogenes? a. transcription factors b. growth factors c. signal-transduction proteins d. ion channels e. growth factor receptors 11. Which of the following statements is FALSE? a. Tumors usually show a clonal evolution. b. Cervical cancer is associated with an animal virus. c. Some cancers are associated with reduced DNA repair. d. Epigenetic changes in somatic cells may be associated with some cancers. e. Most tumors arise from germ-line mutations that accumulate during our life span.

12. Which of the following statements about human papilloma virus (HPV) is FALSE? a. HPV is found in approximately 95% of women with cervical cancer. b. HPV causes benign tumors of epithelial cells. c. HPV indirectly affects regulation of the cell cycle. d. The effort in the United States at HPV vaccination is directed exclusively toward girls. e. HPV can promote high levels of expression of growth-promoting proteins leading to cancer.

13. Mutations in proto-oncogenes are generally _____ whereas mutations in tumor-suppressor alleles are generally _____. a. deletions; duplications b. recessive; dominant c. duplications; deletions d. dominant; recessive e. deletions; base-pair substitutions

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 23_7e 14. A genetics student recently learned of the connection between telomerase and cancer. Curious as to whether he is at risk for developing cancer, he sneaks into a lab at night to analyze his DNA. He isolates DNA from his skin cells and applies PCR using primers specific to the telomerase gene. He runs a gel on his PCR reaction and sees that he has a PCR product on the gel. He is terrified because he knows this means he has the telomerase gene in his skin cells and thinks he will develop skin cancer. He takes his findings to his genetics professor. What does she tell him (besides that he shouldn't be sneaking into a lab!)? a. The telomerase gene should not be present in somatic cells; thus, he is at an increased risk of developing skin cancer and should take the results to his physician. b. His results are normal because the telomerase gene should be present in all cells, including somatic cells. However, its expression should only be found in stem cells and germ cells. He should have looked at the expression levels of the gene, not the presence of the gene itself (and he should study more!). c. His results are inconclusive because they cannot determine if the telomerase gene he amplified has a mutation in its coding sequence, which is what leads to cancer. d. Telomerase mutations that result in cancer behave in a dominant manner. Therefore, rather than going through the effort of a genetic analysis, the student should have constructed a family pedigree to determine if members of his family had skin cancer, which could then determine his risk. e. He should not be concerned about his cancer risk because transcriptionally active telomerase should be present in somatic cells to maintain the length of the chromosomes and thus prevent cancer.

15. Which of the following chromosomal abnormalities is associated with chronic myelogenous leukemia? a. a deletion b. an inversion c. a duplication d. a reciprocal translocation e. an aneuploidy involving one of the shorter autosomal chromosomes

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 23_7e A new form of kidney cancer was identified in a family in which many of the family members were affected.

This family's pedigree is shown below.

* 16. Upon genetic analysis of the cancerous tissue, a mutated form of a gene, named the kid gene, was discovered in all affected individuals but not in the unaffected individuals. Because so many members of the family are affected, the female (age 60), indicated with a "*", is concerned that although she has not developed the disease yet she still may develop the cancer later in life. What would a genetic counselor tell her? a. All members of the family carry one mutant version of the kid gene, but those with cancer gained a second mutation of the kid gene. Therefore, it is highly likely that she will develop the kidney cancer. b. While it is unlikely that she will develop the form of kidney cancer seen in her family, the mutant form of the kid gene she carries puts her at an increased risk of all types of cancer. c. Because her parents did not develop kidney cancer, it is highly unlikely that they had a mutant form of the kid gene, and thus it is highly unlikely that she has a mutant kid gene. Therefore, she is not at risk of developing this form of kidney cancer. d. While she carries a mutant copy of the kid gene, it is highly unlikely that she will develop kidney cancer because two mutant copies are required to be affected with the cancer. e. Her father was a carrier of the mutant kid gene, and thus she is also a carrier of the mutant kid gene.

She will likely not be affected, but her children will be at a high risk of developing the cancer.

17. Evidence that the development of cancer is a multistep process includes the: a. observation that certain tumor-suppressor genes and oncogenes are involved in a sequential manner in the development of colon cancer b. fact that proto-oncogenes are widely conserved in evolution c. usual occurrence of retinoblastoma at a young age d. development of a cancer as a result of activation of a single oncogene by any of a variety of mechanisms e. fact that there are many genetic and epigenetic mechanisms that lead to inactivation of the same tumorsuppressor gene

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 23_7e 18. Consider the Ras signal-transduction pathway. Suppose that a mutation occurred within the gene that encodes the transmembrane receptor protein. Which of the following types of mutations in this receptor could lead to cancer? a. a mutation that allows the receptor to become phosphorylated in the absence of a growth factor bound to it b. a mutation in the extracellular domain that prevents the binding of a growth factor c. a mutation in the transmembrane domain that prevents the signal from being transduced to the interior of the cell d. a mutation within the intracellular domain that prevents interaction with the adaptor molecules e. All of these are types of receptor mutations that could lead to cancer.

19. The normal activity of the retinoblastoma (Rb) protein in the cell is to: a. inhibit p53 activity. b. suppress transcription of tumor-suppressor genes. c. regulate the progression from G1 to S in the cell cycle. d. induce cyclin–CDK complex formation. e. block the initiation of anaphase during the cell cycle.

20. Consider the Ras signal-transduction pathway. Suppose a mutation occurred in the gene that encodes the transmembrane receptor to which a growth factor binds. The mutation no longer allows the growth factor to bind to the receptor. Which of the following methods could be used to overcome this mutation and allow the pathway to move forward? a. Inject extremely high levels of growth factor into the cell. b. Inject normal levels of nonmutated growth factor into the cell. c. Inject only the growth factor binding domain of the transmembrane receptor into the cell. d. Inject the activated form of Ras into the cell. e. Inject nonmutated tumor-suppressor genes into the cell.

21. The process by which genetic changes occur in tumors and allow them to become increasingly aggressive over time is called: a. clonal evolution. b. metastasis. c. loss of heterozygosity. d. epigenetic evolution. e. signal transduction.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 23_7e 22. Many viruses that are associated with cancers in animals are _____ that use reverse transcriptase. a. papilloma viruses b. Epstein-Barr viruses c. retroviruses d. hepatitis B viruses e. None of the answers is correct.

23. A patient's cells from a colorectal polyp were collected, and the APC and ras genes were analyzed. The APC gene was found to be mutated, and the ras gene was found to be up-regulated. The physician classifies the polyp as an adenoma and advises the patient to have the polyp removed. The patient claims that the physician is incorrect in his advice to have the polyp removed. Is the physician's advice logical? Why or why not? a. no, because adenomas are benign tumors and thus are unlikely to progress to a malignant tumor b. no, because the up-regulation of the ras gene will counter the effects of the mutant APC gene, and the benign tumor is unlikely to progress to a malignant state c. no, because until the specific function of the mutated APC gene is known, it cannot be determined that the cells are in a proliferative state d. yes, because the cells of the polyp have already begun to divide inappropriately, and further mutations may allow the tumor to invade other tissues and metastasize e. yes, because up-regulation of the ras gene is causing the cells to divide more slowly, and thus they will be unable to repair future DNA damage that will likely occur in cells surrounding the polyp 24. Most cancers are assumed to arise through which of the following? a. errors in transcription b. the production of unbalanced gametes because of nondisjunction during meiosis c. genetic or epigenetic changes in somatic cells d. delayed cell division during early embryogenesis e. No correct answer is provided.

25. Many tumors exhibit: a. abnormally high levels of telomerase expression. b. abnormally high levels of tumor-suppressor gene expression. c. translocations that move a tumor-suppressor gene to a new location that increases its normal expression. d. deletions that remove an oncogene from the genome. e. DNA replication that is inhibited when it would normally occur.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 23_7e 26. A routine colonoscopy detects a small colorectal polyp on the colon wall that has not reached adenoma status.

Which of the following genes would you most likely predict to be mutated at this stage? a. APC b. p53 c. ras d. antimetastasis gene 27. Which of the following statements BEST characterizes many cancers such as colon cancer? a. They result from the activation of one critical tumor-suppressor gene. b. They result when the transition from G2 to M in the cell cycle is inhibited. c. They result when DNA replication during the S period of the cell cycle is inhibited. d. They result from a series of sequential mutations in a number of genes. e. They result from decreased expression in a series of cellular oncogenes.

28. Which of the following types of cancer is associated with a defect in nucleotide-excision repair? a. retinoblastoma b. xeroderma pigmentosum c. cervical cancer d. chronic myelogenous leukemia e. Bloom syndrome 29. The overexpression of certain microRNAs, called oncomiRs, is associated with some types of cancer. What type of gene is the likely target of these oncomiRs in cancer cells? a. proto-oncogenes b. tumor-suppressor genes c. genes that promote angiogenesis d. genes that allow the G1/S transition e. genes involved in DNA repair 30. The Philadelphia chromosome is: a. an example of an aneuploidy. b. the result of a translocated chromosome involving parts of chromosomes 9 and 22. c. a lengthened version of chromosome 22 that results from a recombination event with chromosome 12. d. a shortened version of chromosome 22 that results from a deletion. e. a lengthened version of chromosome 22 that results from a duplication.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 23_7e 31. In a certain cell line, a tumor-suppressor gene is translocated to a different location under the control of a strong promoter. Which of the following would be TRUE of this cell line compared to a cell line without this translocation? a. The cells will have an increase in angiogenesis. b. The cells will be at an increased risk of transitioning from benign to malignant. c. The cells will show a lower amount of proliferation. d. The cells will be at a slightly increased risk of cancer. e. It depends where the gene was translocated.

32. Which of the following is not a typical mechanism by which a proto-oncogene is converted to an oncogene? a. a chromosomal translocation resulting in the bringing together of two different genes that make a fusion protein b. a chromosomal translocation resulting in enhanced regulation of the proto-oncogene c. complete deletion of the proto-oncogene d. a point mutation in the proto-oncogene e. All of these are mechanisms of converting proto-oncogenes to oncogenes.

33. _____ and _____ are chromosomal mutations that may activate cellular oncogenes by moving them to new regulatory sequences where they become overexpressed. a. Duplications; deletions b. Inversions; translocations c. Duplications; inversions d. Deletions; translocations e. Aneuploidy; deletions 34. Which of the following statements is TRUE concerning the function of proto-oncogenes? a. They make products that act as signals to initiate cellular apoptosis. b. Their products are components of cell growth pathways. c. Proto-oncogenes make products that act as cell checkpoint regulators. d. Proto-oncogenes make products that scan the genome for DNA damage. e. Proto-oncogenes make products that repair DNA at sites of lesions.

35. Certain viruses are instrumental in converting proto-oncogenes to oncogenes. This conversion most commonly results because: a. viruses specifically infect cells that contain proto-oncogenes. b. only viruses contain genes that can convert proto-oncogenes into oncogenes. c. the proto-oncogenes are more likely to undergo mutation or recombination within a virus. d. viruses contain the remainder part of the DNA that is added to the proto-oncogene to form the oncogene.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 23_7e 36. Many tumor cells are able divide uncontrollably in cell culture in the laboratory. Some of these cell lines have genomes with a mutant tumor-suppressor gene. Adding a wild-type copy of this tumor-suppressor gene to the genomes of these cells reduces their ability to divide in culture. In contrast, adding a wild-type copy of a proto-oncogene to the genomes of a tumor cell line with an activated oncogene doesn't have any effect on the uncontrolled cell division shown by this line. Which of the following statements BEST explains why adding a wild-type proto-oncogene doesn't affect cell division in these cell lines? a. Most point mutations that produce activated oncogenes are dominant gain-of-function mutations. b. Many proto-oncogenes encode proteins that act to promote apoptosis. c. Many proto-oncogenes encode proteins involved in DNA repair. d. Many proto-oncogenes are inherited as recessive mutations. e. Most protein products of proto-oncogenes don't have a role in cell division.

37. Mutations in tumor suppressor genes leading to cancer always act in a recessive manner. a. True b. False 38. Which cell cycle checkpoint is most responsible for the decision of the cell to commit to dividing? a. the S/G2 checkpoint b. the G1/S checkpoint c. the spindle-assembly checkpoint d. the G2/M checkpoint e. the G1/G2 checkpoint 39. A rare situation where only one of the two copies of a particular tumor-suppressor gene needs to be inactivated before there is a progression toward cancer is called: a. clonal evolution. b. loss of heterozygosity. c. signal transduction. d. aneuploidy. e. haploinsufficiency.

40. What specifically do most people inherit when they inherit a predisposition to a particular cancer such as retinoblastoma? a. a mutation that causes the overexpression of a DNA repair gene b. a mutation that causes telomerase to have reduced expression in somatic cells c. a deleterious mutation in one copy of a tumor-suppressor gene d. a deletion that removes one copy of an oncogene e. an extra X chromosome from the mother

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 23_7e 41. In Burkitt lymphoma patients, despite translocation, the oncogene c-MYC remains nearly intact in its new location. Yet c-MYC is believed to be responsible for the lymphoma because the c-MYC: a. DNA sequence undergoes hypermethylation b. DNA sequence is nearly intact but is inverted in the new position c. gene is released from inhibition by miRNAs d. gene is placed under the control of B-cell-specific gene regulatory sequences and is highly expressed e. DNA sequence undergoes several point mutations 42. Normal cellular genes whose products are involved in facilitating cell division to occur under appropriate conditions are called: a. proto-oncogenes. b. tumor-suppressor genes. c. passenger genes. d. inhibitor genes. e. driver genes.

43. A cell is unable to progress into the S phase of the cell cycle. Further studies indicate that the RB–E2F complex is unable to dissociate in this cell and thus remains together. Which of the following genes is MOST likely mutated? a. a gene that encodes a cyclin-dependent kinase b. a gene that encodes an acetylase c. a gene that encodes an enzyme that removes phosphate groups from RB d. a gene that encodes an enzyme that phosphorylates E2F e. a gene that encodes the Ras protein 44. Which of the following is a process whereby cancer cells travel to other sites in the body and establish secondary tumors? a. oncogenesis b. angiogenesis c. malignancy d. secondary tumorigenesis e. metastasis

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 23_7e 45. Mutant forms of tumor-suppressor genes typically behave in a recessive manner. However, a recent study of patients with multiple myeloma revealed that approximately 10% of these patients have one normal p53 gene and one mutant p53 gene. If it is the mutant p53 gene that is responsible for this group's cancer, which of the following statements would most likely be TRUE? a. The p53 gene can act in a haploinsufficient manner. b. The p53 gene is not a tumor-suppressor gene in these patients but rather an oncogene. c. The mutant p53 gene most likely is involved in DNA-repair pathways. d. The mutant p53 gene is having epigenetic effects in these patients. e. The mutant p53 gene is being overexpressed in these patients.

46. Which of the following would be the result of a mutation that inactivates cyclin-E? a. inactivation of RB b. inactivation of cyclin-D c. inactivation of Ras d. failure of the cell to progress through the G2/M checkpoint e. failure of E2F to bind to DNA and stimulate transcription 47. The increased levels of telomerase associated with many tumor cells likely promote cancer by: a. enhancing levels of DNA repair so that cells remain normal and have stable genomes and thus would be able to replicate their DNA and divide more often. b. promoting the efficiency of the spindle-assembly checkpoint. c. reducing the expression of several oncogenes. d. allowing cells to continue to divide when normally chromosomes should shorten beyond a point where division would be no be longer possible. e. decreasing the number of epigenetic changes that would promote cancer.

48. Consider the Ras signal-transduction pathway. Suppose a mutation occurred in the gene that encodes MEK that renders it nonfunctional. Which of the following effects would you see in this situation? a. Binding of the growth factor to the receptor would occur, but the receptor's conformational change that is needed for the pathway to go forward would not occur. b. Activated Ras would no longer have the ability to activate Raf, and the signal-transduction pathway would be halted at this step. c. Ras would lose the ability to bind GTP and thus could no longer become activated, and the signaltransduction pathway would be halted at this step. d. MAP kinase would not be activated and thus could not move into the nucleus to activate transcription factors. e. Inactive Ras would move into the nucleus and inactivate tumor-suppressor transcription factors, which could ultimately result in cancer.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 23_7e 49. Retinoblastoma (RB) protein is important in regulating the cell cycle. It must be completely phosphorylated before the cell can move from the G1 phase to the S phase of the cell cycle. Retinoblastoma is a type of cancer. Based on this information, what do you think goes wrong in retinoblastoma?

50. What are two processes that determine how quickly mutations accumulate within a cell?

51. List three observations consistent with the idea that cancer arises through an accumulation of mutations in several genes that promotes cellular proliferation in single cells.

52. A headline in the June 12, 2009, issue of Science Daily proclaimed that "MicroRNA Replacement Therapy May Stop Cancer in Its Tracks." The article described a study in which a virus was used to deliver a microRNA gene to cancerous and noncancerous liver cells in mice. Aggressive liver tumors were stopped in their tracks in 8 out of 10 treated mice. Explain how delivery of a gene encoding microRNA might be able to reduce the growth of cancer cells.

53. What is one line of evidence supporting the idea that cancer is influenced by environmental factors?

54. Briefly describe the role that the genes APC, p53, and Ras may have in the development of colon cancer.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 23_7e 55. Briefly describe the relationship among cyclins, cyclin-dependent kinases (CDKs), and cancer.

56. Human papilloma virus (HPV) is a DNA virus that is associated with cervical cancer. Despite the high prevalence of HPV in the population, the incidence of cervical cancer in the United States has decreased 75% in the past 40 years. a. What are some reasons that cervical cancer has decreased in the United States? b. How can cervical cancer rates be lowered in countries where the rates are still high? c. Based on what you know about some different types of cancers and their genetic bases, can you suggest ways in which some of these other cancers might be addressed with respect to treatment and prevention?

57. Why is it more difficult for researchers to identify tumor-suppressor genes than oncogenes?

58. How can defects in DNA-repair mechanisms lead to the development of cancer?

59. What is angiogenesis? Why are mutations involving angiogenesis associated with tumors?

60. In cancer, mutations in stimulatory genes (such as in proto-oncogenes) are often dominant, whereas mutations in inhibitory genes (such as tumor-suppressor alleles) are often recessive. Why?

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 23_7e 61. What are oncogenes and tumor-suppressor genes? How are they involved in carcinogenesis?

62. Distinguish between driver and passenger mutations in cancer genomes. Propose a strategy whereby "knockin" technology (Chapter 19) could be used to distinguish between driver and passenger mutations. What might be a major limitation to this strategy?

63. What are the important checkpoints in the cell cycle?

64. Explain how DNA sequencing studies can aid in our understanding of cancer formation.

65. For each important checkpoint in the cell cycle, what must happen for the cycle to continue past the checkpoint?

66. Describe an example in which environmental factors interact with a genotype to produce cancer.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 23_7e 67. Chromosome mutations are common in tumor cells. Do these mutations cause cancer, or are they a result of cancer? What evidence is there for your conclusion?

68. The development of colon cancer is relatively well understood. Describe the steps of the development of colon cancer, including any genetic components.

69. Explain how hypermethylation and hypomethylation may be associated with cancers.

70. Explain briefly why changes in oncogenes result in more rapid progression of a cancer compared to changes in tumor-suppressor genes. Mention one situation in which changes in a tumor-suppressor gene have a similar likelihood of causing cancer as changes in an oncogene.

71. What is unique to the DNA changes observed in the Apaf-1 gene in malignant melanoma cells? What are such changes to DNA called, and why do you think these types of changes in cancer have caught researchers' attention?

72. What is autophagy?

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 23_7e 73. List three ways in which proto-oncogenes can be converted to oncogenes by viruses.

74. Mutations in the gene TP53 encoding the protein p53 are found in over 50% of human cancers. What type of protein is p53 at a molecular level? Describe three cellular functions that p53 plays and how losses of those functions can lead to tumorigenesis.

75. Most of the inherited forms of cancer involve a germline mutation in a tumor-suppressor gene while only a very few inherited cancers involve a mutation in an oncogene. What is the most reasonable explanation for this difference?

76. Explain how retroviruses may cause cancers.

77. If you were assigned a new form of cancer to study, you might want to determine whether it has a strong genetic basis or whether it is caused primarily by environmental factors. Propose some ways in which you could attempt to determine which hypothesis is correct. Which methods would be the easiest to use first?

78. What is a possible effect of mutations in genes that influence chromosome segregation during division?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 23_7e 79. With respect to cancer, autophagy has been described as both "friend and foe." Explain why.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 23_7e Answer Key 1. b 2. e 3. b 4. c 5. b 6. b 7. b 8. b 9. d 10. d 11. e 12. d 13. d 14. b 15. d 16. c 17. a 18. a 19. c 20. d 21. a 22. c 23. d 24. c 25. a 26. a Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 23_7e 27. d 28. b 29. b 30. b 31. c 32. c 33. b 34. b 35. c 36. a 37. b 38. b 39. e 40. c 41. d 42. a 43. a 44. e 45. a 46. e 47. d 48. d 49. Retinoblastoma is one example of several cancers that involve problems with the G1/S checkpoint. In this form of cancer, the RB protein does not work properly, and the cell continues past the checkpoint regardless of whether it is ready to do so.

50. DNA repair and apoptosis are two processes that can affect how quickly mutations accumulate within a cell. Active DNA-repair processes usually limit the number of mutations that are created through mistakes in DNA replication and other sources. Often DNA repair is limited in cancer cells and mutations accumulate. When mutations accumulate and the cell becomes unable to maintain normal activities, it may undergo apoptosis and die. In many cancers the ability to undergo apoptosis is compromised and cells continue to grow and accumulate mutations.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 23_7e 51. (1) When single tumors are dissected and carefully analyzed, tumor cells typically contain anomalies in several different chromosomes.

(2) Several different tumor-suppressor genes have been shown to encode enzymes necessary for DNA repair.

(3) Early stages of colon cancer have a limited number of mutated tumor-suppressor genes and mutated oncogenes while later stages have several additional mutated genes.

52. MicroRNAs are small RNAs that pair with complementary sequences on mRNA and degrade the RNA or otherwise inhibit its translation and thereby reduce or eliminate expression of the corresponding gene product.

Many tumor cells exhibit widespread reduction in the expression of many microRNAs. The evidence suggests that these microRNAs are expressed at high levels in normal cells and probably inhibit the expression of oncogenes.

Genetic changes that reduce the abundance of microRNAs can release expression of oncogenes and bring about tumor formation and progression. Presumably the microRNA used in the study enters the tumor cells and shuts down expression of one or more oncogenes and stops the progression of the cancer.

53. There are significant differences in the incidences of specific cancers around the world, and studies show that immigrants to a country will, over time, tend to exhibit the cancer incidence of their host country, regardless of differences in the cancer rate in their country of origin. Many students will surely answer this question with the association of cigarette smoking with lung cancer.

54. All three of these genes are often mutated in colon cancer. The APC mutation often occurs early in the development of the cancer. Among the several functions of APC, a role in ensuring normal segregation of chromatids at mitosis may be particularly important in the initiation of colon cancer. Mutations in p53 are found in many cancers, and this gene also has many functions that include delaying cell division if defects involving DNA are found and inducing apoptosis if repair of damage is not possible. The Ras gene product is involved in signal transduction, and its alteration may result in aberrant cell signaling.

55. Cyclins and CDKs play critical roles in regulating the cell cycle. Cyclin concentration oscillates during the cell cycle, whereas the concentration of CDKs remains relatively constant. Cyclins bind and then activate CDKs to initiate key events in the cell cycle. Thus genes that code for cyclins and for factors that stimulate the formation of activated CDKs are often oncogenes. Genes that inhibit the formation of activated CDKs are good candidates for tumor-suppressor genes. Mutated cyclin genes have been shown to be associated with various forms of cancer, and genes that encode inhibitors of CDKs are mutated or missing in many cancer cells.

56. a. A major reason has been the increased use of Pap tests to detect early precancerous and cancerous growth, allowing time for effective treatment. Vaccination may further lower the rate in future years. b. Ideally, increased use of Pap tests, combined with vaccination, would be used. Realistically, vaccination may provide an easier method for effectively lowering cervical cancer rates in areas where medical care is lacking and/or difficult to implement. c. For cancers with genetic predispositions, the most effective early screening may be genetic testing and analyses of family histories. For cancers with strong environmental components, early-screening programs for those with risk factors, avoidance of risk factors, and similar approaches may be effective. For cancers associated with viruses, vaccinations may be effective, combined with screening programs (especially for those cancers known to be affected with a particular virus).

Students may come up with specific approaches for specific forms of cancer discussed in the text.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 23_7e 57. Tumor-suppressor genes inhibit cancer and are therefore recessive—both alleles must be mutated before the inhibition of cell division is reversed. Because it is the loss of function in tumor-suppressor genes that promotes cell proliferation and cancer, these genes cannot be identified by the simple gain-of-function tests used for identifying oncogenes (introducing the gene into normal cells and screening for a cancerous response).

58. Defects in DNA-repair mechanisms increase mutation rates, increasing the probability of a mutation that could lead to cancer.

59. Angiogenesis is the formation of blood vessels. Mutations associated with angiogenesis can allow excessive blood vessel formation, providing oxygen and nutrients to tumor cells.

60. Many mutations in proto-oncogenes are gain-of-function mutations resulting from base-pair substitutions. They carry out a new function such as stimulating cell division without the need for a normal growth factor, and only one mutant copy of the gene is needed for this new phenotype. For tumor-suppressor genes, most mutations are lossof-function mutations, and the remaining normal copy of the gene provides sufficient gene product to carry out its function, which is often some inhibitory aspect to cell division.

61. The control of the cell division cycle is regulated by stimulatory genes (oncogenes) and inhibitory genes (tumorsuppressor genes), and mutations in either type can lead to cancer by deregulating the cell cycle. Mutations in an oncogene can lead to an overactive cellular process, resulting in unregulated cellular proliferation. A defective tumor-suppressor gene can lead to the same result by blocking the activation of a critical "braking" process necessary for the appropriate control of cell-cycle progression.

62. Driver mutations contribute directly to the development of cancer. Passenger mutations might result from genetic instability in cancer cells, but they do not contribute to the cancer process. "Knock-in" technology is used to replace a section of a normal gene with a mutant sequence. Candidate mutations identified in a cancer genome sequencing project could be introduced into noncancerous cells by knock-in and assessed for their ability to cause the cells to become cancerous. Those mutations that cause normal cells to become cancerous would be identified as driver mutations. Those that do not cause cancer might be passenger mutations. However, some driver mutations might be misidentified as passenger mutations if they depend on the presence of other "codriver" mutations to contribute to the cancer process.

63. The G1/S checkpoint is at the end of the G1 or G0 phase of the cell cycle.

The G2/M checkpoint is at the end of the G2 phase of the cell cycle.

The spindle-assembly checkpoint is in metaphase of mitosis.

There is also a checkpoint within S that some students may mention.

64. DNA sequencing technology has been used to completely sequence the DNA of tumor cells at various stages of cancer development. Tumor DNA sequences can be compared to DNA of normal cells to identify genetic changes that might contribute to cancer formation. Also comparison of tumor DNA at various stages of cancer development (e.g., metastasized versus not metastasized) can allow scientists to better understand mechanisms of tumor evolution and cancer development.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 23_7e 65. The G1/S checkpoint is at the end of the G1 phase of the cell cycle. The cell cannot pass this checkpoint until retinoblastoma (RB) protein is completely phosphorylated, inactivating it. This releases E2F protein, formerly bound to RB protein. E2F protein is a transcription factor needed for the expression of genes required for DNA replication. These genes are required for DNA to be replicated in the S phase of the cell cycle. (Students may add that cyclin D and cyclin E increase in concentration and associate with cyclin-dependent kinases during G1; cyclinD-CDK and cyclin-E-CDK phosphorylate RB.) The G2/M checkpoint is at the end of the G2 phase of the cell cycle. Mitosis-promoting factor (MPF) is activated and gradually increases in concentration. MPF phosphorylates other proteins, leading to changes associated with mitosis (breakdown of the nuclear membrane, etc.). (Students may add that cyclin B combines with CDK to form MPF.) The spindle-assembly checkpoint does not allow mitosis to continue past metaphase until all of the chromosomes are properly lined up on the metaphase plate. (Students may add that cyclin B is not destroyed until the chromosomes are properly lined up. This causes MPF to remain active. MPF levels increase and promote mitosis through metaphase and then decrease as the cell returns toward interphase.)

66. Although lung cancer is clearly associated with smoking, recent genome-wide association studies show that variation at several genes predisposes some people to smoking-induced lung cancer, either by increasing their likelihood of becoming addicted to smoking or by affecting metabolism of carcinogens in cigarette smoke. Another example is skin cancer in individuals who are genetically defective in DNA-repair pathways, which depends on DNA damage from exposure to sunlight. Such individuals likely would be susceptible to a variety of cancers that result from exposure to carcinogens that cause damage to DNA. An example of such a disorder is xeroderma pigmentosum, where affected individuals have defective nucleotide-excision repair and are unable to properly repair DNA damage caused by the UV rays in sunlight.

67. Chromosome mutations may be both causes and effects of cancer. Some particular mutations are always associated with a particular cancer, suggesting that the mutation causes that type of cancer. For example, the translocation between chromosomes 9 and 22 (Philadelphia chromosome) is often associated with chronic myelogenous leukemia. However, there are also varied chromosomal mutations that are not specifically linked with certain cancers. This is often the case with solid-tissue tumors. This suggests that genetic instability is a general feature of these cells and is an effect, rather than a cause, of the disease, although increased instability will likely promote a more complete metastatic cancer outcome by affecting the expression of additional tumor-suppressor genes and oncogenes. Mutations in DNA-repair genes or cell-cycle checkpoint genes that may increase the frequency of chromosomal mutations are found in many cancers.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 23_7e 68. There can be a genetic basis to colon cancer, but most cases are sporadic. Familial adenomatous polyposis coli is an example of a hereditary colon cancer.

In many sporadic forms of colon cancer a mutation inactivates the APC gene early on. This increases the rate of cell division and causes polyps to form.

Later, mutations develop in the ras oncogene. This leads to a continual stimulatory signal for cell division (the ras proto-oncogene is involved in signal transduction). Mutations in p53 also occur. These result in accumulations of mutations because cell division occurs when cells are already damaged. p53 normally prevents division of genetically damaged cells and aids in chromosome segregation.

69. Hypermethylation of DNA correlates with decreased expression of genes in the methylated region. Tumorsuppressor alleles may be inactivated. Hypomethylation is less clearly understood but may contribute to higher mutation rates by causing increased chromosome instability. It may also increase expression of oncogenes.

70. Oncogenes are dominant-acting cell-division stimulatory genes, while tumor-suppressor genes are recessive celldivision inhibitory genes. A single mutation in DNA is sufficient to give rise to an oncogene, but two mutations in the same tumor-suppressor gene are usually required to remove inhibition of cell division. Therefore, changes in oncogenes result in a more rapid progression of cancer.

If an organism is predisposed to cancer by inheriting one mutant copy of a tumor-suppressor gene, then it has to acquire only one mutation, similar to an oncogene, before resulting in cancer.

(The following answer is also OK for the second part of the question.)

In some cases, haploinsufficiency of a tumor-suppressor gene cannot be tolerated, or organisms with this condition are more likely to develop cancer because other factors may combine with the lowered tumor-suppressor product to cause cancer.

71. The unique change is that the promoter of this gene is hypermethylated, yet the DNA sequence and the location of Apaf-1 in the chromosome remain unchanged. Reversibly modifying DNA without altering its sequence is called an epigenetic change. Because an epigenetic change to DNA involves no sequence or gross chromosomal changes, the cancerous alteration may be amenable to drug therapy, a possibility that has inspired interest among researchers who aim to control this type of cancer.

72. Autophagy is a cellular process in which proteins, cytoplasm, damaged organelles, and other cellular components are incorporated into intracellular vesicles and transported to lysosomes. Within lysosomes, the cellular components are degraded and recycled.

73. i. The sequence of the proto-oncogene may be altered or truncated when the proto-oncogene is incorporated into the viral genome. The mutated gene could then produce an aberrant protein product that triggers uncontrolled cell proliferation when the virus invades another cell. ii. Through recombination, a proto-oncogene may become juxtaposed next to a new promoter or enhancer, which can then cause the gene to be overexpressed. iii. The function of a proto-oncogene in the host cell may be altered when viral DNA inserts into the gene. The mutated gene could then produce an aberrant protein product that triggers uncontrolled cell proliferation.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 23_7e 74. p53 is a transcription factor with multiple domains including a DNA binding domain. (It also has two N-terminal transactivation domains, a Pro-rich domain, an NLS, an oligomerization domain and a C-terminal regulatory domain.) p53 acts to regulate the cell cycle. If a cell fails the G1/S checkpoint because of DNA damage, p53 arrests the cell in G1 allowing time for DNA repair. A loss of that role can cause a mutated cell to move forward inappropriately to cell division which may lead to tumorigenesis. p53 acts to promote apoptosis in cells with DNA damage or abnormal chromosomes. A loss of that role will allow for survival of mutated cells which may then become tumorigenic. p53 also acts to maintain genome integrity. A loss of that role can lead to aneuploid cells that are potentially tumorigenic.

75. It is likely that a zygote that receives a mutated oncogene would not develop normally and would abort before birth.

Such mutations are dominant and would be expressed in the heterozygous condition and these genes have normal essential functions during development when not mutated. Mutations in tumor-suppressor genes are usually recessive, so a heterozygous individual would be phenotypically normal unless the second copy of the gene also becomes inactivated.

76. Retroviruses are RNA viruses that use reverse transcriptase to make a DNA copy that can integrate into the host chromosome. Retroviruses can (1) mutate host genes when inserted and (2) alter the expression of host genes (example.g., by inserting promoters).

77. Initially, it may be helpful to look at family histories to see whether the disease is more common among relatives than in the general population. If there is evidence that the disease runs in families, then more time-consuming molecular genetic analyses can be used to attempt to identify specific genes that may be associated with the disease. Specific molecular techniques can also be discussed. It may also be helpful to look at incidences of the disease in different populations, such as migrant populations, to see whether populations with similar genetic profiles show different disease profiles in different environments. Surveys may be useful to attempt to determine whether any suspected environmental factors differ significantly between groups being studied.

78. This can lead to aneuploidies. Aneuploid cells likely contain losses of tumor-suppressor genes and gains of oncogenes, which would help promote cancer.

79. As a friend to cell health (and foe to cancer), autophagy acts to prevent cancer by maintaining cellular homeostasis and the quality of cellular organelles and other cellular components. As a foe to cell health (and friend to cancer), autophagy can help cancer cells survive periods of environmental stress, such as during tumor development and cancer treatments.

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 24_7e Indicate the answer choice that best completes the statement or answers the question.

1. Suppose that researchers estimate that broad-sense heritability of IQ in Sweden is 0.8. Which of the following conclusions is valid, assuming that the estimate is accurate? a. Most variance for IQ in Sweden is due to total genetic variance. b. In Sweden, the environment does not play a large role in determining the IQ of individuals. c. Enriching the environment of poor children in Sweden would not lead to a big improvement on IQ tests. d. Broad-sense heritability of IQ in the United States is probably also close to 0.8. e. If another country has a lower broad-sense heritability for IQ, then this difference must be genetically based.

The height and weight for three dogs are shown below.

Weight in kilograms 7 4 10

Height in centimeters 22 12 30

2. What is the value of the covariance? a. 0 b. 1 c. 3 d. 9 e. 27 3. You are hired as a consultant for a new venture: pet squirrel breeding! Because customers prefer bushy tails on their pet squirrels, you are asked to help produce bushy-tailed squirrels from a local population as quickly and efficiently as possible. Of course, you know that bushy tails are a quantitative genetic trait in squirrels.

What information would you need to obtain in order to accomplish this task? a. the response to selection in the local squirrel population b. the narrow-sense heritability in the local squirrel population c. the selection differential in the local squirrel population d. the dominance variance in the local squirrel population e. the environmental variance in the local squirrel population

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 24_7e 4. Quantitative characters often exhibit a _____ distribution. a. skewed b. normal c. bimodal d. covariance e. tangential The height and weight for three dogs are shown below.

Weight in kilograms 7 4 10

Height in centimeters 22 12 30

5. What is the mean of the height? Round to the nearest whole number. a. 7 b. 12 c. 21 d. 22 e. 80 6. In a population of geese, the narrow-sense heritability for wing span is 0.5. The phenotypic variance is 0.8, and the environmental inheritance is 0.2. What is the additive genetic variance? a. 0.25 b. 0.4 c. 0.5 d. 0.75 e. 1.5 7. The estimated broad-sense heritability for milk production in a herd of dairy cattle is high. Which of the following statements is a valid inference? a. Most of the variation in milk production in the herd is due to additive genetic variance within the herd. b. Most of the variation in milk production in the herd is due to dominant genetic variance. c. Little of the variation in milk production is due to environmental variation within the herd. d. Selective breeding for high milk production in this herd will be effective. e. Broad-sense heritability for milk production will be high in most other herds of dairy cattle.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 24_7e 8. Flower diameter in sunflowers is a quantitative trait. A plant with 6-cm flowers, from a highly inbred strain, is crossed to a plant with 30-cm flowers, also from a highly inbred strain. The F1 have 18-cm flowers. F1 × F1 crosses yield F2 plants with flowers ranging from 6 to 30 cm in diameter, in approximately 4-cm intervals (6, 10, 14, 18, 22, 26, 30). An 18-cm F1 plant is crossed to a 6-cm plant. What is the probability of an offspring with one additive allele, if all genes that influence this trait are unlinked? a. 1/3 b. 1/4 c. 1/6 d. 3/8 e. 1/16 The height of a type of bean plant is determined by five unlinked genes called A, B, C, D, and E with additive alleles. The shortest plants are 130 cm. The tallest plants are 220 cm. The genotypes are known for two bean plants. Plant 1 is genotype AABbccDdEE. Plant 2 is genotype aaBBCcDdEE. 9. What are the heights of Plant 1 and Plant 2? a. 9 cm, 9 cm b. 54 cm, 130 cm c. 130 cm, 184 cm d. 184 cm, 184 cm e. 202 cm, 220 cm 10. Recently, genome-wide association studies that locate genes influencing quantitative traits have been facilitated by the use of: a. pedigrees. b. controlled crosses. c. clonal lineages. d. RFLPs. e. SNPs.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 24_7e The height of a type of bean plant is determined by five unlinked genes called A, B, C, D, and E with additive alleles. The shortest plants are 130 cm. The tallest plants are 220 cm. The genotypes are known for two bean plants. Plant 1 is genotype AABbccDdEE. Plant 2 is genotype aaBBCcDdEE. 11. All the genes are unlinked. Determine the probability of plant 1 (AABbccDdEE) producing an AbcdE gamete and plant 2 (aaBBCcDdEE) prodoucing an aBcdE gamete. a. 1/2 chance from plant 1; 1/2 chance from plant 2 b. 1/4 chance from plant 1; 1/4 chance from plant 2 c. 1/2 chance from plant 1; 1/4 chance from plant 2 d. 1/4 chance from plant 1; 1/2 chance from plant 2 e. 1/4 chance from plant 1; 3/4 chance from plant 2 The height and weight for three dogs are shown below.

Weight in kilograms 7 4 10

Height in centimeters 22 12 30

12. What is the variance of the weight? Round to the nearest whole number. a. 7 b. 9 c. 21 d. 22 e. 81 13. In a group of 10 black bears, the length of the longest claw was measured in each bear. These lengths are listed below in centimeters. What is the mean of these lengths?

2.8, 3.1, 3.8, 3.2, 3.3, 2.9, 3.2, 3.1, 3.5, 3.0 a. 3.19 b. 3.08 c. 3.11 d. 3.30 e. 3.16

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 24_7e The height and weight for three dogs are shown below.

Weight in kilograms 7 4 10

Height in centimeters 22 12 30

14. What is the standard deviation of the height? Round to the nearest whole number. a. 3 b. 9 c. 21 d. 42 e. 81 15. If a QTL is linked to a genetic marker, which of the following statements will be TRUE with respect to the recombination frequency between the two? a. It must be greater than 50%. b. It must be 50%. c. It must be less than 50%. d. It must be less than 20% e. It must be less than 5%.

16. _____ can cause a single genotype to produce a range of potential phenotypes. a. Epistasis b. Genetic variance c. Threshold effects d. Environmental effects e. Heritability

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 24_7e The height and weight for three dogs are shown below.

Weight in kilograms 7 4 10

Height in centimeters 22 12 30

17. What is the variance of the height? Round to the nearest whole number. a. 7 b. 9 c. 21 d. 22 e. 81 18. _____ is measured in terms of the original units squared. a. Mean b. Standard deviation c. Variance d. Correlation coefficient e. Narrow-sense heritability The height and weight for three dogs are shown below.

Weight in kilograms 7 4 10

Height in centimeters 22 12 30

19. What is the value of the correlation coefficient? a. 0 b. 1 c. 3 d. 9 e. 27

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 24_7e 20. What is the standard deviation of the weight? Round to the nearest whole number. a. 3 b. 9 c. 21 d. 42 e. 81 21. A series of experiments shows that oil content in a diploid grain is influenced by five genes (a through e) with additive alleles. The highest-producing strain has 20% oil content; the lowest has close to 0%. A plant of unknown genotype has an oil content of 12%. What is a possible genotype for this plant (+ = additive alleles)? a. a+a+b+b+c+c+d+de+e b. a+a+b+b+ ccddee c. a+a+b+bccddee d. aab+bc+cd+d+e+e+ e. a+a+b+b+c+c+d+dee 22. Knowing the _____ of a trait has great practical importance because it allows statistical predictions regarding the phenotypes of offspring to be made on the basis of the parents' phenotypes. a. variance b. inbreeding coefficient c. heritability d. genotype e. mean 23. Phenotypic variation in tail length of unicorns has the following components: Additive genetic variance Dominance genetic variance Genic interaction variance Environmental variance Genetic–environmental interaction variance

= 0.2 = 0.4 = 0.1 = 0.2 = 0.5

What is the broad-sense heritability for tail length in these unicorns? a. 0.14 b. 0.3 c. 0.5 d. 0.7 e. 1.4

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 24_7e 24. A QTL: a. is one of the genes that influences a trait. b. is a chromosomal region containing genes that influence a quantitative trait. c. will not contain any genes other than the ones influencing a trait. d. is a measure of the phenotypic variation in a quantitative trait. e. is a measure of the genetic variation in a quantitative trait.

25. Two highly inbred tobacco plants are crossed. One has dark green leaves. The other has yellow leaves. The F1 have light green leaves. Five hundred progeny from F1 × F1 crosses are analyzed. Their leaves show continuous variation in color, but none has dark green or yellow leaves. What do these data suggest about the number of genes determining this trait? a. There are two genes that determine this trait. b. There are three genes that determine this trait. c. There are four genes that determine this trait. d. There are more than four genes that determine this trait. e. There is not enough information to estimate the number of genes.

26. Do the principles discovered by Mendel for discontinuous traits also apply to the inheritance of traits that exhibit continuous variation? Why or why not? a. Yes. The environmental factors that control continuous traits also control discontinuous traits according to Mendel's principles. b. Yes. Quantitative characteristics can be explained by additive effects of multiple genes, but the behavior of each gene can be determined by Mendel's principles. c. No. Continuous traits are controlled by both genes and the environment and, therefore, Mendel's principles would not apply. d. No. Discontinuous traits are the result of multiple genes that have additive effects and the behavior of continuous traits as defined by Mendel is the result of just a few loci. e. No. Mendel's principles of inheritance cannot be applied to quantitative traits because they do not involve dominance relationships.

27. _____ measures the effect of alleles at a locus interacting in a nonadditive fashion. a. Additive genetic variance b. Dominance genetic variance c. Genetic variance d. Genetic–environmental interaction variance e. Phenotypic variance

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 24_7e 28. Estimates of heritability made by comparing related individuals assume that the environments of related individuals: a. are more similar than those in which unrelated individuals are raised. b. are not any more similar than those in which unrelated individuals are raised. c. are manipulated so that some siblings within each family experience either one of the extreme ends of the environmental range. d. have no effect at all on phenotypic variance in the family. e. are identical for every sibling in the family.

29. Phenotypic correlation may be due to genetic correlations such as _____, which is when one gene influences two or more traits. a. broad-sense heritability b. dominance genetic variance c. pleiotropy d. QTL e. thresholds 30. Phenotypic variation in a trait is often represented as a _____, which graphs the number of each phenotypic class in a sample. a. bimodal distribution b. frequency distribution c. regression line d. correlation e. variance plot 31. Total phenotypic variance CANNOT be broken down into which component? a. genetic–environmental interaction variance b. genetic variance c. environmental variance d. heritability e. genic interaction variance

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 24_7e The height of a type of bean plant is determined by five unlinked genes called A, B, C, D, and E with additive alleles. The shortest plants are 130 cm. The tallest plants are 220 cm. The genotypes are known for two bean plants. Plant 1 is genotype AABbccDdEE. Plant 2 is genotype aaBBCcDdEE. 32. If these two plants are crossed, could they produce a plant that is taller than either parent? How? a. No. Even if the progeny plants had seven additive alleles, it is not possible for it to be taller than either of the parents. b. No. Environmental factors would prevent the offspring of the cross from inheriting additional additive alleles. c. Yes. It is possible for these plants to produce a plant that has seven or eight additive alleles, making it taller than either parent. d. Yes. It is possible for the progeny plants to be taller than either parent, but only if a nondisjunction event occurred in either of the parents when making gametes. e. No. Because the parent plants are homozygous for most of the genes, they will be taller than any progeny plants they produce.

33. Imagine a novel animal species, G. geneticus, that shows polygenic inheritance and additive effects with respect to eye pigmentation. White-eyed animals have no pigmentation while black-eyed animals have full pigmentation. A cross is carried out at the P0 generation between a pure-breeding (homozygous) white-eyed and black-eyed animal. F1 progeny are interbred and produce F2 animals. The brood size of G. geneticus is large, and 768 progeny are counted. Six F2 progeny have white eyes, and six F2 progeny have black eyes.

How many loci affect eye color in G. geneticus? a. one b. two c. three d. four e. five The height of a type of bean plant is determined by five unlinked genes called A, B, C, D, and E with additive alleles. The shortest plants are 130 cm. The tallest plants are 220 cm. The genotypes are known for two bean plants. Plant 1 is genotype AABbccDdEE. Plant 2 is genotype aaBBCcDdEE. 34. What will be the genotype and phenotype of a plant created from an AbcdE gamete from plant 1 (AABbccDdEE) and an aBcdE gamete from plant 2 (aaBBCcDdEE)? What is the probability of a plant with this genotype? What will be the height of this plant? a. There is a 1/4 chance of an AaBbccddEE genotype, and it will be 166 cm in height. b. There is a 1/2 chance of an AaBbccddEE genotype, and it will be 148 cm in height. c. There is a 1/16 chance of an AaBbccddEE genotype, and it will be 148 cm in height. d. There is a 1/2 chance of an AaBbccddEE genotype, and it will be 115 cm in height. e. There is a 1/16 chance of an AaBbccddEE genotype, and it will be 166 cm in height.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 24_7e 35. When _____ is high for a particular trait, offspring tend to resemble their parents for that trait. a. phenotypic variance b. the number of genes controlling a trait c. narrow-sense heritability d. genetic–environmental interaction e. environmental variance 36. Since the entire population is too large to work with, a sample is often used to characterize the population.

This sample should: a. be at most 5% of the population size to minimize the computational efforts. b. be randomly selected so that it represents the entire population. c. include individuals that are selected to fully represent the extreme values in the distribution. d. be small enough to mitigate against chance events skewing the distribution of the sample.

The height of a type of bean plant is determined by five unlinked genes called A, B, C, D, and E with additive alleles. The shortest plants are 130 cm. The tallest plants are 220 cm. The genotypes are known for two bean plants. Plant 1 is genotype AABbccDdEE. Plant 2 is genotype aaBBCcDdEE. 37. If the progeny plant resulting from an AbcdE gamete from plant 1 (AABbccDdEE) and an aBcdE gamete from plant 2 (aaBBCcDdEE) was only 60 cm because of an allele in a gene other than A–E, what is the most likely genetic phenomenon is at work? a. A mutation occurred that is epistatic to the height genes because, no matter what genotype the plant has for the genes A–E, the phenotype is still 60 cm. b. Crossing over occurred within the A–E region that created recombination in the progeny. c. There must have been a change in the environment, such as low water, that caused the stunted growth in the plant. d. The genes A–E do not actually work in an additive manner but rather in a dominant mode of inheritance. e. Multiple alleles must be present in at least two of the loci involved.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 24_7e 38. Two varieties—X (susceptible) and Y (resistant)—of eggplant were used to characterize the genetics of resistance to a necrotizing plant pathogen. Eight thousand plants from each variety were inoculated, and the average number of necrotic lesions produced per leaf was recorded. Variety X averaged 52 lesions per leaf, whereas variety Y averaged six lesions per leaf. The F1 progeny from a cross between variety X and variety Y plants averaged 21 lesions per leaf, with a standard deviation of 3.7. Of the F2 progeny, 8000 were again tested by inoculation, and a total of four plants produced were phenotypically similar to the original parental plants—two plants for each extreme (i.e., parental) phenotype. Overall, the F2 plants averaged 18 lesions per leaf with a standard deviation of 7.5. Determine the broad-sense heritability of resistance to this particular pathogen among the F2 plants, assuming that there is no genetic–environmental interaction variance for this trait. a. 0.426 b. 0.563 c. 0.500 d. 0.757 e. 0.832 39. Suppose a new species of shark is discovered, and the dorsal fin length is found to be determined by additive alleles at three independently assorting loci (A and a, B and b, C and c). The dorsal fin length of aabbcc sharks is 12 cm, and the dorsal fin length of AABBCC sharks is 30 cm. Assuming that the alleles contribute equally, which of the following genotypes would represent a shark with a dorsal fin length of 18 cm? a. aaBbcc b. AAbbcc c. AaBbCc d. AABBcc e. AaBbcc 40. Flower diameter in sunflowers is a quantitative trait. A plant with 6-cm flowers, from a highly inbred strain, is crossed to a plant with 30-cm flowers, also from a highly inbred strain. The F1 have 18-cm flowers. F1 × F1 crosses yield F2 plants with flowers ranging from 6 to 30 cm in diameter, in approximately 4-cm intervals (6, 10, 14, 18, 22, 26, 30). The number of different genes influencing flower diameter in this plant is: a. three. b. four. c. five. d. six. e. seven.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 24_7e The height and weight for three dogs are shown below.

Weight in kilograms 7 4 10

Height in centimeters 22 12 30

41. What is the mean of the weight? Round to the nearest whole number. a. 7 b. 12 c. 21 d. 22 e. 80 42. In a group of 10 black bears, the length of the longest claw was measured in each bear. These lengths are listed below in cm. What is the variance of these lengths?

2.8, 3.1, 3.8, 3.2, 3.3, 2.9, 3.2, 3.1, 3.5, 3.0 a. 3.19 b. 3.08 c. 3.11 d. 3.30 e. 3.16 43. In a normal distribution, 99% of the measurements fall within: a. 1% of the mean. b. 5% of the mean. c. plus or minus one standard deviation of the mean. d. plus or minus two standard deviations of the mean. e. plus or minus three standard deviations of the mean.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 24_7e 44. Distributions A and B in the figure below have:

a. the same mean and the same variance. b. different means and different variances. c. the same mean but different variances. d. different means but the same variance. e. different means but are correlated.

45. Two varieties—X (susceptible) and Y (resistant)—of eggplant were used to characterize the genetics of resistance to a necrotizing plant pathogen. Eight thousand plants from each variety were inoculated, and the average number of necrotic lesions produced per leaf was recorded. Variety X averaged 52 lesions per leaf, whereas variety Y averaged six lesions per leaf. The F1 progeny from a cross between variety X and variety Y plants averaged 21 lesions per leaf, with a standard deviation of 3.7. Of the F2 progeny, 8000 were again tested by inoculation, and a total of four plants produced were phenotypically similar to the original parental plants—two plants for each extreme (i.e., parental) phenotype. Overall, the F2 plants averaged 18 lesions per leaf with a standard deviation of 7.5. How many gene pairs control resistance to this pathogen? a. two b. four c. five d. six e. nine

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 24_7e 46. A population may eventually stop responding to artificial selection because: a. individuals in the population have become homozygous for the alleles responsible for the trait. b. natural selection opposes any further change in the characteristic. c. natural selection can no longer enhance the results of artificial selection. d. individuals in the population have become homozygous for the alleles responsible for the trait and natural selection opposes any further change in the characteristic. e. individuals in the population have become homozygous for the alleles responsible for the trait and natural selection can no longer enhance the results of artificial selection.

47. Heritability indicates the: a. degree to which a characteristic is genetically determined. b. proportion of phenotypic variation in a trait that is due to genetic differences. c. degree to which a characteristic is environmentally determined. d. proportion of phenotypic variation in a trait that is due to the environment. e. extent to which identical twins are phenotypically similar.

The height of a type of bean plant is determined by five unlinked genes called A, B, C, D, and E with additive alleles. The shortest plants are 130 cm. The tallest plants are 220 cm. The genotypes are known for two bean plants. Plant 1 is genotype AABbccDdEE. Plant 2 is genotype aaBBCcDdEE. 48. If the progeny plant from an AbcdE gamete from plant 1 (AABbccDdEE) and an aBcdE gamete from plant 2 (aaBBCcDdEE) was a few centimeters taller than your prediction, what is the most likely reason? a. Crossing over occurred within the A–E region that created recombination in the progeny. b. The height trait is likely also influenced by environmental conditions as well. c. The genes A–E do not actually work in an additive manner but rather in a dominant mode of inheritance. d. Multiple alleles must be present in at least two of the loci involved. e. There are actually more loci involved than just genes A–E. 49. Estimate how many centimeters each allele contributes to the height difference of 90 cm. a. 8 cm b. 9 cm c. 11.25 cm d. 22.5 cm e. 55 cm

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 24_7e 50. If broad-sense heritability for a trait is very high and narrow-sense heritability is very low, which of the following statements must be TRUE? a. Additive variance must be high relative to total phenotypic variance. b. Dominance variance must be high relative to total phenotypic variance. c. Environmental variance must be high relative to total phenotypic variance. d. Genotypic–environmental covariance must be high relative to total phenotypic variance. e. No conclusions can be drawn without additional information.

51. _____ is a way to measure the strength of the association between two variables. a. Regression b. Covariance c. Correlation d. Variance e. Standard deviation 52. You are hired as a consultant for a new venture: pet squirrel breeding! Because customers prefer bushy tails on their pet squirrels, you are asked to help produce bushy-tailed squirrels from a local population as quickly and efficiently as possible. Of course, you know that bushy tails are a quantitative genetic trait in squirrels.

Suppose you are hired by a breeder in a remote location to help set up a breeding program. You want to avoid any unnecessary trips to this remote location. Can you take what you have learned about your local population of squirrels and apply it to the population in the remote location? Why or why not? a. Yes. You can compare the two populations as long as you are studying the same trait in both populations. b. Yes. You can compare the two populations because their narrow-sense heritabilities will be very similar. c. Yes. You can compare the two populations because heritability calculations require environmental differences between populations to be omitted. d. No. Heritability only applies to a particular population in a particular environment. e. No. While heritability can apply across different populations, it would not apply here because the populations vary too greatly in their environments.

53. A _____ allows us to predict the value of one variable from the value of a correlated variable. a. regression b. covariance c. correlation d. variance e. standard deviation

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 24_7e 54. The _____ is the statistic that measures the spread of a distribution around the mean. a. covariance b. correlation coefficient c. regression coefficient d. variance e. average 55. Which model BEST explains inheritance of complex traits such as height, weight, and IQ? a. control of the traits by a single gene with dominant and recessive alleles b. control of the traits by two genes with independent assortment c. control of the traits by more than one gene with no effect of the environment d. control of the traits by more than one gene plus environmental effects e. control of the traits by environmental factors and not genetic factors 56. Unlike most examples of this trait, the height characteristic that Mendel studied in pea plants exhibited _____ variation. a. continuous b. discontinuous c. meristic d. threshold e. quantitative 57. In a population of pigs, the average body weight is 178 kg. Suppose you select the 15 largest pigs, whose average body weight is 205 kg, and interbreed them. The average body weight of the offspring of the selected pigs is 191 kg. If you were to select pigs that have an average weight of 240 kg, what would be the predicted weight of the progeny produced by the selected pigs? a. 30 b. 62 c. 115 d. 208 e. 270

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 24_7e 58. The graph below shows how yield in two varieties of corn respond to different environmental conditions.

Which of the following statements is TRUE? a. Under tested conditions, variety 1 has a higher yield than variety 2. b. Under tested conditions, variety 2 has a higher yield than variety 1. c. Under tested conditions, variety 2 is more sensitive to environment quality than variety 1. d. Variety 1 is genetically superior to variety 2. e. Variety 2 is genetically superior to variety 1.

59. The regression coefficient measures the _____ of the line depicting the relationship between two variables. a. variance b. slope c. distribution d. covariance e. mean

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 24_7e 60. Phenotypic variation in tail length of unicorns has the following components: Additive genetic variance Dominance genetic variance Genic interaction variance Environmental variance Genetic–environmental interaction variance

= 0.2 = 0.4 = 0.1 = 0.2 = 0.5

What is the broad-sense heritability for tail length in these unicorns? a. 0.14 b. 0.3 c. 0.5 d. 0.7 e. 1.4 Indicate one or more answer choices that best complete the statement or answer the question.

61. Genetic variance is comprised of all of the following components EXCEPT: (Select all that apply.) a. additive genetic variance b. dominance genetic variance c. broad-sense heritability d. genic interaction variance e. genetic–environmental interaction variance 62. Which of the following might limit the response of a characteristic to selection? (Select all that apply.) a. Genetic variation for a characteristic may become exhausted after selection for many generations. b. Genetic variation for a characteristic is typically exhausted after 20 generations of selection. c. There may be a threshold limit to further changes in the characteristic under selection. d. Natural selection will supersede artificial selection over time. e. There may be a correlated response between multiple characteristics that acts to limit the response of one characteristic to selection.

63. QTL mapping requires which of the following? (Select all that apply.) a. genetic markers b. offspring c. a genetic map d. a controlled cross e. an estimate of homozygosity in the population

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 24_7e 64. Which of the following statement(s) about heritability is/are TRUE? (Select all that apply.) a. Measures of heritability are specific to a defined population in a given environment. b. The heritability of an individual cannot be estimated. c. If heritability for a trait is high in a particular population, improving the environment is unlikely to result in a significant improvement in the trait over time. d. Heritability does not indicate the degree to which a characteristic is genetically determined. e. Heritability says nothing about the nature of differences between populations in a characteristic.

65. Is it possible for a trait to be discontinuous and yet determined by multiple genetic and environmental factors?

Explain.

66. Garter snakes eat poisonous newts. Selection favors increased resistance to the poison. Garter snakes must also escape their own predators, and they do so by slithering away (fast snakes are more likely to escape than slow snakes are). After studying snakes for some time, you discover that (1) both resistance and speed are highly heritable and (2) selection on resistance and speed is strong. Nonetheless, you find that neither trait increases very much each generation. Explain this result.

67. In a natural population of outbreeding plants, the variance of the total number of seeds per plant is 16. From the natural population, 20 plants are taken into the laboratory and developed into separate true-breeding lines by self-fertilization—with selection for high, low, or moderate number of seeds—for 10 generations. The average variance in the tenth generation in each of the 20 sets is about equal and averages 5.8 across all the sets. Estimate the broad-sense heritability for seed number in this population.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 24_7e 68. Suppose that a study of college students in Sweden estimates heritability of IQ to be 0.8. Based on this evidence, a school board in a struggling urban school district in the United States concludes that improving teaching methods and the physical conditions of schools is useless because intellectual achievement is determined almost entirely by genes and relatively little by the educational environment. Critique this conclusion based on your understanding of heritability.

69. Narrow-sense heritability for IQ scores has been estimated as 0.4. If the mean IQ score in the population is 100, what is the predicted IQ for a group of children whose parents had a mean IQ score of 150?

70. Frankenstein tells Dracula that he thinks that fast-flying bats tend to suck blood faster than slow-flying bats do.

Because flight speed is easier to measure than blood sucking is, Dracula considers how he might use this information in his breeding program. Could Dracula select on flight speed and get an increase in blood sucking? What information would Dracula need to be able to answer this question, and how might he determine this information?

71. Two different varieties of pumpkin have the same mean fresh weight of 22.5 lb. However, one variety (LV) has a very low variance, while the other (HV) has a much higher variance in fresh weight. a. What are some possible reasons for the differences in the variance between these two varieties? b. If you were a pumpkin breeder and wanted to develop a heavier variety of pumpkin, which of these two varieties would you select to start your breeding program? Explain why.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 24_7e 72. A study was conducted to investigate the levels of calcium and growth hormone excreted in chicken dung on a major chicken farm. The variances (V) were calculated and are shown below.

Variance VP VE VA VD

Calcium 321.2 119.7 107.4 94.1

Growth hormone 95.8 69.4 10.1 16.3

a. Calculate the narrow-sense heritability for both traits (calcium and growth hormone levels). b. Which trait, if any, will more likely respond to artificial selection?

73. Count Dracula is interested in raising vampire bats on his Transylvanian ranch. For obvious reasons, he would like his bats to be able to suck blood very quickly, and he needs to have this flock of vampire bats ready for next Halloween. As you might know, blood sucking is a quantitative genetic trait in bats. What information does Count Dracula need in order to know if he will be successful at breeding faster-sucking bats quickly, and how might he determine this information?

74. Cloning is a procedure by which exact genetic duplicates are made. A litter of cloned rabbits was born from a single surrogate mother rabbit. Despite being genetically identical, each of the baby rabbits looked very different from one another. Explain in detail why this might have occurred.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 24_7e 75. A plant breeder has determined the following variances for yield of corn in his fields: Total phenotypic variance Additive genetic variance Dominance genetic variance Environmental variance

100 30 50 20

Assume that there is no genetic–environmental interaction or other variances. a. Calculate the total genetic variance. b. Calculate the broad-sense heritability indicated by the data. c. Calculate the narrow-sense heritability indicated by the data. d. The breeder wishes to improve yield. If the average yield in the starting population is 400 and the breeder selects for breeding plants with an average yield of 500, what will be the expected average yield among the offspring of the selected plants?

76. The table below shows narrow-sense heritability estimates for body weight and egg production in chickens.

Trait Body weight Egg production

h2 0.5 0.2

a. Which chicken trait would respond best to selection? Explain your answer. b. Explain the expected gain if high-egg-producing chickens are selectively bred. c. Explain how and why the h2 would change over generations of selective breeding for high egg production.

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 24_7e 77. A selective breeding program in corn is undertaken to increase oil content. The results are shown below.

a. Based on the results shown in the graph, during which part of the selective breeding program was narrowsense heritability the highest—early, middle, or late? Explain your answer. b. Provide an explanation why the breeding program became ineffective after generation 60.

78. Two varieties—X (susceptible) and Y (resistant)—of eggplant were used to characterize the genetics of resistance to a necrotizing plant pathogen. Eight thousand plants from each variety were inoculated, and the average number of necrotic lesions produced per leaf was recorded. Variety X averaged 52 lesions per leaf, whereas variety Y averaged six lesions per leaf. The F1 progeny from a cross between variety X and variety Y plants averaged 21 lesions per leaf, with a standard deviation of 3.7. Of the F2 progeny, 8000 were again tested by inoculation, and a total of four plants produced were phenotypically similar to the original parental plants—two plants for each extreme (i.e., parental) phenotype. Overall, the F2 plants averaged 18 lesions per leaf with a standard deviation of 7.5. a. How many gene pairs control resistance to this pathogen? b. Determine the broad-sense heritability of resistance to this particular pathogen among the F2 plants, assuming that there is no genetic–environmental interaction variance for this trait.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 24_7e 79. You and a close friend from the University of Tennessee go hiking in the mountains. You are surprised to see a salamander that is bright orange in color. Thinking that some overzealous Tennessee fans have painted this particular amphibian, you walk on. As you continue your hike, you find other salamanders of this species and note that there is variation in the coloration: some individuals are bright orange, but others are drab in color.

You also note that these salamanders often eat bright orange insects. How might you explain the presence of the color polymorphism in this salamander species?

80. You are hired as a consultant for a new venture: pet squirrel breeding! Because customers prefer bushy tails on their pet squirrels, you are asked to help produce bushy-tailed squirrels from a local population as quickly and efficiently as possible. Of course, you know that bushy tails are a quantitative genetic trait in squirrels. a. What information would you need to obtain in order to accomplish this task? b. Given your success at this job, you are hired by a breeder in a remote location to help set up a breeding program. You hate the cold weather and want to avoid any unnecessary trips to the remote location. Can you take what you have learned about your local population of squirrels and apply it to the remote location population? Why or why not?

Using quantitative genetics, researchers identified a gene, DGAT-2, that seems to play a key role in determining the oil content of corn. In high-oil producing strains, DGAT-2 has an insertion of one codon that leads to the insertion of a phenylalanine residue in the protein. In the low-oil producing strain, this insertion is absent.

81. What nongenetic factors would have to be taken into consideration when interpreting the results of the genetic experiments suggested in part (a)?

82. What further genetic tests could you perform to determine whether introducing a genetically modified form of DGAT-2 into low-oil-producing corn strains and/or other plants would lead to increased oil content?

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 24_7e 83. Do the principles discovered by Mendel for discontinuous traits also apply to the inheritance of traits that exhibit continuous variation? Explain your answer.

84. Cleft palate is a birth defect in which the two sides of the roof of the mouth do not fuse before birth. Cleft palate is considered to be a threshold trait. Define "threshold trait" and explain how cleft palate is an example of one.

The height of a type of bean plant is determined by five unlinked genes called A, B, C, D, and E with additive alleles. The shortest plants are 130 cm. The tallest plants are 220 cm. The genotypes are known for two bean plants. Plant 1 is genotype AABbccDdEE. Plant 2 is genotype aaBBCcDdEE. 85. a. Estimate how many centimeters each allele contributes to the height difference of 90 cm. b. The genotypes are known for two bean plants. Plant 1 is genotype AABbccDdEE. Plant 2 is genotype aaBBCcDdEE. What are their heights? c. If the two plants in the previous question are crossed, could they produce a plant that is taller than either parent? How? d. All the genes are unlinked. Determine the probability of an AbcdE gamete from plant 1 and an aBcdE gamete from plant 2. e. What will be the genotype and phenotype of a plant from these gametes, and what is the probability of a plant with this genotype? f. If the progeny plant from the cross in the previous question was only 60 cm because of an allele in a gene other than A–E, what genetic phenomena are at work? g. If the progeny plant with the genotype in part (d) was a few centimeters taller than your prediction, how would you explain it?

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 24_7e 86. Astronomers are able to measure precisely the distance between Earth and Margalit's comet on its approach toward the Sun. During the same time period, the price of oil increases. You measure the correlation coefficient between the price of oil and the distance between Earth and the comet and discover that the correlation coefficient is equal to –1. How would you interpret this correlation?

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 24_7e Answer Key 1. a 2. e 3. b 4. b 5. c 6. b 7. c 8. d 9. d 10. e 11. b 12. b 13. a 14. b 15. c 16. d 17. e 18. c 19. b 20. a 21. d 22. c 23. a 24. b 25. d 26. b Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 24_7e 27. b 28. b 29. c 30. b 31. d 32. c 33. c 34. e 35. c 36. b 37. a 38. d 39. b 40. a 41. a 42. a 43. e 44. b 45. d 46. d 47. b 48. b 49. b 50. b 51. c 52. d 53. a 54. d Copyright Macmillan Learning. Powered by Cognero.

Page 29

Name:

Class:

Date:

Chap 24_7e 55. d 56. b 57. d 58. c 59. b 60. c 61. c, e 62. a, c, e 63. a, b, c, d 64. a, b, d, e 65. Yes, it is. These kinds of traits are described as meristic, meaning that they have a limited number of distinct phenotypes but are quantitative. As noted in the text, litter size in mice is a meristic trait. Other examples of meristic traits would be the number of bristles on fruit-fly abdomens and the number of eggs laid by sea turtles. In these cases, there are limited numbers of distinct phenotypes produced, but the specific phenotype ultimately expressed is controlled by multiple genes (i.e., is polygenic) and by environmental factors. Also, there are threshold traits. A threshold trait is discrete, but underlying the trait is a continuous distribution in a population of genetic and environmental risk associated with the trait that can be explained by multifactorial inheritance. If an individual exceeds a threshold value of risk, that individual will show the trait.

66. Speed and resistance are probably negatively genetically correlated. Positive selection on one trait will cause a negatively genetically correlated trait to decrease. If both speed and resistance are selected to increase, this correlated negative response will dampen the response to selection in both traits.

Note that the question states that both traits are highly heritable and under strong positive selection. Genetic variation should not be exhausted if the heritability is high.

67. An estimate of VP is the variance in seed number in the natural population, or 16. An estimate of VE is the variance in the population in the true-breeding, inbred population, which is presumably homozygous for genes that affect seed number.

VG = VP – VE H2 = VG/VP = (16 – 5.8)/16 = 0.64

Copyright Macmillan Learning. Powered by Cognero.

Page 30

Name:

Class:

Date:

Chap 24_7e 68. Students might note that IQ should not be equated with intellectual achievement, but they must go further than that to demonstrate their understanding of heritability. The school board's decision is based on at least two major misconceptions about heritability. The first misconception is that estimates of heritability are universal, so that an estimate of heritability in one population can be used to estimate it in another population. In fact, an estimate of heritability only applies to the population that was measured in the study. A second major misconception is that high heritability indicates that the environment is not important to the trait of interest. In fact, a high heritability indicates only that there is relatively more genetic variance than environmental variance contributing to differences in the trait.

If the environment is uniformly poor in the school district and there is at least some genetic variation, then heritability will be high. However, this does not mean that performance can't be improved by enriching the environment. Students might also note that the methods used to estimate heritability in humans are not very reliable and that any estimate should be interpreted with caution.

69. The selection differential in this example is 50 IQ units. If the narrow-sense heritability is 0.4, then the increase in IQ over the mean IQ score should be h2 × S = 0.4 × 50 = 20. The predicted IQ, given these assumptions, is 120.

70. He would need to demonstrate that there is a genetic correlation between flight speed and blood-sucking rate. He needs a measure of additive genetic variances for both traits in order to calculate and evaluate an additive genetic correlation.

71. a. VP = VG + VE + VGE. Assuming that the two varieties are grown in the same environment, differences in phenotypic variance could be explained by (i) differential effects of various environmental factors on the different varieties (the varieties have different values of VGE), or (ii) inherent genetic differences between the varieties (the varieties have different values of VG). For example, the variety with higher weight variance may be more genetically diverse. b. In some cases, the variety with higher variance may be better because more variation in weight range would be represented in the progeny; thus, there would be more likelihood of finding significantly heavier individuals to select out. Note, however, that if you choose to start with the higher variance variety (HV), you would hope that a major proportion of the variance in weight observed was due to genetic variation. On the other hand, it is often beneficial to produce progeny with uniform phenotypes. If uniformity has high importance, then the pumpkin variety exhibiting less variance (LV) may be the better one to start with.

72. VP = phenotypic variance VA = additive variance VD = dominance variance VE = environmental variance a. For calcium: h2 = VA /VP = VA /(VE + VA + VD) = (107.4)/(119.7 + 107.4 + 94.1) = 0.334. For growth hormone: h2 = (10.1)/(69.4 + 10.1 + 16.3) = 0.105. b. Artificial selection will more likely alter calcium levels than growth hormone levels.

73. Blood sucking is a quantitative genetic trait. Therefore, Dracula needs to know the narrow-sense heritability of the trait. A good way to calculate the narrow-sense heritability of this trait would be to determine the parent–offspring regression. The slope of the regression line would provide an estimate of the narrow-sense heritability.

Copyright Macmillan Learning. Powered by Cognero.

Page 31

Name:

Class:

Date:

Chap 24_7e 74. Because phenotypic variation is made up of both genetic variation and environmental variation, and we are told that the rabbits are genetically identical, there must be environmental variation that explains the phenotypic differences.

Even within the seemingly "constant" environment of a rabbit's womb, there is environmental variation.

75. a. Total genetic variance VG = VA + VD = 30 + 50 = 80 b. H2 = VG/VP = 80/100 = 0.80 c. h2 = VA /VP = 30/100 = 0.30 d. The proportion of the selection differential (500 – 400 = 100) that shows up as the selection response is given by h2 = 0.30. Therefore, the selection response = h2 × selection differential = 0.30 × 100 = 30 units. The average yield among the offspring will therefore be 430 units.

76. a. The h2 value for body weight is higher, so it would respond better to selection than would egg production. The h2 indicates the proportion of variation in the population that is caused by difference in additive, selectable alleles.

Because h2 is higher for body weight, that trait will respond better to selective breeding using parents with the desired trait. b. The h2 predicts that the expected response to selection will be 0.2. In other words, offspring of the chickens from lines with the highest egg production will differ from the nonselected population by 20%. The mean value of the offspring's population will be 20% higher than the mean value of the nonselected parent population. c. As the additive alleles that give high egg production are selected for, the selected population will become increasingly homozygous for additive alleles in their genotypes. h2 will decrease because less of the variance in the population will be caused by genotypic differences.

77. a. The effectiveness of a selective breeding program depends on h2 . The breeding program was most effective during generations 0–20, indicating that h2 was highest at this time. b. The most likely reason the breeding program became ineffective was that genetic variation for oil content was eventually exhausted by selection for genes that contribute to oil production. Another possibility is that genetic correlations interfered. For example, if higher oil content is associated with lower fertility or viability, then further improvement might not be possible.

78. a. Two of each extreme parental (P1) type plant were produced in the F2 progeny pool. The number of F2 progeny plants expressing either extreme phenotype = 2/8000 = 1/4000. Using the formula (1/4)n and solving for n: 1/4000 1/4096 = (1/4)6, indicating that six gene pairs are involved in the pathogen resistance response. b. The broad-sense heritability (H2) represents the proportion of the phenotypic variance that is due to genetic variance and is calculated by dividing the genetic variance by the phenotypic variance: H2 = VG/VP. If there is not genetic–environmental interaction variance, then VP = VG + VE. Because the F1 plants have identical genotypes, we can use the variance in the F1 as an estimate of environmental variance. Therefore, VE = (3.7)2 = 13.69. Note that the phenotypic variance among the F2 = VG + VE = (7.5)2 = 56.25. Genetic variance among the F2 can be calculated as follows: VG = VP – VE = (56.25 – 13.69) = 42.56. Therefore, among the F2 plants, H2 = (42.56)/(56.25) = 0.757.

Copyright Macmillan Learning. Powered by Cognero.

Page 32

Name:

Class:

Date:

Chap 24_7e 79. An obvious hypothesis is that the variation comes from the fact that there is variation in what salamanders eat. This would imply that the variance is totally environmental: The orange salamanders ate orange bugs and turned orange.

You can test to see if the coloration is environmental by doing a common garden experiment in which you feed an experimental group of salamanders the orange bugs and see if they all turn orange (an environmental effect). If they do not, at least some of the coloration variation is due to genetics.

80. a. The most important thing you need to know is the narrow-sense heritability of tail-bushiness in your local squirrel population. Given the narrow-sense heritability, you can predict the response you would expect from the selection you will impose on tail-bushiness. b. No, you cannot apply what you learned to the remote location population. Genetic variance depends on which genes are present and which particular environment those genes are expressed in. The particular gene combinations present within different populations often differ, and overall phenotypic variance (VP) is composed of both the genetic and environmental variances. Because heritability is specific to a defined population in a given environment, it is not useful to extrapolate heritabilities from one population to another.

81. Environmental influences (temperature, moisture availability, light etc. etc.). To begin with, transgenic or genome edited plants might be grown in the same environment to assess the influence of a DGAT-2 genetic modification in a controlled environment. In follow-up experiments, changing the environmental conditions for growth of potentially high-oil-producing genetically modified plants would be useful to find the conditions that maximize oil production.

82. Answers may vary. A transgene encoding DGAT-2 with the insertion could be introduced into corn or other plants to assess whether it leads to increased oil production. Genome editing techniques such as Crispr/Cas might be used to edit the endogenous DGAT-2 locus of low-oil-producing or other plants, and then test whether this changes oil content.

83. Yes. Over the 18-year period after Mendel's work was rediscovered, work by several eminent geneticists including Nilsson-Ehle, East, Johannsen, and Fisher demonstrated that the inheritance of quantitative characteristics could be explained by the cumulative (i.e., additive) effects of multiple genes, but the behavior of each gene was determined

by Mendel's principles.

84. A threshold trait is a multifactorial trait that has two phenotypes: present or absent. Multiple genes and environmental effects contribute to the development of the trait. If inherited and environmental factors combine to cross a "threshold," the trait is expressed. Babies are born either with cleft palate or with normal palate, so there are only two phenotypes. Inherited factors and events in development combine to determine whether the trait develops or not.

Copyright Macmillan Learning. Powered by Cognero.

Page 33

Name:

Class:

Date:

Chap 24_7e 85. a. The difference between the two heights is 90 cm. There are 10 alleles total (two each for five different genes). If each additive allele contributes equally to increasing height from a baseline of 130 cm, the contribution of each is 9 cm. b. They both have six additive alleles, each contributing 9 cm in height to a baseline of 130 cm, so (130 cm) + (6 × 9 cm) = 184 cm. c. Yes. Because each parent plant contributes one allele of each gene, a progeny plant could inherit all the additive alleles from each parent and potentially be taller than either parent. In this case, a plant could have seven or eight additive alleles (AaBBCcDdEE, AaBbCcDDEE, or AaBBCcDDEE). d. For each gene, there is a 1/2 (50%) chance of either allele, but for the homozygous genes, the two alleles have the same effect on phenotype. The genes are unlinked (independently assorting), so inheritance of each allele is an independent event.

Plant 1: (1A) × 1/2b × (1c) × 1/2d × (1E) = 1/4 Plant 2: (1a) × (1B) × 1/2c × 1/2d × (1E) = 1/4 e. There is a 1/4 × 1/4 = 1/16 probability of an AaBbccddEE plant with four additive alleles, so (130 cm) + (4 × 9 cm) = 166 cm. f. Spontaneous mutation must have occurred in one of the plant germ-lines or in a gamete from one of them. The mutation is epistatic to the height genes because, no matter what genotype the plant has for the genes A–E, the phenotype is still 60 cm. The mutated allele is dominant to the normal allele. g. Multifactorial traits are usually influenced not only by genotype but also by environment. For height, nongenetic factors like infection by a pathogen or availability of sunlight, water, and nutrition can influence the phenotype.

86. At face value, the perfect negative correlation means that a change in the distance is always accompanied by a proportional change in oil prices. However, it should be clear that there is no proven causal relationship between these two variables: Correlation does not imply a cause-and-effect relation.

Copyright Macmillan Learning. Powered by Cognero.

Page 34

Name:

Class:

Date:

Chap 25_7e Indicate the answer choice that best completes the statement or answers the question.

1. One way to define _____ is any change in allelic frequencies within a population. a. mutation b. natural selection c. equilibrium d. evolution e. sampling error 2. The frequency of q in population 1 is 0.8, and the frequency of q in population 2 is 0.3. If the migration rate from population 1 into population 2 is 0.2, what will be the frequency of q in population 2 in the next generation? a. 0.4 b. 0.6 c. 0.5 d. 0.1 e. 0.0 3. If there is random mating in a population and no evolutionary forces are acting on the population, what will be the expected outcome? a. The allelic frequencies will remain the same, but the genotypic distribution will change. b. The genotypic distribution will remain the same, but the allelic frequencies will change. c. Both the genotypic distribution and the allelic frequencies will change. d. Both the genotypic distribution and the allelic frequencies will remain the same. e. No prediction can be made about the genotypic distribution and allelic frequencies from one generation to the next.

4. The A1 allele mutates into the A2 allele at a rate of 4 × 10-5, and the A2 allele mutates into the A1 allele at a rate of 1 × 10-5. Assuming no other evolutionary forces are at work, the frequency of the A2 allele will eventually be: a. 0.8. b. 0.2. c. 0.5. d. 0.25. e. 1.0.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 25_7e 5. _____ has the effect of homogenizing allelic frequencies among populations. a. Migration b. Sampling error c. Directional selection d. Assortative mating e. Selection–mutation equilibrium 6. Huntington's disease is caused by a single dominant allele and results in progressive mental and neurological damage. The disease usually becomes symptomatic when a person is between 30 and 50 years old, and the patient usually dies within 15 years of diagnosis. Approximately 1 in 25,000 Caucasians have this disease.

Huntington's disease has not been associated with any other disease, now or in the past. Why might natural selection not have eliminated such a deleterious allele from the population? a. Natural selection acts through reproduction, and most individuals with Huntington's disease reproduce prior to discovery. b. Diseases tend to remain in populations because of heterozygous carriers. c. Modern health care has acted as an agent against selection. d. Natural selection tends not to work on human diseases and in human populations. e. Natural selection only works on young individuals or newborns. Huntington's disease only works on older people.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 25_7e 7. Use Figure 25.4 to determine the BEST answers for the following question.

When a population is in Hardy–Weinberg equilibrium, what is the significance of two alleles with equal frequencies ( a. These are the allelic frequencies found in most populations. b. These allelic frequencies maximize the proportion of homozygotes in the population. c. These allelic frequencies maximize the proportion of heterozygotes in the population. d. These allelic frequencies guarantee that neither allele will become fixed in a population. e. These allelic frequencies minimize the proportion of heterozygotes in the population.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 25_7e 8. Which of the following evolutionary forces does NOT normally change allelic frequencies? a. nonrandom mating b. mutation c. selection d. drift e. migration 9. When an allele has a frequency of 1.0 in a population, it is _____ in the population. a. lost b. dominant c. overdominant d. fixed e. recessive 10. Human blood type is determined by three alleles, IA , IB, and IO. The alleles IA and IB are co-dominant to each other, and both are dominant to IO. Within a large, randomly mating population (540,000 individuals), the frequencies for the blood type alleles are 0.3 for the IA allele, 0.6 for the IO allele, and 0.1 for the IB allele.

Calculate the expected numbers of people in the population having each of the blood types A, B, AB, and O. a. A = 243,000 people, B = 70,200 people, AB = 32,400 people, and O = 194,400 people b. A = 162,000 people, B = 54,000 people, AB = 3,240 people, and O = 324,000 people c. A = 48,600 people, B = 5,400 people, AB = 32,400 people, and O = 194,400 people d. A = 194,400 people, B = 64,800 people, AB = 32,400 people, and O = 194,400 people e. A = 243,000 people, B = 5,400 people, AB = 32,400 people, and O = 194,400 people 11. Which of the following does NOT bring about evolution in a population? a. small population size b. migration of individuals from a population with a different genetic structure c. mutation d. selection e. random mating 12. _____ is when the heterozygote has a higher fitness than either of the two homozygotes (W11 < W12 > W22).

Allelic frequencies will change in this population until _____. a. Underdominance; an allele is fixed b. Homeostasis; Hardy–Weinberg equilibrium is achieved c. Dominance; a tipping point is passed d. Recombination; linkage equilibrium is obtained e. Overdominance; a stable equilibrium is reached Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 25_7e 13. Suppose that in a population the frequency of a particular recessive condition is 1/400. Assume the presence of only a dominant allele (A) and a recessive allele (a) in the population and that the population is at Hardy– Weinberg equilibrium. What is the frequency of the recessive allele that causes the condition? a. 0.000625 b. 0.0025 c. 0.025 d. 0.05 e. 0.95 14. Rapid changes in allelic frequencies by _____ take place in populations that are small. a. mutation b. natural selection c. inbreeding d. outbreeding e. genetic drift 15. _____ increases the frequency of homozygotes in a population compared to the results of random mating. a. Inbreeding b. Migration c. Outcrossing d. Genetic drift e. Directional selection 16. You are studying cannibals in Borneo and want to determine if a specific village population is in Hardy– Weinberg equilibrium with respect to the two codominant M and N blood type alleles (LM, LN) segregating at the single-gene locus. Choose the answer below that gives the correct expected number of M, MN, and N individuals and the critical value to which you will compare your chi-square value (see Table 3.7 for critical values).

Phenotypes a. M = 357, MN = 525, N = 193; chi-square critical value = 5.991 b. M = 0.331, MN = 0.488, N = 0.180; chi-square critical value = 3.814 c. M = 357, MN = 525, N = 193; chi-square critical value = 3.814 d. M = 0.267, MN = 0.619, N = 0.114; chi-square critical value = 3.814 e. M = 287, MN = 665, N = 123; chi-square critical value = 7.815

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 25_7e 17. In overdominance, the HIGHEST fitness is found in which genotype? a. homozygote for the dominant allele b. homozygote for the recessive allele c. heterozygote d. mutant e. inbred recessive 18. _____ ultimately produces all new genetic variation in a population. a. Outcrossing b. Migration c. Evolution d. Mutation e. Equilibrium 19. The evolutionary force of _____ tends to increase genetic variation within a population but decrease genetic variation between populations, while the evolutionary force of _____ tends to decrease genetic variation within a population but increase genetic variation among populations. a. migration; genetic drift b. mutation; some types of natural selection c. genetic drift; migration d. some types of natural selection; mutation e. positive assortative mating; negative assortative mating 20. The _____ measures the probability of alleles being identical by descent. a. selection coefficient b. inbreeding coefficient c. gene pool d. Hardy–Weinberg law e. migration rate 21. The only way for evolution to take place is if there is _____ within a population. a. nonrandom mating b. natural selection c. genetic drift d. Hardy–Weinberg equilibrium e. genetic variation

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 25_7e 22. The _____ is all of the genetic information within a Mendelian population. a. effective population size b. Hardy–Weinberg equilibrium c. fitness d. genotypic frequency e. gene pool 23. Suppose a disease is caused by a recessive allele. Individuals with the disease produce only about 60% of the offspring of individuals with the normal phenotype. The mutation rate from the dominant to the recessive allele is 1 × 10-5. At mutation–selection equilibrium, what will be the frequency of the recessive allele? a. 0.005 b. 0.000025 c. 0.004 d. 0.00002 e. 0.0 24. _____ occurs when one allele or trait is favored over another. a. Directional selection b. Mutation–selection equilibrium c. Founder effects d. Stabilizing selection e. Overdominance 25. DNA typing is used to compare evidence DNA (E) left at a crime scene to two suspects (S1 and S2).

Suspect 1 is excluded by the evidence, but suspect 2 remains included. What is the frequency of suspect 2's genotype if the allelic frequencies in the population are f(A1) = 0.1, f(A2) = 0.2, and f(A3) = 0.7, and the population is at Hardy–Weinberg equilibrium?

a. 0.01 b. 0.02 c. 0.04 d. 0.28 e. 0.49

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 25_7e 26. By the early 1980s, the population of Florida panthers had dwindled to fewer than 30. Within this population, there was a high frequency of detrimental traits, including low sperm count and undescended testicles in males and kinked tails in both sexes. The future for this iconic population was bleak at best. Which of the following is the BEST solution that would both increase the likelihood that the population would survive and preserve the genetic identity of this unique population? a. Genetic rescue could be performed, in which panthers from another population would be introduced in order to bring in new genetic variation and to reduce the genetic load. b. The population could be removed and relocated to a better habitat, which should help the population recover. c. Individuals with the genetic abnormalities could be removed from the population. d. Some individuals could be moved so they will mate with specific other, healthier animals, which would assure more mixing of nondetrimental alleles. e. There is no viable solution to saving this population.

27. If the frequency of a recessive disease causing allele (q) is 0.005, what is the frequency of individuals with the disease in a population with an inbreeding coefficient of 0.25? a. 0.00127 b. 0.000025 c. 0.991 d. 0.00125 e. 0.0025 28. Through _____, alleles may disappear from a population simply by chance. a. negative directional selection b. genetic rescue c. genetic drift d. fixation e. underdominance 29. By the early 1980s, the population of Florida panthers had dwindled to fewer than 30. Within this population, there was a high frequency of detrimental traits, including low sperm count and undescended testicles in males and kinked tails in both sexes. The future for this iconic population was bleak at best. Which of the following provides the BEST explanation for the observed detrimental traits? a. The mutation rate in this population is very high due to the small number of individuals. b. Natural selection is acting in this population and resulting in more individuals with detrimental traits. c. Migration of individuals into this population has introduced more detrimental traits. d. The population is experiencing a genetic bottleneck, where the genetic variation is declining and harmful alleles are drifting to high frequencies. e. Recurrent mutation and natural selection are acting as opposing forces on this population, and an equilibrium has not yet been reached.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 25_7e 30. _____ refers to the situation where the heterozygote has a lower fitness than both homozygotes (W11 > W12 < W22). a. Overdominance b. Hardy–Weinberg equilibrium c. Underdominance d. Heterozygote superiority e. Genetic rescue 31. When considering the gene pool of a population, Hardy–Weinberg equilibrium may apply to: a. only the entire genome but not individual loci. b. one locus but not necessarily another. c. dominant and recessive alleles but not codominant alleles. d. only to the effective number of breeders. e. only autosomal but not X-linked loci.

32. A Mendelian population is defined by individuals that are: a. interbreeding. b. inbreeding. c. evolving. d. segregating. e. migrating.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 25_7e 33. Use Figure 25.4 to determine the BEST answers for the following question.

When a population is in Hardy–Weinberg equilibrium, what is the approximate frequency of the aa homozygote and the when the frequency of AA is approximately 0.5? a. aa = 0.25, Aa = 0.25 b. aa = 0.25, Aa = 0.5 c. aa = 0.02, Aa = 0.48 d. aa = 0.1, Aa = 0.4 e. aa = 0.4, Aa = 0.1

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 25_7e 34. _____ is the movement of genes between populations. a. Evolution b. Migration c. Positive assortative mating d. Outcrossing e. Genetic drift 35. You are studying cannibals in Borneo and collect the following data with the two codominant M and N blood type alleles (LM, LN) segregating at the single-gene locus. What are the genotypic and allelic frequencies for this population?

Phenotypes a. LM LM = 0.287, LM LN = 0.665, LN LN = 0.123; LM = 0.885, LN = 0.115 b. LM LM = 0.267, LM LN = 0.619, LN LN = 0.114; LM = 0.5, LN = 0.5 c. LM LM = 0.267, LM LN = 0.619, LN LN = 0.114; LM = 0.267, LN = 0.114 d. LM LM = 0.267, LM LN = 0.619, LN LN = 0.114; LM = 0.576, LN = 0.424 e. LM LM = 0.287, LM LN = 0.665, LN LN = 0.123; LM = 0.576, LN = 0.424 36. A new kind of tulip is produced that develops only purple or pink flowers. Assume that flower color is controlled by a single-gene locus and that the purple allele (C) is dominant to the pink allele (c). A random sample of 1000 tulips from a large cultivated field yields 847 purple flowers and 153 pink flowers. What is the frequency of the purple and pink alleles in this field population? a. purple allele frequency = 0.847, pink allele frequency = 0.153 b. purple allele frequency = 0.153, pink allele frequency = 0.847 c. purple allele frequency = 0.82, pink allele frequency = 0.18 d. purple allele frequency = 0.61, pink allele frequency = 0.39 e. purple allele frequency = 0.39, pink allele frequency = 0.61

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 25_7e 37. Use Figure 25.4 to determine the BEST answers for the following question.

When a population is in Hardy–Weinberg equilibrium and p = 0.1 and q = 0.9, what are the approximate genotypic freq a. AA = 0.01, Aa = 0.18, aa = 0.81 b. AA = 0.81, Aa = 0.18, aa = 0.01 c. AA = 0.25, Aa = 0.5, aa = 0.25 d. AA = 0.2, Aa = 0.6, aa = 0.2 e. AA = 0.05, Aa = 0.4, aa = 0.55

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 25_7e 38. If there are two alleles, A and a, in a population and the population is at Hardy–Weinberg equilibrium, which frequency of A would produce the greatest frequency of heterozygotes? a. 0.1 b. 0.25 c. 0.5 d. 0.75 e. 1 39. Genetic diseases in humans are usually rare and recessive. Why are the frequencies of alleles that cause rare, recessive diseases (or other recessive traits, for that matter) generally much higher than the frequency of the diseases (or traits) themselves? a. Diseases caused by dominant alleles are generally lethal, and so most diseases are caused by recessive alleles. b. Most of the rare, recessive alleles within the population are "hidden" within heterozygote carriers, which do not manifest the disease (or express the trait). c. Mutation rates are very low, and so recessive alleles are rare, which results in few recessive disease traits. d. Most recessive mutations are lethal; as a result there are few recessive diseases. e. Detrimental alleles are always being removed from a population due to natural selection.

40. Differential reproduction of genotypes leads to evolution via what process? a. inbreeding b. natural selection c. genetic rescue d. genetic drift e. mutational load 41. Human blood type is determined by three alleles, IA , IB, and IO. The alleles IA and IB are co-dominant to each other, and both are dominant to IO. Within a large, randomly mating population (540,000 individuals), the frequencies for the blood type alleles are 0.3 for the IA allele, 0.6 for the IO allele, and 0.1 for the IB allele.

What percentage of the type B people are heterozygotes (IBIO)? a. 12% b. 13% c. 92.3% d. 6% e. 60%

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 25_7e 42. A population that goes through a dramatic reduction in size will experience: a. an effective population size. b. Hardy–Weinberg equilibrium. c. a genetic rescue. d. a genetic bottleneck. e. inbreeding depression.

43. Suppose that in a population the frequency of a particular recessive condition is 1/400. Assume the presence of only a dominant allele (A) and a recessive allele (a) in the population and that the population is at Hardy– Weinberg equilibrium. What is the frequency of heterozygotes in the population? a. 0.0025 b. 0.05 c. 0.095 d. 0.9025 e. 0.0475 44. What is genetic rescue? a. The introduction of a transgene into an inbred population to increase fitness. b. Genome editing of individuals within an inbred population to increase fitness. c. Culling (killing) individuals with potential lower fitness levels within an inbred population. d. The reproductive isolation individuals with potential lower fitness levels within an inbred population. e. The introduction of new genetic variation into an inbred population.

45. A new kind of tulip is produced that develops only purple or pink flowers. Assume that flower color is controlled by a single-gene locus and that the purple allele (C) is dominant to the pink allele (c). A random sample of 1000 tulips from a large cultivated field yields 847 purple flowers and 153 pink flowers. Estimate the proportion of all purple flowering plants that are heterozygotes and homozygotes. a. heterozygotes = 694, homozygotes = 153 b. heterozygotes = 565, homozygotes = 282 c. heterozygotes = 476, homozygotes = 372 d. heterozygotes = 424, homozygotes = 423 e. heterozygotes = 0, homozygotes = 847 46. Mutation has what effect on a population? a. It creates or increases genetic variation. b. It promotes the fixation of alleles. c. It reduces the amount of genetic variation. d. It homogenizes genetic variation across populations. e. It increases rates of outcrossing.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 25_7e 47. If the relative fitness of the A1A1 genotype is 0.6, A1A2 is 1.0, and A2A2 is 0.9, eventually the frequency of the A2 allele will be: a. 0.8. b. 0.2. c. 0.0. d. 0.4. e. 0.6.

48. Which agent of evolution is most likely responsible for the decrease in the frequency of a recessive allele as shown in the accompanying chart?

a. genetic drift b. natural selection c. mutation d. assortative mating e. inbreeding 49. Which agent of evolution tends to reduce genetic variation between populations and increase genetic variation within each population? a. natural selection b. mutation c. migration d. inbreeding e. genetic drift

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 25_7e 50. _____ is the product of sampling errors and chance events that may result in changes in allelic frequencies. a. Mutation b. Genetic drift c. Directional selection d. Inbreeding e. Evolution Indicate one or more answer choices that best complete the statement or answer the question.

51. What molecular factors can lead to genetic variation in populations beyond that observable by phenotype?

(Select all that apply.) a. Synonymous codons. b. Sequence changes in introns that don't affect a gene product. c. Sequence changes in codons that lead to only marginal differences in a gene product. d. Sequence changes in intergenic DNA. e. Sequence changes in SNPs that don't affect a gene product.

52. In which of the following species might you see F, the inbreeding coefficient, reach a value of 1 after approximately six generations or more? (Choose all that apply.) a. D. melanogaster b. H. sapiens c. C. elegans d. A. thaliana e. E. coli 53. A population consists of 100 individuals of the following genotypes:

a. What is the frequency of the A allele? b. What is the frequency of the a allele? c. Is the population at Hardy–Weinberg equilibrium? Explain your answer. d. What agent of evolution would systematically produce this genotypic distribution?

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 25_7e 54. Like the ABO blood type antigens, the co-dominant M-N antigens are also present in human red blood cells.

A sample of 5631 individuals in a population was examined for M-N antigens, and the results follow. a. Calculate the frequency of each allele in the population. b. Are the M-N genotypes in Hardy–Weinberg equilibrium?

- Blood type M (LMLM): 1245 individuals - Blood type MN (LMLN): 3421 individuals - Blood type N (LNLN): 965 individuals

55. A newspaper story on AIDS reports that an allele (R) confers complete resistance to HIV when homozygous and that 1% of the human population is resistant (RR). The story also states that 20% of the population carries one copy of the resistance allele (i.e., is heterozygous). Is this a guess, an approximation, or an exact frequency, assuming that the population is at Hardy–Weinberg equilibrium?

56. You are studying a very large population of crocodiles on the Nile River in Africa and have identified a newly arisen (by mutation) allele at the C locus. The initial allelic frequency of the mutant allele c is 0.01. You have also determined that the allele acts additively. On a moonless night, you genotype nesting females and count the number of eggs they lay. You find that, on average, CC females produce 98 eggs, Cc females produce 99 eggs, and cc females produce 100 eggs. In this species of crocodile, all eggs hatch and survive to maturity. a. Will the c allele increase, decrease, or stay the same in the next generation? b. Will you likely be able to observe the change in the allelic frequency over the next couple of generations?

Why? c. In the long run, what will happen to the frequencies of the C and c alleles in this population?

57. The Hardy–Weinberg law (equation) is a mathematical model in which allelic frequencies in populations remain constant from generation to generation. Given all the conditions that must be met for the Hardy–Weinberg equation to be valid, why is this equation useful for studying population genetics?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 25_7e 58. Huntington's disease is caused by a single dominant allele and results in progressive mental and neurological damage. The disease usually becomes symptomatic when a person is between 30 and 50 years old and the patient usually dies within 15 years of diagnosis. Approximately 1 in 25,000 Caucasians have this disease.

Huntington's disease has not been associated with any other disease, now or in the past. Why might natural selection not have eliminated such a deleterious gene from the population?

59. The fitness for a particular species of South American bats is determined by a single gene locus with two segregating alleles, (W) and (w), which determines echolocation ability. The dominant allele W causes normal echolocation, whereas the recessive allele w impairs echolocation ability. In the large bat population that you are studying, you determine initial frequencies of the W and w alleles to be 0.7 (p) and 0.3 (q), respectively. If the genotypes are in Hardy–Weinberg equilibrium on fertilization, and the selection coefficient (s) is 0.4, determine the effects of natural selection on the allelic frequencies after one generation.

60. A new kind of tulip is produced that develops only purple or pink flowers. Assume that flower color is controlled by a single-gene locus and that the purple allele (C) is dominant to the pink allele (c). A random sample of 1000 tulips from a large cultivated field yields 847 purple flowers and 153 pink flowers. a. Determine the frequency of the purple and pink alleles in this field population. b. Estimate the proportion of all purple flowering plants that are heterozygotes and homozygotes.

61. Explain how inbreeding can have a positive effect on population fitness.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 25_7e 62. You are studying a single-gene locus with two alleles in a population that is in Hardy–Weinberg equilibrium.

Examination of a large sample of individuals from the population reveals there are six times as many heterozygotes as there are homozygote recessive individuals in this population. What is the frequency of the recessive allele?

63. You are studying cannibals in Borneo and want to determine if a specific village population is in Hardy– Weinberg equilibrium with respect to the two codominant M and N blood type alleles (LM, LN) segregating at the single-gene locus. The following are sample data from this population of the cannibals:

M 287

MN 665

Phenotypes N 123

Total # genotypes 1075

a. Calculate the genotypic frequencies. b. Calculate frequencies for the LM and LN alleles. c. Determine whether this population of Borneo cannibals is in Hardy–Weinberg equilibrium.

64. You discover a certain species of weed growing in soil contaminated with toxic PCBs and later determine that the PCB resistance is due to a single dominant allele. a. If 45% of the seeds from a randomly mating population of resistant weeds will germinate in contaminated soil, what is the frequency of the PCB-resistance allele? b. Among all the plants that germinate, what proportion will be heterozygous? c. What proportion will be homozygous dominant?

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 25_7e 65. Evolutionary geneticists carefully genotype a population of saber-toothed tigers and find the following genotypes: 250 tigers are A1/A1, 500 tigers are A1/ A2, and 250 tigers are A2/A2. Tragically, an asteroid lands in the middle of the population, killing 50% of each genotype. What will the genotypic frequencies be in the next generation?

66. You are studying the population in Iceland for X-linked alleles, and sampling experiments indicate that about 6% of the men have red–green color blindness (caused by a recessive, X-linked allele c). Assume that the population in Iceland mates randomly. a. What percentage of men will carry the allele c? What percentage of women will carry at least one copy of the allele? b. What percentage of women is expected to be color blind? c. What percentage of the total population is color blind? d. What percentage of color-blind individuals are men? e. What percentage of individuals in the population is expected to be normal carriers for the color-blind allele? f. After two generations, what percentage of men in the population is expected to be color blind? To have normal vision?

67. Because real-life populations are, of course, not infinitely large, why is the Hardy–Weinberg condition of an "infinitely large population" usually met for natural populations?

68. Human blood type is determined by three alleles, IA , IB, and IO. The alleles IA and IB are co-dominant to each other, and both are dominant to IO. Within a large, randomly mating population (540,000 individuals), the frequencies for the blood type alleles are 0.3 for the IA allele, 0.6 for the IO allele, and 0.1 for the IB allele. a. Calculate the expected numbers of people in the population having each of the blood types A, B, AB, and O. b. Determine the percentage of type B people that are heterozygotes (IBIO).

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 25_7e 69. If a population is in Hardy–Weinberg equilibrium, the allelic and genotypic frequencies will not change. Prove the preceding statement for a pair of alleles in population X, which is in Hardy–Weinberg equilibrium.

70. Evolutionary biologists usually define population size using the effective population size rather than the census number. What is the effective population size, and what factors can affect it?

71. Genetic diseases in humans are usually rare and recessive. Why are the frequencies of alleles that cause rare, recessive diseases (or other recessive traits, for that matter) generally much higher than the frequency of the diseases (or traits) themselves?

72. By the early 1980s, the population of Florida panthers had dwindled to fewer than 30. Furthermore, within this population was a high frequency of detrimental traits, including low sperm count and undescended testicles in males and kinked tails in both sexes. The future for this iconic population was bleak at best. Provide an explanation for the observed detrimental traits. Provide a possible solution that would both increase the likelihood that the population would survive and preserve the genetic identity of this unique population.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 25_7e Answer Key 1. d 2. a 3. d 4. a 5. a 6. a 7. c 8. a 9. d 10. a 11. e 12. e 13. d 14. e 15. a 16. c 17. c 18. d 19. a 20. b 21. e 22. e 23. a 24. a 25. e 26. a Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 25_7e 27. a 28. c 29. d 30. c 31. b 32. a 33. d 34. b 35. d 36. d 37. a 38. c 39. b 40. b 41. c 42. d 43. c 44. e 45. c 46. a 47. a 48. b 49. c 50. b 51. a, b, d, e 52. c, d

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 25_7e 53. a. p = f(A) = [2(55) + 1(20)]/2(100) = 130/200 = 0.65 b. q = f(a) = [2(25) + 1(20)]/2(100) = 70/200 = 0.35 c. If the population were at Hardy–Weinberg equilibrium, the genotypic distribution would be as follows: E(AA) = 100 × p2 = 42.25 E(Aa) = 100 × 2pq = 45.5 E(aa) = 100 × q2 = 12.25 The observed distribution is obviously very different from that expected from Hardy–Weinberg equilibrium.

Therefore, the population is not at Hardy–Weinberg equilibrium. d. Notice that there is an excess of both types of homozygous individuals in the observed distribution compared to that expected of Hardy–Weinberg equilibrium. This distribution would be expected from either inbreeding or positive assortative mating. Also underdominance could lead to selection against heterozygotes and produce such a distribution.

54. a. The total number of alleles sampled = 2(1245) + 2(3421) + 2(965) = 2(5631) = 11,262. The total number of LM alleles in the sample = p = 2(1245) + (3421) = 5911; and the frequency of the LM allele = 5911/11,262 = 0.525. The total number of LN alleles = q = 2(965) + (3421) = 5351; and the frequency of the LN allele = 5351/11,262 = 0.475. Note that p + q = 1. b. p2 (LMLM ) + 2pq (LMLN ) + q2 (LNLN ) = 1.

For (LMLM): p2 = (0.525)2 = 0.275625. Therefore, (5631) x (0.275625) = 1552.

For (LMLN): 2pq = 2(0.525) x (0.475) = 0.49875. Therefore, (5631) x (0.498) = 2808.

For (LNLN): q2 = (0.475)2 = 0.225625. Therefore, (5631) x (0.225) = 1270.

Test for equilibrium using the chi-square test: (1245 – 1552)2/(1552) + (3421 – 2808)2/(2808) + (965 – 1270)2/(1270) = 60.7 + 133.8 + 73.2 = 267.7 The chi-square statistic is clearly so large that our hypothesis must be rejected, and thus we assume that the population is not in Hardy–Weinberg equilibrium with respect to the M-N blood type antigens.

55. It is a close approximation. If we assume the population is in Hardy–Weinberg equilibrium, the allelic frequency of R is 0.1 (the genotypic frequency of RR is p2, or 1% or 0.01, therefore p = 0.1) and since p + q = 1, the frequency of the r allele is 0.9. The genotypic frequency of the heterozygote is 2pq, which is 0.18 or 18%.

56. a. The cc genotype has the highest fitness, and its relative fitness indicates that c will increase in the next generation. b. Because the absolute difference in fitness is just one egg, the change in allelic or genotypic frequencies is likely to be very slow and, therefore, hard to observe. c. If the directional selection is consistent, even though the differences in fitness are very small, the C allele eventually will be eliminated from the population, and the c allele will be fixed. Of course, we assume that the allele is not lost from the population through random drift (which should not be common, given the stated large population size) and that the fitnesses remain constant.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 25_7e 57. By specifying the ideal conditions that must be met for allelic frequencies to remain constant in populations, the Hardy–Weinberg law can identify evolutionary forces that can cause changes in allelic frequencies in the real world.

Remember, the Hardy–Weinberg law is a model and thus, by definition, provides a simplistic view of reality, which serves as a starting point and basis for comparison for further examination and description of real populations.

58. Natural selection acts through reproduction, and most individuals with Huntington's disease reproduce before discovering they have the disease (although genetic testing may have an impact on this).

59. (1) Determine the initial status of the population.

For WW: relative frequency (p2) = 0.49; relative fitness = 1.0 For Ww: relative frequency (2pq) = 0.42; relative fitness = 1.0 For ww: relative frequency (q2) = 0.09; relative fitness = 1 – 0.4 = 0.6 (2) Determine contributions of each genotype to the next generation: For WW: relative contribution to next generation = 0.49 × 1 (relative fitness) = 0.49 For Ww: relative contribution to next generation = 0.42 × 1 (relative fitness) = 0.42 For ww: relative contribution to next generation = 0.09 × 0.6 (relative fitness) = 0.054 (3) To determine the relative contributions of each genotype to the next generation, divide the individual relative contributions (just calculated) by the sum of all of their contributions (i.e., 0.49 + 0.42 + 0.054 = 0.964): For WW: 0.49/0.964 = 0.508 For Ww: 0.42/0.964 = 0.436 For ww: 0.063/0.964 = 0.065 (4) It is now possible to calculate the frequency of the w allele after one generation of selection because all the genes transmitted by the ww homozygotes and half the genes transmitted by the Ww heterozygotes are w. In the next generation, the frequency of the w allele (q1) will be q1 = 0.065 + 1/2 (0.436) = 0.283. Note that the frequency of the w allele in the bat population decreased from an initial frequency of 0.3 to 0.283 after one generation.

60. a. Assuming the population is in Hardy–Weinberg equilibrium, let p = the dominant purple allele (C) and q = the recessive allele (c). Applying the Hardy–Weinberg equation, the frequency of the pink allele = q2 = 153/1000 = 0.153 and q = (0.153)1/2 = 0.39. Because p + q = 1, p = 1 – q = 0.61. b. If the population is in Hardy–Weinberg equilibrium, p2 + 2pq + q2 = 1. For the heterozygotes, 2pq = 2(0.61) (0.39) = 0.476. Therefore, 47.6%, or (0.476)(1000) = 476 of all the purple flowered plants in this population are expected to be heterozygotes (Cc), and consequently, (p2)(1000) = (0.61)2 (1000) = 372 plants are expected to be homozygous dominants (CC).

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 25_7e 61. Although inbreeding is generally harmful for outcrossing species, a number of plants and animals regularly inbreed and are successful. Inbreeding increases homozygosity, and eventually all individuals in the population become homozygous for the same allele. If a species undergoes inbreeding for a number of generations, many deleterious recessive alleles are weeded out by natural or artificial selection so that the population becomes homozygous for beneficial alleles. In this way, the harmful effects of inbreeding may eventually be eliminated, leaving a population that is homozygous for beneficial traits.

62. The Hardy–Weinberg equation is p2 + 2pq + q2 = 1 (and remember that p + q = 1). The sample results from this population indicate that the ratio of heterozygotes to homozygote recessives is 6:1, respectively. Transfer these data into the relevant Hardy–Weinberg equation elements to get 2pq/q2 = 6. Therefore, 2p = 6q, 2(1 – q) = 6q, and 2 – 2q = 6q, and finally, q = 0.25.

63. a. Genotypic frequencies: LM LM = 287/1075 = 0.267 LM LN = 665/1075 = 0.619 LN LN = 123/1075 = 0.114 b. Allelic frequencies: - For the LM allele: 2(287) + (665) = 1239. Total number of all alleles = 1239 + 665 + 2(123) = 2150.

Therefore, the frequency of the LM allele = 1239/2150 = 0.576.

- For the LN allele: (665) + 2(123) = 911. Therefore, the frequency of the LN allele = 911/2150 = 0.424. c. Let p = frequency of LM (0.576), and q = frequency of LN (0.424). At equilibrium, p2 + 2pq + q2 = 1. The observed number of LMLM individuals = 287; the expected number is p2 (1075) = (0.576)2 (1075) = 357. The observed number of LNLN individuals = 123; the expected number = q2 = (0.424)2(1075) = 193. The observed number of LMLN individuals = 665; the expected number = 2pq (1075) = 2(0.576)(0.424)(1075) = 525.

Perform a chi-square test to test for Hardy–Weinberg equilibrium: (287 – 357)2/(357) + (123 – 193)2/(193) + (665 – 525)2/(525) = 13.7 + 25.4 + 37.3 = 76.4. Degrees of freedom = 3 – 1 – 1 = 1. Chi-square critical value = 3.814, which is much less than 76.4. Therefore, this is a highly significant chi-square statistic, and the hypothesis must therefore be rejected. We conclude that the M-N allelic frequencies are not in equilibrium in this population.

64. a. Because 45% of the seeds germinate, the 55% that do not germinate must be homozygous recessives (q2). The frequency of the recessive allele (q) is therefore (0.55)1/2 = 0.742; the frequency of the dominant resistance allele = p = 1 – q = 1 – 0.742 = 0.258. b. The expected frequency of heterozygotes = 2pq = 2(0.258)(0.742) = 0.383, and the overall frequency of surviving genotypes that are heterozygotes = (0.383)/(0.45) = 0.85 (85%). c. The expected frequency of homozygous dominants = p2 = (0.258)2 = 0.067, and the overall frequency of surviving genotypes that are homozygote dominant = (0.067)/(0.45) = 0.15 (15%).

65. The genotype frequencies are 0.25 A1/ A1, 0.5 A1/ A2, and 0.25 A2/ A2. Because the asteroid affected all the genotypes in proportion to their numbers, this is not a selection event. If the population was in Hardy–Weinberg equilibrium, the asteroid would not change that. Therefore, there would be no change in the frequencies of the alleles or genotypes.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 25_7e 66. a. Let p and q equal the frequencies of the dominant and recessive alleles, respectively. The frequency of the colorblind allele in males = q (hemizygous) = 0.06, or 6%; the frequency of the color-blind allele in females = q2 (homozygotes) + 2pq (heterozygotes) = (0.06)2 + 2(0.06)(1 – 0.06) = (0.0036) + 2(0.06)(0.94) = 0.116, or 11.6%. b. The expected frequency of color blindness among females (cc) in the population = q2 = (0.06)2 = 0.0036, or 0.36% of the females. c. Of the total population, 3% are color-blind males (1/2 of 6%) and 0.18% are color-blind females (1/2 of 0.36%). Therefore, 3.18% of the population would be color blind. d. Of all of the individuals who are color blind, 3%/3.18% = 0.943 or 94.3% are males. e. The population is assumed to be in Hardy–Weinberg equilibrium with q = 0.06 and p = 0.94; therefore, the frequency of heterozygote carriers among females = 2pq = 2(0.94)(0.06) = 0.113, or 11.3%. However, since only half of the population are females, the frequency of carriers in the whole population is 11.3%/2 = 5.65%. f. For a population in Hardy–Weinberg equilibrium, the allelic (gene) and genotypic (zygotic) frequencies will NOT change from generation to generation. Therefore, because the frequency of the recessive allele (q) = 0.06 (or 6%), the frequency of color-blind males (XY) in the population will be 6%. Similarly, after two (or three, or four, etc.) generations, (1 – 0.06) or 94% of the men in the population are expected to have normal vision.

67. It is usually met because most natural populations are large enough that chance factors (i.e., random genetic drift) have negligible effects on allelic frequencies.

68. The Hardy–Weinberg law can be applied for multiple allelic series by simply adding another variable (r) to the normal equation (p + q)2 to make (p + q + r)2, which equals p2 + 2pr + q2 + 2qr + 2pq + r2. a. First, you must calculate the expected frequencies for all genotypes: IA IA = p2 = (0.3)(0.3) = 0.09 IB IB = q2 = (0.1)(0.1) = 0.01 IO IO = r2 = (0.6)(0.6) = 0.36 IA IB = 2pq = 2(0.3)(0.1) = 0.06 IA IO = 2pr = 2(0.3)(0.6) = 0.36 IB IO = 2qr = 2(0.1)(0.6) = 0.12 b. Next, calculate the expected number of individuals having each blood type: Blood type A → 0.09 (IA IA ) + 0.36 (IA IO ) = 0.45 (540,000) = 243,000 people Blood type B → 0.01 (IB IB ) + 0.12 (IB IO ) = 0.13 (540,000) = 70,200 people Blood type AB → 0.06 (IA IB )(540,000) = 32,400 people Blood type O → 0.36 (IO IO )(540,000) = 194,400 people The total number of (IBIO) heterozygotes in the population is 0.12 (540,000) = 64,800; and the total number of type B individuals in the population = 70,200. Therefore, the percentage of type B individuals who are heterozygotes = 64,800/70,200 = 0.923 = 92.3%.

Copyright Macmillan Learning. Powered by Cognero.

Page 27

Name:

Class:

Date:

Chap 25_7e 69. (1) Determine the genotypic frequencies using the two alleles R and r. p2 (RR) + 2pq (Rr) + q2 (rr) = 1. Note that the total proportion of dominant R alleles in population X equals the sum of all the gametes produced by the RR homozygotes + half the gametes produced by the Rr heterozygotes, that is, p2 (RR) + pq (Rr).

(2) Calculate the frequency of R in the gene pool of the next generation. Note that the percentage of all the R and r alleles in the population is 100%. For these alleles (i.e., the alleles of interest), p + q = 1 (100%), and q = 1 – p.

Therefore, p2 (RR) + pq (Rr) = p2 + p (1 – p) = p2 + p – p2 = p. Because the frequency of the dominant allele (R) remains constant in the next generation, it follows that neither the frequency of the recessive allele nor the frequencies of the relevant genotypes (RR, Rr, rr) will change in subsequent generations for population X. In other words, p2 (RR) + 2pq (Rr) + q2 (rr) = 1 = p + q.

70. Population size is usually defined as the number of individuals in a group. But the evolution of a gene pool depends only on those individuals who contribute genes to the next generation. Population geneticists usually define population size as the equivalent number of breeding adults, the effective population size (Ne). Several factors determine the equivalent number of breeding adults, including sex ratio, variation between individuals in reproductive success, fluctuations in population size, age structure of the population, and whether mating is random.

71. Most of the rare, recessive alleles within the population are "hidden" within heterozygote carriers, which, of course, do not manifest the disease (or express the trait).

72. The population is experiencing a high genetic load due to its small size. This is an example of the bottleneck effect in which genetic variation declines and harmful alleles can drift to high frequencies or even fixation. A possible solution would be genetic rescue, in which panthers from another population would be introduced in order to bring in new genetic variation and to reduce the genetic load. However, care should be taken in this approach that the genes from the introduced animals do not "swamp" the gene pool of the Florida panthers. Therefore, the number of introduced animals should be small compared to the number of native animals, and the population should be monitored to ensure that the introduced animals and their offspring are not reproducing at a rate that far exceeds that of the native animals.

Copyright Macmillan Learning. Powered by Cognero.

Page 28

Name:

Class:

Date:

Chap 26_7e Indicate the answer choice that best completes the statement or answers the question.

1. The microsatellite study of bighorn sheep described in Section 26.1 was a study of: a. anagenesis. b. cladogenesis. c. speciation. d. proteins. e. variation but not evolution.

2. _____ is BEST known for developing the concept of biological species. a. Mendel b. Darwin c. Mayr d. Fisher e. Linnaeus 3. Which of the following would NOT be part of the process of flowering plant speciation through polyploidy? a. formation of a hybrid b. nondisjunction of chromosomes c. allopatric speciation d. fusion of gametes e. the formation of an allopolyploid 4. _____ are sets of genes that are similar in sequence but encode different products. a. Codons b. Genomes c. Exons d. Pseudogenes e. Multigene families 5. The _____ concept defines a species as a group of organisms that are capable of exchanging genes. a. morphospecies b. phylogenetic species c. biological species d. cladogenic species e. evolutionary species 6. In Darwin's finches (Figure 26.7, below), which genera have the LATEST common ancestor in common?

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chap 26_7e

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chap 26_7e

a. Geospiza and Certhidea b. Pinaroloxias and Camarhynchus c. Geospiza and Camarhynchus d. Geospiza and Pinaroloxias e. Pinaroloxias and Platyspiza 7. Considering the strengths and weaknesses of molecular data, what could one NOT argue to be true? a. Molecular methods can be used in all organisms since they all have proteins and DNA. b. Molecular data are difficult to quantify. c. Molecular methods can access a large amount of data. d. Molecular data can provide direct information on the process of evolution. e. Molecular variation is clear and interpretable, even if the relationship between some traits and their underlying genes is complex.

8. Which region of a gene should have the HIGHEST rates of substitution? a. first position of a codon b. second position of a codon c. third position of a codon d. 5′ flanking region e. 3′ untranslated region 9. _____ characteristics evolved from the same characteristic in a common ancestor. a. Similar b. Paralogous c. Homologous d. Phylogenetic e. Parsimonious 10. Evolution can be defined as _____ change that takes place in a(n) _____. a. any; individual b. a genetic; individual c. any; group of organisms d. a genetic; group of organisms e. All of the answers are correct.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chap 26_7e 11. The construction of a phylogeny from DNA sequence data requires: a. the identification of monophyletic groups prior to analysis. b. a proper alignment of nucleotide bases from homologous genes. c. the use of the fossil record to calibrate rates of nucleotide substitution. d. removing any sequences with insertions or deletions from the analysis. e. the use of genes that influence traits associated with reproductive isolating mechanisms.

12. A phylogenetic tree for three species, A, B, and C, is shown below.

A particular DNA region that is shared by all three species is cloned from each species and sequenced. There are about 100 nucleotide substitutions between A and C and between B and C. There are only 20 differences between A and B. The fossil record indicates that the ancestor to A and B diverged from the ancestor to C around 10 million years ago. What is the best estimate of the divergence between A and B based on this information, assuming a constant rate of mutation and the same rate of mutation on all branches? a. 1 million years ago b. 2 million years ago c. 5 million years ago d. 10 million years ago e. 20 million years ago 13. Which reproductive isolating mechanism is exhibited by two different species of flowers, where one is pollinated by butterflies and another by bees? a. prezygotic ecological b. postzygotic c. prezygotic behavioral d. prezygotic temporal e. prezygotic mechanical

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chap 26_7e 14. Which reproductive isolating mechanism is exhibited by two different species of tree frogs when their males vocalize their distinct mating songs at the same time in the same forest? a. prezygotic ecological b. postzygotic c. prezygotic behavioral d. prezygotic temporal e. prezygotic mechanical 15. Based on the gene tree in Figure 26.11 (below), which of the following pairs of genes are MOST similar?

a. tilapia PRL1 and cherry salmon SOMA b. human SOMA and human PRL c. alligator PRL2 and chicken PRL d. catfish PRL and chicken SOMA e. whale PRL and catfish PRL

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chap 26_7e 16. Which of the following statements about the concept of “evolution in biology” is FALSE? a. It involves genetic change. b. It takes place in groups. c. It involves the process of mutation. d. It can involve genetic changes within one individual over time. e. It can involve anagenesis and cladogenesis.

17. The _____ approach to inferring relationships among species selects the phylogeny that minimizes the number of evolutionary changes. a. maximum likelihood b. Bayesian c. maximum parsimony d. distance e. rooted tree 18. For most protein-coding genes, the rate of substitution in _____ is considerably higher than in _____. a. synonymous sites; nonsynonymous sites b. nonsynonymous sites; synonymous sites c. nonsynonymous sites; introns d. 3′ untranslated regions; synonymous sites e. 3′ untranslated regions; introns 19. According to the molecular clock data in Figure 26.17 (below), approximately what proportion of differences (per amino acid site) has accumulated between the common ancestor of all vertebrates (including fish) and the common ancestor of humans and kangaroos?

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chap 26_7e

a. 0.1 b. 0.3 c. 0.5 d. 0.7 e. 0.9

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chap 26_7e 20. _____ speciation arises in the absence of any geographic barrier to gene flow. a. Autocratic b. Allopatric c. Sympatric d. Parapatric e. Peripatric 21. Which of the following statements about rates of nucleotide substitutions is FALSE? a. Rates are the same for nonsynonymous substitutions within one gene. b. Rates are higher for synonymous substitutions than nonsynonymous substitutions. c. Rates can vary between different genes in one species. d. Rates can vary between homologous genes in different species. e. Rates are high in pseudogenes.

22. A(n) _____ tree is one that contains a node representing the common ancestor to all other nodes in the tree. a. gene b. phylogenetic c. outgroup d. most parsimonious e. rooted 23. One example of _____ speciation is when a small group of individuals colonizes an island. a. autocratic b. allopatric c. sympatric d. parapatric e. peripatric 24. The underlying assumptions of using a molecular clock to date evolutionary events are based on: a. the neutral mutation hypothesis. b. the balance hypothesis. c. rates of exon shuffling. d. the model of allopatric speciation. e. maximum parsimony analyses.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chap 26_7e 25. Before the advent of molecular techniques, researchers indirectly assessed genetic variation in populations through the study of _____ diversity. a. protein b. phenotypic c. mutational d. behavioral e. fossil 26. All of the following are examples of postzygotic reproductive isolating mechanisms EXCEPT: a. a sheep adapted to mountains mating with a sheep adapted to desert producing offspring maladapted to either habitat. b. two plant species flowering at different times. c. the hybrid offspring of two species dying before birth. d. the hybrid offspring of two species being sterile. e. the hybrid offspring of two species being fertile but their offspring being inviable or sterile.

27. Alfred Roca and his colleagues used DNA sequencing to reassess the genetic relationships among African elephants. What did they conclude? a. Genetic variation was uncommon in African elephants. b. Cladogenesis occurred within African elephants. c. There was extensive gene flow between forest and savannah elephants. d. The genetic variation correlates with differences in phenotype that had been previously observed. e. There was extensive difference in the mitochondrial genome between savannah and forest elephants.

28. The observation that some genes are mosaics of other genes can be explained by: a. evolution. b. horizontal gene transfer. c. exon shuffling. d. gene duplication. e. pseudogenes.

29. _____ represent the evolutionary relationships among groups of organisms. a. Alignments b. Nodes c. Synonymous rates of substitution d. Phylogenetic trees e. Species concepts

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chap 26_7e 30. The relatively small amount of genetic differentiation among some species suggests that many phenotypic differences reflect changes in _____, rather than the evolution of new genes. a. gene expression b. gene duplication c. exon shuffling d. whole-genome duplication e. horizontal gene transfer 31. Which of the following statements with respect to the neutral-mutation hypothesis is FALSE? a. The evolution of most genetic variation in a population is influenced by genetic drift. b. When natural selection is at work on protein variants, it will lead to little variation in the population. c. Most molecular variation is adaptively neutral—for example, most protein variants are functionally equivalent. d. Natural selection is still an important evolutionary force. e. Genetic variation in a population is maintained mainly by natural selection.

32. Of the various evolutionary forces, _____ is responsible for the origin of new genetic variation in a species. a. natural selection b. genetic drift c. mutation d. migration e. nonrandom mating 33. In Darwin's finches (Figure 26.7, below), the species Geospiza magnirostris is MOST closely related to which species?

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chap 26_7e

a. Geospiza scandens b. Geospiza fortis c. Geospiza conirostris d. Geospiza difficilis e. Certhidea olivacea

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chap 26_7e 34. All of the following are examples of prezygotic reproductive isolating mechanisms EXCEPT: a. a cat trying to mate with a lion. b. a male fiddler crab waving its claw to attract a mate. c. the hybrid offspring of two species dying before birth. d. fireflies flashing their light-emitting organs. e. two plant species flowering at different times.

35. Which of the following is NOT a problem with the biological species concept? a. It is sometimes difficult to determine if reproductive isolation exists. b. Reproductive isolation does not apply to asexual organisms. c. The concept of reproductive isolation does not work for fossils. d. Reproductive isolation is not a mechanism that ensures that species are evolving independently. e. All of these are problems with the biological species concept.

36. Which reproductive isolating mechanism is exhibited by two different Drosophila species where one species breeds in the early morning and the other in the afternoon? a. prezygotic ecological b. postzygotic c. prezygotic behavioral d. prezygotic temporal e. prezygotic mechanical 37. The biological species concept distinguishes species based on a. genetic differences. b. phenotypic differences. c. time since anagenesis. d. reproductive isolating mechanisms. e. differences in ploidy level.

38. Horizontal gene transfer is most common: a. among bacteria. b. among eukaryotes. c. among viruses. d. from eukaryotes to bacteria. e. from plants to animals.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chap 26_7e 39. Can different species have very few visible morphological differences or none at all? a. no, because different species must have distinct, visible morphological differences b. yes, if they are not under selection to evolve such differences (cryptic species) c. yes, if they are hybrids of other species d. yes, when their DNA is identical e. no, because their DNA is not identical, and phenotype must always reflect genotype 40. In the phylogeny of Darwin's finches (Figure 26.7, below), which species serves as an outgroup?

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chap 26_7e

a. Pinaroloxias inornata b. Geospiza fortis c. Geospiza conirostris d. Geospiza difficilis e. Certhidea olivacea 41. _____ is the splitting of one lineage into two distinct lineages. a. Genesis b. Anagenesis c. Phylogenesis d. Cladogenesis e. Neogenesis 42. _____ is the idea that, if postzygotic isolating mechanisms exist between two species, then natural selection will favor traits that lead to the evolution of prezygotic isolating mechanisms. a. Hybrid breakdown b. Reinforcement c. Gametic isolation d. Phylogenetic species e. Hybrid inviability

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chap 26_7e 43. Which of the following statements about exon shuffling is FALSE? a. It can lead to the evolution of new genes. b. It involves exchange of exons between different genes. c. It often brings together distinct protein domains. d. The mechanism by which it occurs is well understood. e. It has been observed only a very few times by evolutionary geneticists.

44. Which of the following are homologous? a. bat wings and fly wings b. bat wings and bird legs c. bat wings and mouse legs d. bat wings and human feet e. bat wings and bird wings 45. Support for evolution is NOT found in a. comparative anatomy. b. the fossil record. c. the distributions of species (biogeography). d. direct observation such as the evolution of pesticide resistance. e. the change an organism undergoes as it ages.

46. Which of the following markers is NOT used to investigate genetic variation in natural populations? a. DNA sequence b. RFLP c. microsatellite d. isotope e. allozyme

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chap 26_7e 47. Based on Figure 26.11 (below), which pair of genes was the EARLIEST to diverge?

a. sea bass PRL and tilapia PRL2 b. sea bass PRL and eel PRL c. sea bass PRL and tilapia PRL1 d. eel PRL and catfish PRL e. catfish PRL and carp PRL Indicate one or more answer choices that best complete the statement or answer the question.

48. Through which of the following mechanisms can horizontal gene transfer occur? (Select all that apply.) a. conjugation in bacteria b. transduction in bacteria c. transformation in bacteria d. endosymbiosis of prokaryotic cells by eukaryotic cells e. self-fertilization of a C. elegans hermaphrodite

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chap 26_7e 49. (a) What are pseudogenes? (b) Why might homologous pseudogenes in a group of closely related species be more useful than "real" genes for elucidating phylogenetic relationships between those species?

50. Discuss the role of the availability of whole-genome sequences in understanding the evolutionary process.

51. (a) What is a molecular clock? (b) In what situation(s) would it be especially useful? (c) Explain how a molecular clock is calibrated. (d) What are some known problems with using molecular clocks?

52. Why is sympatric speciation thought to be so rare in nature? Specifically, which evolutionary forces act against sympatric speciation, and what evolutionary forces might act to allow sympatric speciation?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chap 26_7e 53. The diagram below represents the ranges of three closely related species (A, B, and C), which are separated by a mountain range and river. As an expert geologist, you hypothesize that the area was originally occupied by a single species. Its range was split first by the mountain range. Later, the river formed, separating populations on the eastern side of the mountains. If your hypothesis is correct, what should the phylogenetic relationships be among the three species? Draw a phylogeny as your answer.

54. How does speciation by polyploidy occur?

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chap 26_7e 55. Suppose that there are three species that are closely related as indicated by the phylogenetic tree below.

Each species contains only one copy of a particular gene. The version in Species A is designated GeneA, and so on. The relationships among the three versions of the gene in the different species are shown below.

Explain in terms of duplications and/or deletions the incongruity between the phylogenetic (species) tree and the gene tree.

56. How might gene duplication provide a mechanism for the addition of new genes with novel functions?

57. What is the first step in cladogenesis? Give an example of how this might occur.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chap 26_7e 58. How much genetic differentiation is required for reproductive isolation to take place?

59. Is the buildup of reproductive barriers in two allopatric populations the result of natural selection for reproductive isolation?

60. Phylogenies can be used to reconstruct the evolutionary history of a group of species. The DNA sequences of genes are frequently used to construct such phylogenetic trees. Will the phylogenies of genes always match the history of the organisms that the genes come from?

61. How has molecular data helped resolve the evolutionary relationships among distantly related taxa?

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chap 26_7e 62. Based on careful inspection of the genetic code, why might the second position of a codon have the lowest rate of substitutions in a gene?

63. For most protein-encoding genes, the synonymous rate of change is considerably higher than the nonsynonymous rate. In comparing two taxa, how might you interpret the evolutionary history of a gene if the nonsynonymous rate of change was higher than the synonymous rate?

64. What two factors are thought to play a critical role in sympatric speciation? Explain how they lead to sympatric speciation.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chap 26_7e 65. What is the traditional definition of a species? How do studies of the genomes of chimpanzees and bonobos, and of H. sapiens, Neanderthals, and Denisovans challenge this definition?

66. (a) What is horizontal gene transfer? (b) How does horizontal gene transfer generate incongruity between phylogenetic (species) trees and gene trees?

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chap 26_7e Answer Key 1. a 2. c 3. c 4. e 5. c 6. c 7. b 8. c 9. c 10. d 11. b 12. b 13. e 14. c 15. c 16. d 17. c 18. a 19. c 20. c 21. a 22. e 23. b 24. a 25. b 26. b Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chap 26_7e 27. b 28. c 29. d 30. a 31. e 32. c 33. b 34. c 35. d 36. d 37. d 38. a 39. b 40. e 41. d 42. b 43. d 44. c 45. e 46. d 47. b 48. a, b, c, d 49. (a) Pseudogenes are nonfunctional genes that arise from duplication of a gene followed by one or more mutations that destroy the function of one copy. (b) Because pseudogenes are nonfunctional, they accumulate mutations freely and rapidly compared to functional genes. Therefore, one is more likely to find a suitable number of mutation events when comparing the sequences of homologous pseudogenes, even in closely related species. Furthermore, mutations in pseudogenes are selectively neutral, and analysis based on such mutations will not be confounded by natural selection.

Copyright Macmillan Learning. Powered by Cognero.

Page 24

Name:

Class:

Date:

Chap 26_7e 50. Whole-genome sequences are providing new information about how genomes evolve and about the processes that shape the size, complexity, and organization of genomes. By examining different genes in a genome, we have been able to observe that exons of different genes have been exchanged, creating genes that are mosaics of other genes.

Whole-genome sequences make apparent the duplication of individual genes, chromosomal segments, and even entire genomes. Genome sequences from different organisms also might provide evidence for horizontal transfer of genes.

51. (a) A molecular clock is the amount of change that a protein or DNA sequence has undergone during the evolution of two divergent species. If the rate of change is constant and known, the amount of change can be used to estimate the time since the divergence. (b) A molecular clock is most useful when the fossil record is incomplete and does not allow for an estimation of the time since divergence between two species. (c) The molecular clock can be calibrated by determining the number of differences in a protein or DNA sequence between species whose time since divergence is known from the fossil record. (d) Some studies show that molecular changes do not always occur at a constant rate or at the same rate within different lineages so that a calibration might not be valid for the species in question.

52. In sympatry, gene flow will overwhelm the forces of natural selection, drift, and assortative mating. If natural selection and assortative mating are strong enough, you could get sympatric speciation.

53.

54. A common mechanism for polyploidy is for two species to hybridize, followed occasionally by a spontaneous doubling of chromosome numbers. Such a polyploid is called an allopolyploid. Each chromosome will have a pairing partner so that chromosome segregation can proceed in a normal way. The allopolyploid might hybridize with the two parent species, but the differences in chromosome number will likely cause reproductive isolation between them. Thus, the allopolyploid will be a new species, almost instantaneously formed.

55. One explanation is that the common ancestor to all three species contained two copies of this gene due to a duplication event. Call these copies 1 and 2. Suppose that copy 1 was lost (deleted) during evolution of Species C, and both 1 and 2 were retained by the common ancestor of Species A and B. After diverging from each other, Species A retained copy 1 and lost copy 2, and Species B retained copy 2 and lost copy 1. Therefore, Species B and C would both have descendants of copy 2, and Species A would have a descendant of copy 1.

56. After a gene duplicates, there are two copies of the sequence, one of which is free to change and potentially take on a new function.

Copyright Macmillan Learning. Powered by Cognero.

Page 25

Name:

Class:

Date:

Chap 26_7e 57. The first step is genetic isolation of subpopulations, usually by some geographical barrier like a river or mountain range.

58. There is no universal answer, but some species are differentiated at many loci, while others differ in just a few genes.

59. No. Selection has no means of acting because the populations are isolated and there is no gene flow between them.

Reproductive isolation results "accidentally" from the accumulation of genetic changes due to random events such as mutation and genetic drift and due to adaptation to different environments.

60. Trees constructed from different genes can result in very different phylogenies. Each gene has its own evolutionary history. On average, we expect the phylogeny reconstructed from many genes to reflect accurately the actual phylogenetic history of the species, but if we sample just one of a few genes, we might get a different tree for each gene.

There are some common ways for this incongruence to occur. There could have been horizontal gene transfer at a gene in a particular species that would make that species look as though it was quite unrelated to the others. A gene could become polymorphic in a population—a new allele could arise by mutation. That polymorphism could be maintained during a speciation event, and different alleles could subsequently be fixed in the new species. The gene tree would group species that shared that same allele, even if those were not the most closely related taxa.

Finally, if gene duplication occurs within a species, there may be several very similar sequences in each species.

After a speciation event, it would be critical to compare the appropriate members of the gene family (orthologs).

Comparison of different members of the multigene family would lead to incongruities between the gene phylogeny and species phylogeny.

61. Trying to assess the evolutionary history of distantly related organisms is often difficult because the organisms have few characteristics in common. All organisms have certain molecular traits in common, such as ribosomal RNA sequences and some fundamental proteins. These molecules offer a valid basis for comparisons among all organisms.

62. The highest rates of substitutions occur in regions of the gene that have the least effect on function, such as the third position of a codon. The lowest rates of substitution are seen in nonsynonymous changes in the coding region because these substitutions always alter the amino acid sequence of the protein and are often deleterious. A change in the second position of a codon always alters the amino acid sequence of a protein and is therefore always a nonsynonymous change.

63. The expectation that the synonymous rate of change is higher than the nonsynonymous rate is based on the assumption that synonymous changes are tolerated by natural selection and nonsynonymous changes are eliminated by selection. If, however, those amino acid changes are beneficial to the organism, selection would lead to an increase in their frequencies. A nonsynonymous rate of change higher than the synonymous rate would imply positive selection on those changes.

64. Sympatric speciation can be accomplished by strong disruptive selection along with positive assortative mating. In this model, one homozygous genotype is favored on one resource and the other homozygous genotype is favored on a different resource. Heterozygotes do not do well on either resource and are selected against. In this situation, natural selection will favor genotypes at other loci that favor positive assortative mating, which will produce more of the favored homozygous individuals.

Copyright Macmillan Learning. Powered by Cognero.

Page 26

Name:

Class:

Date:

Chap 26_7e 65. The traditional definition of a species is a reproductively isolated group. However, the studies of chimpanzees and bonobos indicate extensive gene flow (genetic exchange) took place between these two groups after they had become "distinct" species. The same phenomenon is observed between hominin species: H. sapiens, Neanderthals, and Denisovans.

66. (a) Horizontal gene transfer is the transfer of genetic material from one species to another. It occurs commonly in nature between different species of bacteria but is thought to be rare between eukaryotic cells. (b) Horizontal gene transfer generates incongruities between species and gene trees because distantly related species can end up carrying very similar DNA fragments that have been transferred horizontally. The genes will appear close together on a gene tree even though the species in which they reside are far apart on a species tree.

Copyright Macmillan Learning. Powered by Cognero.

Page 27