Principles of Genetics, Binder Ready Version, 7th Edition Test Bank

Type:

Test Bank

Resource:

Principles of Genetics

Edition:

7th Edition

Author(s):

D. Peter Snustad Michael J. Simmons

Chapter 01 Test Bank Question Type: Multiple Choice 1) Which of the following requirements must be met by a substance in order to be considered hereditary material?

1. The material must be able to replicate or be replicated 2. The material must encode information regarding the structure and function of the organism.

3. The material must be able to change over time. a) 1 b) 2 c) 3 d) 1 and 2 only e) All of these Answer: e Section: 1.1 An Invitation Difficulty: Easy 2) In 1953 the structure of _____ was determined, as well as its role as the hereditary material for various organisms. a) Amino acids b) DNA c) Protein d) Carbohydrates e) Lipids Answer: b Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 3) Which of the following persons contributed greatly to the understanding of the field of modern genetics? a) Gregor Mendel b) James Watson c) Francis Crick d) B and C only e) All of these Answer: e Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy

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4) Which of the following is considered a milestone in the field of genetics? a) The Human Genome Project b) The discovery of the rules of inheritance c) The discovery of the structure of DNA d) All of these e) None of these Answer: d Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 5) The heritable factors that Mendel studied are now known as: a) Nucleic acids b) Amino acids c) Genes d) Unit factors e) Peptides Answer: c Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 6) How did Mendel discover the rules of inheritance? a) He interbred plants that showed different traits b) He interbred animals that showed similar traits c) He crossbred protists that showed different traits d) All of these e) None of these Answer: a Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 7) Different forms of the same gene are known as: a) Peptides b) Amino acids c) Proteins d) Alleles e) Gene differences Anwer: d Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy

8) Mendel proposed that each individual organism (i.e. pea plants) carried how many alleles, or copies of a gene? a) 1 b) 2 c) 3 d) 4 e) 5 Answer: b Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 9) Which of the following is not a principle of inheritance discovered by Mendel’s garden pea experiments?

1. Each individual organism carries 2 copies of a gene in its normal state 2. Each individual organism randomly passes down one copy of each gene in its gametes (egg or sperm cell) 3. Each gene is inherited independently of the others and is a discrete entity a) 1 b) 2 c) 3 d) 2 and 3 only e) All of these Answer: e Section: 1.2 Three Great Milestones in Genetics Difficulty: Medium 10) The building blocks of genes are: a) Proteins b) Amino acids c) Nucleic acids d) Lipids e) Carbohydrates Answer: c Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 11) Nucleic acids are composed of a series of: a) Amino acids b) Nucleotides c) Nucleic chromosomes d) Peptide chains e) Sugars Answer: b Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy

12) Nucleotides are composed of which of the following? a) Pentose sugar, phosphate molecule, and a nitrogenous base b) Pentose sugar, nucleic acid, and a nitrogenous base c) Hexane molecule, muramic acid, and a phosphorylated base d) Amino acid, phosphate molecule, and a nitrogenous base e) DNA, RNA, and amino acid chain Answer: a Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 13) Which of the following is the sugar found in RNA? a) Deoxyribose b) Ribose c) Sucrose d) Fructose e) Glucose Answer: b Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 14) Which of the following is not a nitrogenous base found in DNA? a) Cytosine b) Adenine c) Guanine d) Uracil e) Thymine Answer: d Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 15) Which of the following individuals are credited with the discovery of the structure of DNA? a) James Watson and Arthur Kornberg b) Francis Crick and Arthur Kornberg c) Paul Erlich and James Watson d) James Watson and Francis Crick e) Francis Crick and Paul Erlich Answer: d Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy

16) Which of the following correctly describes the structure of DNA as determined in 1953?

1. A phosphate backbone with sugars linked together by nitrogenous base bonds.

2. A sugar-phosphate backbone with nitrogenous bases linked together by weak bonds 3. A nitrogenous base backbone with sugars linked together by phosphate bonds. a) 1 b) 2 c) 3 d) 1 and 3 e) None of these Answer: c Section: 1.2 Three Great Milestones in Genetics Difficulty: Medium 17) Which of the following is an example of a correct pairing of nitrogenous bases as is found in the structure of dsDNA? a) A pairs with C b) A pairs with T c) A pairs with G d) A pairs with A Answer: b Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 18) Which of the following is different between RNA and DNA? 1. RNA is double stranded and DNA is single stranded 2. RNA is single stranded and DNA is double stranded 3. RNA contains the nitrogenous base Uracil and DNA contains the base Thymine a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: e Section: 1.2 Three Great Milestones in Genetics Difficulty: Medium 19) The collection of DNA molecules that is characteristic of an organism is known as its: a) Allelic makeup b) Genome c) Phenotype d) DNA array e) None of these Answer: b Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy

20) The Human Genome Project's purpose is/was: a) To determine the sequence of the nucleotide pairs in human beings b) To determine the cause of genetic diseases in human beings c) To determine the rate of mutation in the human population d) All of these e) None of these Answer: a Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 21) This man ran a privately funded project that was similar to the federally funded Human Genome Project. a) Gregor Mendel b) James Watson c) Craig Venter d) Francis Crick e) Anton Van Leeuwenhoek Answer: c Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 22) The study of genes on a molecular level is often referred to as: a) Proteomics b) Genomics c) Amino acid array technology d) Classical Genetics e) None of these Answer: b Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy 23) The National Center for Biotechnology (NCBI) is an organization that: a) Maintains updated databases regarding genome sequences b) Maintains updated databases regarding information about proteins c) Is a source to obtain publications regarding genomic research d) All of these e) None of these Answer: d Section: 1.2 Three Great Milestones in Genetics Difficulty: Easy

24) Which of the following best describes the flow of information in a biological system? a) DNA → RNA → Protein b) Protein → RNA → DNA c) DNA → Protein → RNA d) RNA → DNA → Protein e) Protein → DNA → RNA Answer: a Section: 1.3 DNA as the Genetic Material Difficulty: Easy 25) Why is DNA considered to be the genetic material in cellular organisms? a) It can replicate b) It can direct the function and behavior of an organism c) It can change over time d) It can replicate and it can direct the function and behavior of an organism e) All of these Answer: e Section: 1.3 DNA as the Genetic Material Difficulty: Easy 26) Coding units composed of triplets of adjacent nucleotides are known as: a) Proteins b) Peptides c) Codons d) Tripods e) None of these Answer: c Section: 1.3 DNA as the Genetic Material Difficulty: Easy 27) The process in which DNA is used as a template to produce RNA is known as: a) Replication b) Transcription c) Translation d) Codon usage e) None of these Answer: b Section: 1.3 DNA as the Genetic Material Difficulty: Easy

28) The process in which RNA is used as a template to produce a polypeptide chain is known as: a) Replication b) Transcription c) Translation d) Codon usage e) None of these Answer: c Section: 1.3 DNA as the Genetic Material Difficulty: Easy 29) What substance is produced via the process of transcription? a) ssDNA b) dsDNA c) proteins d) mRNA e) enzymes Answer: d Section: 1.3 DNA as the Genetic Material Difficulty: Easy 30) The collection of all the different proteins in an organism is known as: a) Genome b) Proteome c) Polypeptide sequence d) All of these e) None of these Answer: b Section: 1.3 DNA as the Genetic Material Difficulty: Easy 31) The study of all the proteins in cells is known as: a) Genomics b) Allelomics c) Proteomics d) Gene array technology e) None of these Answer: c Section: 1.3 DNA as the Genetic Material Difficulty: Easy

32) If the flow of hereditary information follows the pathway RNA→DNA, the process of _____________ ____________ must be involved. a) Congruent Transcription b) Inverse Replication c) Reverse Transcription d) Reverse Replication e) Reverse Translation Answer: c Section: 1.3 DNA as the Genetic Material Difficulty: Easy 33) Changes that are introduced into the DNA strand during replication are known as: a) Codons b) Phenotypes c) Genotypes d) Mutations e) Errors Answer: d Section: 1.4 Genetics and Evolution Difficulty: Easy 34) Which of the following proposed that variation makes it possible for a species to change over time? a) Alfred Wallace b) Charles Darwin c) Gregor Mendel d) Alfred Wallace and Charles Darwin e) All of these Answer: d Section: 1.4 Genetics and Evolution Difficulty: Easy 35) Established historical relationships between organisms are known as a: a) Phylogeny b) Genealogy c) Genomics d) Proteomics e) None of these Answer: a Section: 1.4 Genetics and Evolution Difficulty: Easy

36) If a geneticist pursues his science by analyzing the outcomes of crosses between different strains of an organism, he is practicing: a) Molecular genetics b) Classical genetics c) Population genetics d) Conservation genetics e) Environmental genetics Answer: b Section: 1.4 Genetics and Evolution Difficulty: Easy 37) By analyzing patterns of inheritance, geneticists can localize genes to specific chromosomes and sometimes localize genes to specific positions within chromosomes. This is known as: a) Chromosome counting b) Chromosome mapping c) Chromosome positioning d) Genome analysis e) Proteome analysis Answer: b Section: 1.5 Levels of Genetic Analysis Difficulty: Easy 38) Studies that emphasize the passing down of genes and chromosomes from one generation to another are known as studies in: a) Population genetics b) Hereditary genetics c) Transmission genetics d) Molecular genetics e) None of these Answer: c Section: 1.5 Levels of Genetic Analysis Difficulty: Easy 39) If a geneticist is studying how DNA replicates, is expressed and how mutations occur in the DNA strand, perhaps by studying a particular DNA sequence, he is practicing: a) Molecular genetics b) Classical genetics c) Transmission genetics d) Population genetics e) Conservation genetics Answer: a Section: 1.5 Levels of Genetic Analysis Difficulty: Easy

40) If a geneticist is studying the frequency of sickle cell anemia in a group of individuals, to see how often the sickle cell allele is present within the group as a whole, he is practicing: a) Molecular genetics b) Classical genetics c) Transmission genetics d) Population genetics e) Conservation genetics Answer: d Section: 1.5 Levels of Genetic Analysis Difficulty: Easy 41) An organism that has been altered by the introduction of a foreign gene is known as: a) GMO b) GMP c) FDA d) SOP e) PCR Answer: a Section: 1.6 Genetics in the World: Applications of Genetics to Human Endeavors Difficulty: Easy 42) An example of a commonly produced agricultural GMO is: a) Philodendron b) Corn c) Bacillus thuringiensis d) Canker worm e) Butterfly Answer: b Section: 1.6 Genetics in the World: Applications of Genetics to Human Endeavors Difficulty: Easy 43) A professional who is trained to advise individuals about their risks of inheriting or transmitting a genetic disease is known as: a) Molecular geneticist b) Population geneticist c) Genetic counselor d) Physician e) None of these Answer: c Section: 1.6 Genetics in the World: Applications of Genetics to Human Endeavors Difficulty: Easy

44) Which of the following is a disease whose treatment or diagnosis has been influenced by the field of genetics? a) Cystic Fibrosis b) Breast Cancer c) Heart Disease d) Diabetes e) All of these Answer: e Section: 1.6 Genetics in the World: Applications of Genetics to Human Endeavors Difficulty: Easy 45) Which of the following is a way that the study of genetics influences society? a) The use of DNA evidence in a paternity suit b) The use of DNA to create new pharmaceutical drugs c) The use of DNA to create drought resistant and pest resistant crops d) All of these e) None of these Answer: d Section: 1.6 Genetics in the World: Applications of Genetics to Human Endeavors Difficulty: Easy

Question Type: Essay 46) Briefly describe the process of DNA replication.

Answer: The process of DNA replication is based on the complementary nature of the strands that make up duplex DNA molecules. These strands are held together by hydrogen bonds between specific base pairs—A paired with T, and G paired with C. When these bonds are broken, the separated strands can serve as templates for the synthesis of new partner strands. The new strands are assembled by the stepwise incorporation of nucleotides opposite to nucleotides in the template strands. This incorporation conforms to the base-pairing rules. Thus, the sequence of nucleotides in a strand being synthesized is dictated by the sequence of nucleotides in the template strand. At the end of the replication process, each template strand is paired with a newly synthesized partner strand. Thus, two identical DNA duplexes are created from one original duplex.

Section: 1.3 DNA as the Genetic Material Difficulty: Medium 47) Briefly explain how genetic information is expressed.

Answer: The expression of genetic information to form a polypeptide is a two-stage process. First, the information contained in a gene's DNA is copied into a molecule of RNA. The RNA is assembled in stepwise fashion along one of the strands of the DNA duplex. During this assembly process, A in the RNA pairs with T in the DNA, G in the RNA pairs with C in the DNA, C in the RNA pairs with G in the DNA, and U in the RNA pairs with A in the DNA. Thus, the nucleotide sequence of the RNA is determined by the nucleotide sequence of a strand of DNA in the gene.

The process that produces this RNA molecule is called transcription, and the RNA itself is called a transcript. The RNA transcript eventually separates from its DNA template and, in some organisms, is altered by the addition, deletion, or modification of nucleotides. The finished molecule, called the messenger RNA or simply mRNA, contains all the information needed for the synthesis of a polypeptide.

The second stage in the expression of a gene's information is called translation. At this stage, the gene's mRNA acts as a template for the synthesis of a polypeptide. Each of the gene's codons, now present within the sequence of the mRNA, specifies the incorporation of a particular amino acid into the polypeptide chain. One amino acid is added at a time. Thus, the polypeptide is synthesized stepwise by reading the codons in order. When the polypeptide is finished, it dissociates from the mRNA, folds into a precise three-dimensional shape, and then carries out its role in the cell. Some polypeptides are altered by the removal of the first amino acid, which is usually methionine, in the sequence.

Section: 1.3 DNA as the Genetic Material Difficulty: Medium

48) Briefly explain two ways the study of genetics has influenced agriculture Answer: Selective breeding programs—now informed by genetic theory—continue to play important roles in agriculture. High- yielding varieties of wheat, corn, rice, and many other plants have been developed by breeders to feed a growing human population. Selective breeding techniques have also been applied to animals such as beef and dairy cattle, swine, and sheep, and to horticultural plants such as shade trees, turf grass, and garden flowers. Plant and animal breeders are also employing the techniques of molecular genetics to introduce genes from other species into crop plants and livestock. This process of changing the genetic makeup of an organism was initially developed using test species such as fruit flies. Today it is widely used to augment the genetic material of many kinds of creatures. Plants and animals that have been altered by the introduction of foreign genes are called GMOs—genetically modified organisms Section: 1.6 Genetics in the World: Applications of Genetics to Human Endeavors Difficulty: Medium 49) How have physicians benefited from the study of genetics?

Answer: Classical genetics has provided physicians with a long list of diseases that are caused by mutant genes. From this work, physicians have learned to diagnose genetic diseases, to trace them through families, and to predict the chances that particular individuals might inherit them, such as is the case with the BRCA1 gene which is diagnostic for breast cancer. Advances in molecular genetics are providing new ways of detecting mutant genes in individuals. Diagnostic tests based on analysis of DNA are now readily available. Molecular genetics is also providing new ways to treat diseases, such as with the advent of the production of human insulin being produced on a large scale by using the bacterium E.coli. Human gene therapy is another way in which molecular genetic technologies are used to treat diseases, such as Cystic fibrosis Section: 1.6 Genetics in the World: Applications of Genetics to Human Endeavors Difficulty: Medium 50) How does the study of genetics impact society as a whole?

Answer: Discoveries from genetics raise deep, difficult, and sometimes disturbing existential questions. Modern societies rely on technology to provide food and health care. Discoveries from genetic research have initiated countless business ventures in the biotechnology industry.

Companies that market pharmaceuticals and diagnostic tests, or that provide services such as DNA profiling, have contributed to worldwide economic growth. Another way is legal. DNA sequences differ among individuals, and by analyzing these differences, people can be identified uniquely.

Such analyses are now routinely used in many situations—to test for paternity, to convict the guilty and to exonerate the innocent of crimes for which they are accused, to authenticate claims to inheritances, and to identify the dead. Evidence based on analysis of DNA is now commonplace in courtrooms all over the world.

Section: 1.6 Genetics in the World: Applications of Genetics to Human Endeavors Difficulty: Medium

Chapter 02 Test Bank Question Type: Multiple Choice 1) The most abundant molecule in the cell is: a) Carbohydrate b) Lipid c) Water d) Protein e) Nucleic acid Answer: c Section: 2.1 Cells and Chromosomes Difficulty: Easy 2) A molecule that interacts easily with water is known as: a) Hydrophonic b) Hydrophilic c) Hydrophobic d) Lipid based e) Protein based Answer: b Section: 2.1 Cells and Chromosomes Difficulty: Easy 3) A molecule that does not interact well with water is known as: a) Hydrophonic b) Hydrophilic c) Hydrophobic d) Water loving e) Protein based Answer: c Section: 2.1 Cells and Chromosomes Difficulty: Easy 4) The internal liquid portion of the cell is known as: a) Cytoplasmic membrane b) Nucleus c) Cytoplasm d) Mitochondria e) Lysosome Answer: c Section: 2.1 Cells and Chromosomes Difficulty: Easy

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5) The function of carbohydrates in a cell is to: a) Serve as an energy source b) Serve as a major constituent of the cell membrane c) Act as the genetic information inside the cell d) None of these e) All of these Answer: a Section: 2.1 Cells and Chromosomes Difficulty: Easy 6) In cells, lipids function as: The function of lipids in a cell is to: 1. An energy source 2. A major constituent of many cellular structures like the cell membrane 3. The genetic information a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: d Section: 2.1 Cells and Chromosomes Difficulty: Easy 7) The function of proteins in a cell is to: a) Act as the genetic information inside the cell b) Serve as a constituent of many cellular structures c) Catalyze chemical reactions d) Choice B and C are correct e) All choices are correct Answer: e Section: 2.1 Cells and Chromosomes Difficulty: Easy 8) Proteins that catalyze biochemical reactions are known as: a) Proteosomes b) Lysosomes c) Chemosomes d) Enzymes e) Anions Answer: d Section: 2.1 Cells and Chromosomes Difficulty: Easy

9) The main components of a membrane in a cell are: a) Lipids b) Proteins c) Carbohydrates d) Lipids and Proteins e) Lipids and Carbohydrates Answer: d Section: 2.1 Cells and Chromosomes Difficulty: Easy 10) Which of the following is a function of the cell membrane? a) Separates the contents of the cell from the outside environment b) Controls the passage of substances into and out of the cell c) Helps the cell maintain its shape d) Contain substances that interact with the external environment e) All of these Answer: e Section: 2.1 Cells and Chromosomes Difficulty: Easy 11) Prokaryotes can be characterized by: 1. The lack of a true nucleus or compartment in which the DNA is located 2. The unique cell walls composed of murein 3. The lack of mitochondria 4. All of these a) 1 b) 2 c) 3 d) 4 e) 1 and 3 Answer: d Section: 2.1 Cells and Chromosomes Difficulty: Easy 12) Eukaryotes can be characterized by: 1. The presence of a true nucleus or compartment in which the DNA is located 2. The presence of mitochondria 3. The presence of membrane bound organelles a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: e Section: 2.1 Cells and Chromosomes Difficulty: Easy

13) In eukaryotes, where is the extranuclear DNA contained? a) Mitochondria b) Cell membrane c) Chloroplasts d) Mitochondria and cell membrane e) Mitochondria and chloroplasts Answer: e Section: 2.1 Cells and Chromosomes Difficulty: Easy 14) Which of the following structures is found in both prokaryotic and eukaryotic cells? a) Mitochondria b) Murein cell wall c) Ribosome d) Chloroplast e) Lysosome Answer: c Section: 2.1 Cells and Chromosomes Difficulty: Easy 15) The ability of a cell to move through its environment is known as: a) Cell trafficking b) Cell motility c) Flagellar motion d) Phagocytosis e) None of these Answer: b Section: 2.1 Cells and Chromosomes Difficulty: Easy 16) The ability of the cell to move materials to specific locations within the cell is known as: a) Cell trafficking b) Cell motility c) Flagellar motion d) Phagocytosis e) None of these Answer: a Section: 2.1 Cells and Chromosomes Difficulty: Easy

17) Which structure in the eukaryotic cell is responsible for the motion of the cell’s ability to move through its environment and its ability to move substances within the cell? a) Nucleus b) Lysosome c) Cell membrane d) Cytoskeleton e) Endoplasmic reticulum Answer: d Section: 2.1 Cells and Chromosomes Difficulty: Easy 18) Which of the following structures is only found in eukaryotic cells? a) Ribosomes b) Cell membrane c) Nucleus d) Cell wall e) None of these Answer: c Section: 2.1 Cells and Chromosomes Difficulty: Easy 19) Which of the following is not an example of a difference between eukaryotic and prokaryotic chromosomes? a) Eukaryotic chromosomes are linear and the prokaryotic chromosome is circular b) Eukaryotes have more than one chromosome and prokaryotes have only one chromosome c) Prokaryotic chromosomes are larger than eukaryotic chromosomes d) All of these are examples of how eukaryote and prokaryote chromosomes differ e) None of these are examples of how eukaryote and prokaryote chromosomes differ Answer: c Section: 2.1 Cells and Chromosomes Difficulty: Easy 20) Eukaryotic cells that possess two copies of each chromosome are said to exist in a ____________ state. a) Haploid b) Diploid c) Aneuploid d) Polyploid e) None of these Answer: b Section: 2.1 Cells and Chromosomes Difficulty: Easy

21) Body cells (i.e. not sex cells) are known as: a) Germ cells b) Gametes c) Somatic cells d) spermatozoa e) oocytes Answer: c Section: 2.1 Cells and Chromosomes Difficulty: Easy 22) Sex cells typically possess only one copy of each chromosome and are said to exist in a ___________ state. a) Haploid b) Diploid c) Anueploid d) Polyploid e) None of these Answer: a Section: 2.1 Cells and Chromosomes Difficulty: Easy 23) The central point that connects the two rod-like portions of the chromosome is known as: a) Centrosome b) Centromere c) Central element d) Central spindle e) Connecting element Answer: b Section: 2.1 Cells and Chromosomes Difficulty: Easy 24) The process of cell division in a prokaryotic cell is known as: a) Fusion b) Fruition c) Fission d) Mitosis e) Meiosis Answer: c Section: 2.1 Cells and Chromosomes Difficulty: Easy

25) Eukaryotic cells divide during which phase of the cell cycle? a) G1 b) S c) G2 d) M e) K Answer: d Section: 2.1 Cells and Chromosomes Difficulty: Easy 26) The process that physically separates eukaryotic daughter cells from each other following nuclear division is known as: a) Mitosis b) Meiosis c) Cytokinesis d) Karyokinesis e) None of these Answer: c Section: 2.1 Cells and Chromosomes Difficulty: Easy 27) The process through with eukaryotic cells distribute their genetic material equally and exactly to their offspring is known as: a) Mitosis b) Meiosis c) Binary fission d) Cytokinesis e) None of these Answer: a Section: 2.1 Cells and Chromosomes Difficulty: Easy 28) During which phase of the cell cycle does chromatin replication take place? a) G1 b) S c) G2 d) Prophase e) Anaphase Answer: b Section: 2.1 Cells and Chromosomes Difficulty: Easy

29) Which of the following cellular components is responsible for executing the distribution of chromosomes during the process of mitosis? a) Nucleus b) Mitochondria c) Microtubules d) Chloroplasts e) Flagella Answer: c Section: 2.2 Mitosis Difficulty: Easy 30) Which of the following events characterizes prophase of mitosis?

1. Initial formation of spindle fibers 2. Condensation of chromosomes 3. Movement of chromosomes to the equatorial plane of the cell a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: d Section: 2.2 Mitosis Difficulty: Easy 31) Which of the following events characterizes metaphase of mitosis?

1. Attachment of spindle fibers to the kinetochores of the chromosomes 2. Movement of chromosomes to the equatorial plane of the cell 3. Separation of sister chromatids that are being pulled to the poles of the cell a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: d Section: 2.2 Mitosis Difficulty: Easy

32) Which is the cause of the separation of sister chromatids during anaphase?

1. The kinetochore breaks apart 2. The spindle fibers shorten 3. The materials holding the sister chromatids degrade a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: e Section: 2.2 Mitosis Difficulty: Easy 33) The reformation of the nuclear membrane and the decondensing of the chromosomes are hallmark events in which phase of mitosis? a) Prophase b) Metaphase c) Anaphase d) Telophase e) Interphase Answer: d Section: 2.2 Mitosis Difficulty: Easy 34) If a cell begins interphase (G1) with 4 chromatin molecules, how many chromatin molecules will each daughter cell possess at the end of telophase, after cytokinesis? a) 2 b) 4 c) 8 d) 16 e) 32 Answer: b Section: 2.2 Mitosis Difficulty: Easy 35) If a cell enters prophase with 10 chromosomes how many chromosomes will be in the cell at the middle of anaphase? a) 5 b) 10 c) 20 d) 40 e) 80 Answer: c Section: 2.2 Mitosis Difficulty: Easy

36) At the end of meiosis, each of the four daughter cells, which will be involved in sexual reproduction will exist in a _________ state. a) Haploid b) Diploid c) Triploid d) Polyploid e) None of these Answer: a Section: 2.3 Meiosis Difficulty: Easy 37) Chromosomes that exist as members of a pair and have the same genetic makeup are known as: a) Heterologues b) Homologues c) Homozygous d) Heterozygous e) None of these Answer: b Section: 2.3 Meiosis Difficulty: Easy 38) Which of the following is a characteristic event of Prophase 1?

1. Separation of homologues 2. Synapsis of homologues 3. Crossing over between homologues a) 1 b) 2 c) 3 d) 1 and 3 e) 2 and 3 Answer: d Section: 2.3 Meiosis Difficulty: Easy 39) The points where chromosomes have physically crossed over are known as: a) Centromeres b) Kinetochores c) Chiasmata d) Diplonema e) Pachynema Answer: c Section: 2.3 Meiosis Difficulty: Easy

40)Which of the following occurs in Meiosis 1 but not in Mitosis? a) Homologous chromosomes pair during prophase b) Homologous chromosomes line up along the metaphase plate c) Homologous chromosomes separate from each other during anaphase d) All of these e) None of these Answer: d Section: 2.3 Meiosis Difficulty: Easy 41) Which of the following events takes place during Meiosis 2? a) Homologous chromosomes cross over b) Sister chromatid separate and move to the poles of the cell c) Homologous chromosomes line up along the metaphase plate d) The number of chromosomes is reduced by half e) All of these take place during Meiosis 2 Answer: b Section: 2.3 Meiosis Difficulty: Easy 42) Which of the following is a reason why the daughter cells that result from the process of Meiosis are not genetically identical to the parent cell at the beginning of the process?

1. Crossing over 2. Each homologous chromosome in a pair is inherited from a different parent (i.e. one from mom and one from dad) 3. Each homologue randomly assorts into a daughter cell a) 1 b) 2 c) 3 d) All of these e) 2 and 3 only Answer: d Section: 2.3 Meiosis Difficulty: Easy 43) In plants, during which phase of the life cycle does meiosis occur? a) Gametophytic b) Sporophytic c) Mitotic d) All of these e) None of these Answer: b Section: 2.3 Meiosis Difficulty: Easy

44) An organism that is favored for use in genetic research is known as a/an: a) Model organism b) Specialized organism c) Clone d) Intermediate organism e) Mammal Answer: a Section: 2.4 Life Cycles of Some Model Genetic Organisms Difficulty: Easy 45) Which of the following is considered a model organism for use in genetic research? a) E. coli b) Saccharomyces cerevisiae c) Drosophila melanogaster d) Mus musculus e) All of these Answer: e Section: 2.4 Life Cycles of Some Model Genetic Organisms Difficulty: Easy

Question Type: Essay 46) Briefly compare the differences between eukaryotic and prokaryotic cells.

Answer: Prokaryotic cells are usually less than a thousandth of a millimeter long, and they typically lack a complicated system of internal membranes and membranous organelles. Their hereditary material—that is, the DNA—is not isolated in a special subcellular compartment.

Organisms with this kind of cellular organization are called prokaryotes. Examples include the bacteria, which are the most abundant life forms on earth, and the archaea, which are found in extreme environments such as salt lakes, hot springs, and deep-sea volcanic vents. All other organisms—plants, animals, protists, and fungi—are eukaryotes. Eukaryotic cells are larger than prokaryotic cells, usually at least 10 times bigger, and they possess complicated systems of internal membranes, some of which are associated with conspicuous organelles. The hallmark of all eukaryotic cells is that their hereditary material is contained within a large, membrane-bounded structure called the nucleus. The nuclei of eukaryotic cells provide a safe haven for the DNA, which is organized into discrete structures called chromosomes. Individual chromosomes become visible during cell division, when they condense and thicken. In prokaryotic cells, the DNA is usually not housed within a well-defined nucleus Section: 2.1 Cells and Chromosomes Difficulty: Medium 47) Colchicine is a drug that is commonly used to treat gout, a disease characterized by the accumulation of uric acid crystals in tissues. Colchicine acts by binding and inhibiting the microtubules. What is the role of microtubules in the cell, and what would be the effect would of inhibiting microtubles?

Microtubules are key components of the cytoskeleton, so they control the location of organelles and the transport of vesicles within the cell. They also play a key role in cell division, since microtubules assemble into the spindle apparatus and attach to the chromosomes during mitosis.

Inhibiting the mitotic spindle would prevent proper segregation of the chromosomes during mitosis. This means normal cell division will be inhibited, and the cells that do divide will likely be aneuploid. These cells will be destined for apoptosis, thanks to the presence of the cell cycle checkpoints.

Section: 2.2 Mitosis Difficulty: Medium

48) You are given a set of slides to analyze, but you do not know if you are looking at an example of mitosis or meiosis. What clues can you look for at each of these stages that might help you distinguish between cells that are going through mitosis or cells that are going through meiosis I? Can you distinguish between cells undergoing mitosis and cells in meiosis II? Explain your answer.

There are three visual clues that you are looking at meiosis, and all of them occur during meiosis I. - If you observe synapsis, that would indicate you are looking at a cell in prophase I. - If there are tetrads aligned at the metaphase plate, you will be able to conclude that the cell is going through metaphase I. - If you see homologous chromosomes separating but sister chromatids are still attached at the centromere, you will know you are looking at anaphase I. If you are just looking at the cells and using DNA staining, mitosis should result in a diploid cell at the end of telophase and cytokinesis, whereas meiosis II should result in a haploid cell. In addition, if you were to stain specific DNA sequences using FISH, you could identify that crossing over had occurred, and that would indicate that you are looking at meiosis.

Section: 2.3 Meiosis Difficulty: Medium 49) Why has E.coli become a commonly used model organism in genetic research?

Answer: This organism can be cultured in the laboratory on a simple medium, it is amenable to all sorts of biochemical analyses, and mutant strains with different growth requirements can be isolated easily. E. coli cells and the viruses that infect them are tiny creatures that can be cultured in the laboratory to produce tens of billions of their own kind in a short period of time. These large population sizes allow researchers to screen efficiently for rare events such as the occurrence of a particular kind of mutation.

Section: 2.4 Life Cycles of Some Model Genetic Organisms Difficulty: Medium 50) Compare and contrast gametogenesis in Arabidopsis and mice. What is the difference between a sporophyte and a gametophyte?

In both plants and animals, the egg is produced from a diploid precursor cell, the megaspore mother cell or the oogonium. These cells undergo a reductional division during meiosis, and only one of the four sets of haploid genomes eventually becomes the egg. In plants, the other polar bodies become part of the endosperm, yielding a triploid tissue designed to nourish the developing embryo. In mice, the 3 polar bodies in mice eventually degrade, and only one haploid set form the egg. The sperm is produced by microsporogenesis in plants, so one microspore develops into four haploid pollen grains (sperm). The process of spermatogenesis also begins with one diploid precursor cell, the spermatogonia, which undergo a reductional division to yield four haploid sperm cells, or gametes.

The sporophyte is the mature plant, which is diploid. A sporophyte produces haploid spores through the process of meiosis. The gametophyte is the haploid plant structure that produces haploid gametes by mitosis.

Section: 2.4 Life Cycles of Some Model Genetic Organisms Difficulty: Medium

Chapter 03 Test Bank Question Type: Multiple Choice 1) What does the term “true-breeding” mean? a) The organism displays little genetic variation from one generation to the next b) The organism displays extreme genetic variations from one generation to the next c) The organism displays a blending of genetic variations from one generation to the next. d) The organism cannot be used for fertilization e) The organism cannot be used for genetic study Answer: a Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy 2) Why was Mendel successful at tracking patterns of inheritance whereas other scientists were not? a) Mendel used the garden pea which was a good model organism b) Mendel tracked only one contrasting phenotypic characteristic at a time c) Mendel used an organism that was true-breeding d) All of these e) None of these Answer: d Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy 3) A recessive trait is one that is: a) Masked by a dominant trait, if a dominant trait is present in the genotype b) Not masked by any other trait present in the genotype c) Masked by another recessive trait, if another recessive trait is present in the genotype d) All of these e) None of these Answer: a Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy 4) A dominant trait is one that: a) Will always be expressed if it is present in the genotype b) Is not observed in every generation if it is present in the genotype c) Masked by a recessive trait, if a recessive trait is present in the genotype d) All of these e) None of these Answer: a Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy

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5) A monohybrid cross is one in which: a) Two traits are being studied at the same time b) One trait is being studied c) Two organisms are being studied at the same time d) One organism is being studied e) None of these Answer: b Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy 6) A heritable factor that exists in two forms is also known as a(n): a) Allele b) Gene c) Factor d) Organism e) Heterozygote Answer: b Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy 7) An organism that inherits two different alleles for a single trait, one from its mother and one from its father is known as a(n): a) Homozygote b) Homo sapien c) Heterozygote d) Allelic variation e) Blended individual Answer: c Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy 8) An organism that inherits two identical alleles, one from its mother and one from its father is known as a(n) a) Homozygote b) Homo sapien c) Heterozygote d) Allelic variation e) Blended individual Answer: a Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy

9) An organism’s allelic constitution is also known as its: a) Phenotype b) Mosaic nature c) Genotype d) Genetic code e) Appearance Answer: c Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy 10) The physical appearance of an organism is known as its: a) Genotype b) Genetic code c) Mosaic nature d) Heterozygosity e) Phenotype Answer: e Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy 11) In a monohybrid cross, the P generation consists of plants that are true-breeding (one is homozygous dominant, the other homozygous recessive) and the F1 generation is allowed to selffertilize. The F2 generation will consist of organisms that exhibit what ratio of dominant phenotypes to recessive phenotypes? a) 3:2 b) 1:2 c) 1:3 d) 3:1 e) None of these Answer: d Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Medium 12) In a monohybrid cross, if the P generation consists of plants that are true-breeding (one is homozygous dominant, the other homozygous recessive) and the F1 generation is allowed to selffertilize, then the F2 generation will consist of organisms that exhibit what ratio of homozygous dominant individuals to heterozygous individuals to homozygous recessive individuals? a) 3:2:1 b) 1:3:1 c) 1:2:1 d) 3:1 e) 1:3 Answer: c Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Medium

13) Which of the following is a principle that Mendel discovered using monohybrid crosses?

1. Principle of segregation 2. Principle of dominance 3. Principle of independent assortment a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: d Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy 14) The principle of dominance states that: a) In a heterozygote, one allele can never conceal the presence of another allele. b) In a heterozygote, one allele may conceal the presence of another allele. c) In a homozygote, one allele may never conceal the presence of another allele. d) In a homozygote, one allele may conceal the presence of another allele. e) None of these Answer: b Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy 15) The principle of segregation states that: a) In a heterozygote, two different alleles separate from each other during the formation of gametes. b) In a heterozygote, two of the same alleles separate from each other during the formation of gametes. c) In a heterozygote, two different alleles do not separate from each other during the formation of gametes. d) In a heterozygote, two of the same alleles do not separate from each other during the formation of gametes. e) None of these Answer: a Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy 16) A dihybrid cross is one in which: a) One trait is studied b) Two traits are studied at the same time c) Two organisms are studied at the same time d) Four traits are studied at the same time e) None of these Answer: b Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy

17) In a dihybrid cross, if the P generation consists of plants that are true-breeding (one is homozygous dominant, the other homozygous recessive) and the F1 generation is allowed to selffertilize, then the F2 generation will consist of organisms that exhibit what phenotypic ratio? a) 3 dominant for both traits : 9 dominant for trait #1 and recessive for trait #2 : 3 dominant for trait #2 and recessive for trait #1 : 1 recessive for both traits b) 3 dominant for both traits : 3 dominant for trait #1 and recessive for trait #2 : 3 dominant for trait #2 and recessive for trait #1 : 1 recessive for both traits c) 9 dominant for both traits : 3 dominant for trait #1 and recessive for trait #2 : 3 dominant for trait #2 and recessive for trait #1 : 1 recessive for both traits d) 9 dominant for both traits : 1 dominant for trait #1 and recessive for trait #2 : 3 dominant for trait #2 and recessive for trait #1 : 1 recessive for both traits e) None of these Answer: c Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Medium 18) The principle of independent assortment states that: a) The alleles of different genes segregate independently of each other b) The alleles of different genes segregate dependent upon each other c) The alleles of the same genes do not segregate independently of each other d) None of these e) All of these Answer: a Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy 19) Which of the following is a principle that Mendel discovered using dihybrid crosses?

1. Principle of segregation 2. Principle of dominance 3. Principle of independent assortment a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: c Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy

20) The fact that all seven of the garden pea traits studied by Mendel obeyed the principle of independent assortment means that the a) Haploid number of garden peas is 7. b) Diploid number of garden peas is 7. c) Seven pairs of alleles determining these traits are on the same pair of homologous chromosomes. d) The seven genes determining these traits behave as if they are on different chromosomes. e) Formation of gametes in plants is by mitosis only.

Answer: d Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Medium 21) In tigers, a recessive allele causes a white tiger. If two phenotypically normal tigers are mated and produce a white offspring, what percentage of their remaining offspring is expected to be white? a) 50% b) 25% c) 100% d) 0% e) Unable to determine based upon the information given Answer: b Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Medium 22) A man was born with six fingers on each and six toes on each foot. His wife and their son have a normal number of digits. Having extra digits is a dominant trait. The couple's second child has extra digits. What is the probability that their next child will have extra digits? a) 10% b) 25% c) 50% d) 100% e) 0% Answer: c Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Medium 23) What is a genetic cross that would be expected to yield a 1:2:1 genotypic ratio? a) Self cross b) Monohybrid cross c) Dihybrid cross d) Test cross e) F1 cross Answer: b Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Easy

24) Black fur in mice is dominant to brown fur. Short tails is dominant to long tails. What proportion of the offspring from a cross between an individual with the genotype BbTt and BBtt will have black fur and long tails? a) 1/16 b) 3/16 c) 6/16 d) 8/16 e) 9/16 Answer: d Section: 3.2 Applications of Mendel’s Principles Difficulty Level: Medium 25) In a trihybrid cross between individuals who are heterozygous for all three traits, the expected proportion of offspring showing all three recessive traits is a) 1/16 b) 1/64 c) 27/64 d) 9/16 e) 9/64 Answer: b Section: 3.2 Applications of Mendel’s Principles Difficulty Level: Medium 26) In a cross between AABBCC and aabbcc (P generation), offspring (F1) are allowed to selfcross. What fraction of the offspring will express the dominant phenotype for genes A and C, but recessive for b in the F2 generation? a) 9/16 b) 9/64 c) 3/16 d) 3/64 e) 1/64 Answer: b Section: 3.2 Applications of Mendel’s Principles Difficulty Level: Medium 27) Which of the following is a method for analyzing a cross involving two genes? a) Punnett Square Method b) Forked-Line Method c) Probability Method d) All of the above e) None of the above Answer: d Section: 3.2 Applications of Mendel’s Principles Difficulty Level: Easy

28) In the chi-square test, a critical value is: a) the point where the discrepancies between observed and expected numbers are not likely to be due to chance. b) the point where the difference between observed and expected is attributed solely to chance. c) the point where there is no discrepancy between observed and expected numbers. d) All of these e) None of these Answer: a Section: 3.3 Testing Genetic Hypotheses Difficulty Level: Easy 29) The chi-square statistic is calculated as: a) 2 =  (O-E)/E b) 2 =  (–)/ c) 2 =  (–)/ d) 2 =  (–)/ e) None of these Answer: a Section: 3.3 Testing Genetic Hypotheses Difficulty Level: Easy 30) The degrees of freedom associated with a chi-square test is equal to: a) The number of data categories minus one. b) The number of data categories divided by one. c) The number of data categories plus one. d) The number of measurements minus one. e) The number of measurements plus one.

Answer: a Section: 3.3 Testing Genetic Hypotheses Difficulty Level: Easy 31) You perform a dihybrid cross between two plants that have traits expressing a simple dominance relationship. Then, you self the F1 generation and analyze the progeny. How many degrees of freedom do you have for your chi-square test? a) 1 b) 2 c) 3 d) 4 e) 5 Answer: c Section: 3.3 Testing Genetic Hypotheses Difficulty Level: Medium

32) Biologists tend to reject the conclusion that the deviation is due to chance if the probability is less than: a) 10% b) 5% c) 15% d) 95% e) 100% Answer: b Section: 3.3 Testing Genetic Hypotheses Difficulty Level: Easy 33) Why has progress studying the genetic traits of human beings been slow? a) It is impossible to make controlled crosses with human beings b) Valid family records are difficult to obtain, and would be essential for study validity. c) Human beings do not typically produce large numbers of progeny d) All of these e) None of these Answer: d Section: 3.4 Mendelian Principles in Human Genetics Difficulty Level: Easy 34) A diagram that is used to show the relationships between family members and is used to track genetic traits is known as a(n): a) Punnett Square b) Pedigree c) Forked-Line d) Family Tree e) Probability Diagram Answer: b Section: 3.4 Mendelian Principles in Human Genetics Difficulty Level: Easy 35) In a pedigree analysis which of the following would represent an affected male? a) Colored circle b) Non-colored circle c) Colored square d) Non-colored square e) Non-colored diamond Answer: c Section: 3.4 Mendelian Principles in Human Genetics Difficulty Level: Easy

36) If there are only two possible phenotypic classes, the probabilities associated with the various outcomes is referred to as: a) Binomial probabilities b) Trinomial probabilities c) Phenotypic probabilities d) Genotypic probabilities e) None of these Answer: a Section: 3.4 Mendelian Principles in Human Genetics Difficulty Level: Easy Question Type: Essay 37) In a cross between individuals of the genotype AaBbCc × AaBbCc, what is the probability of producing the genotype AABBCC? Answer: 1/64 (¼ AA × ¼ BB × ¼ CC) Section: 3.1 Mendel’s Study of Heredity Difficulty Level: Medium 38) A normally pigmented man (dominant) marries an albino woman (recessive). They have three children, one of whom is an albino. What is the genotype of the man?

Answer: Heterozygote Section: 3.2 Applications of Mendel’s Principles Difficulty Level: Medium 39) Flower position, stem length, and seed shape were three traits that were studied by Mendel.

Each is controlled by an independently assorting gene. If a plant that is heterozygous for all three traits was allowed to self-fertilize, what proportion of the offspring would be expected to show all three dominant phenotypes?

Answer: 9/64 Section: 3.2 Applications of Mendel’s Principles Difficulty Level: Medium 40) The dominant gene (G) produces green hair in aliens and the recessive gene (g) produces blue hair. Webbed fingers are due to the dominant gene (F) and normal fingers are recessive (f).

An alien who was heterozygous for hair color and who had normal fingers had a father with blue hair and webbed fingers and a mother with green hair and webbed fingers. What are the genotypes of all individuals (as completely as possible)?

Answer: Alien = Ggff; Alien's father = ggFf; Alien's mother = G_Ff Section: 3.2 Applications of Mendel’s Principles Difficulty Level: Medium

41) In Guinea pigs, black hair (B) is dominant over white (b), rough coat texture (R) is dominant over smooth (r), and short hair (S) is dominant over long hair (s). Cross a homozygous black, rough, short-haired Guinea pig and a white, smooth, long-haired one. What would the phenotype(s) of the offspring be? If two of the F1 offspring were crossed, what would the Answer: F1 phenotype = Black, rough, short haired; Dominant for all three traits = 27/64, dominant for two traits and recessive for one = 9/64, dominant for one trait and recessive for two = 3/64, recessive for all three traits = 1/64 Section: 3.2 Applications of Mendel’s Principles Difficulty Level: Medium 42) In sesame, the one-pod condition (P) is dominant to the three-pod condition (p), and normal leaf (L) is dominant over wrinkled leaf (l). The two characters are inherited independently. A cross between two members of the F1 generation produces the following progeny: 318 one-pod normal, 185 one-pod wrinkled, 323 three-pod normal and 184 three-pod wrinkled. Determine, using chi-square analysis, whether the data fits the typical 9:3:3:1 ratio. Show your work Answer: The data does not fit with the 9:3:3:1 dihybrid ratio therefore the hypothesis would fail to be accepted.

Section: 3.3 Testing Genetic Hypotheses Difficulty Level: Medium 43) Assume that a chi-square test was conducted to test the goodness of fit to a 9:3:3:1 ratio and a chi-square value of 10.62 was obtained. Should the null hypothesis be accepted? Why? Answer: No it should not be accepted, because the chi-square value is higher than the critical value thereby leading one to observe that the probability that the deviation is due to chance to be less than 5%.

Section: 3.3 Testing Genetic Hypotheses Difficulty Level: Medium 44) In a cross of tall tomato plants (T) and dwarf tomato plants (t), the F2 generation consisted of 102 tall and 44 dwarf plants. Does this F2 data fit a ratio of 3:1?

Answer: Yes, the chi-square value shows that the F2 data does fit the ratio of 3:1, because the chi-square value is lower than the critical value thereby leading one to observe that the probability that the deviation is due to chance is greater than 5%.

Section: 3.3 Testing Genetic Hypotheses Difficulty Level: Medium

45) Below is a pedigree of a fairly common human hereditary trait where the boxes represent males and the circles represent females. Shading symbolizes the abnormal phenotype. Assuming only one gene pair is involved, is the inheritance pattern recessive or dominant?

Answer: Recessive Section: 3.4 Mendelian Principles in Human Genetics Difficulty Level: Medium 46) Two parents who are not affected by cystic fibrosis have a child who is affected. Assuming that cystic fibrosis is controlled by one gene, what is the probability that their next three children will also be affected by cystic fibrosis?

Answer: 1/64 Section: 3.4 Mendelian Principles in Human Genetics Difficulty Level: Medium 47) Albinism is caused by a recessive autosomal allele. A man and woman, both normally pigmented, have an albino child together. The mother is now pregnant for a third time and her doctor tells her she is having fraternal twins. What is the probability that both children will have normal pigmentation?

Answer: 9/16 (¾ normal pigmentation child #1 × ¾ normal pigmentation child #2) Section: 3.4 Mendelian Principles in Human Genetics Difficulty Level: Medium 48) Curly hair is caused by a dominant gene in humans. This trait is rare among northern Europeans. If a curly-haired northern European marries a person with straight hair, what proportion of their offspring would be expected to have curly hair, assuming the curly haired individual was a heterozygote?

Answer: 1/2 Section: 3.4 Mendelian Principles in Human Genetics Difficulty Level: Medium 49) What is the probability of having 3 girls and 1 boy in a family, assuming that the probability of producing a female offspring is 0.49?

Answer: 0.06 (0.49 girl#1 x 0.49 girl #2 x 0.49 girl #3 x 0.51 boy#1) Section: 3.4 Mendelian Principles in Human Genetics Difficulty Level: Medium

50) Two parents who are not affected with cystic fibrosis are known carriers for the disease, which is controlled by the inheritance of a single gene. What is the probability that these parents will produce a female offspring who is affected with cystic fibrosis, assuming that the probability of producing a female offspring is 0.51?

Answer: 0.1275 (0.51 female × 0.25 CF) Section: 3.4 Mendelian Principles in Human Genetics Difficulty Level: Medium

Chapter 04 Test Bank Question Type: Multiple Choice 1) If an allele has the same phenotypic effect in heterozygotes as in homozygotes—that is, the genotypes Aa and AA are phenotypically indistinguishable then the allele is: a) Recessive b) Dominant c) Incompletely dominant d) Codominant e) Partially dominant Answer: b Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 2) Red snapdragons are crossed with white snapdragons and in the F1 generation this cross produces pink offspring. The allele (W) controlling the red color is considered to be: a) Recessive b) Dominant c) Incompletely dominant d) Codominant e) None of these Answer: c Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 3) When a heterozygous organism shows characteristics found in each of the associated homozygote states because the different alleles contribute equally to the phenotype, the gene exhibits: a) Simple dominance b) Semi-dominance c) Codominance d) Incomplete dominance e) Partial dominance Answer: c Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy

4) Which of the following is the most appropriate representation of the codominant allele controlling the M blood group? a) M b) m c) LM d) LM e) None of these Answer: d Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 5) A medical technician is analyzing a blood sample in a clinical lab. It is determined that the red blood cells have both A antigens and B antigens present on their surface. In this case, the red blood cells are exhibiting a genetic characteristic known as: a) Simple dominance b) Incomplete dominance c) Partial dominance d) Semi-dominance e) Codominance Answer: e Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 6) Which of the following abbreviations denotes a wild type individual? a) cM b) C+ c) c d) C e) None of these represents a wild type individual Answer: b Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 7) Altered forms of the wild type allele are also commonly known as: a) Varieties b) Mutants c) Oddities d) All of these e) None of these Answer: b Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy

8) In genetic nomenclature, genes are typically named after the: a) Dominant allele b) Recessive allele c) Mutant allele d) Wild type allele e) All of these Answer: c Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy Difficulty: 9) The ABO blood system in humans is controlled by three distinguishable alleles. This is an example of what genetic principle? a) Incomplete dominance b) Partial dominance c) Semi-dominance d) Multiple alleles e) Pleomorphism Answer: d Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 10) If an individual has type O blood, what is the correct genotype for this individual? a) IOi b) ii c) IAIO d) IBi e) It is impossible to tell based upon the information given.

Answer: b Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 11) A woman who is blood type AB has a child with a man who is blood type O. Which of the following is not a possible blood phenotype for the child produced from this union? a) AB b) O c) A d) Choice A and Choice B are not possible e) All of these are possible Answer: d Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy

12) A woman who is blood type A has a child who is blood type O. Which of the following individuals could not be the father of this child? a) Type A b) Type B c) Type AB d) Type O e) All of these could be the fether Answer: c Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 13) Non-functional alleles are said to be: a) Polymorphic b) Amorphic c) Hypomorphic d) Heteromorphic e) Homomorphic Answer: b Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 14) Alleles that are partially functional are said to be: a) Polymorphic b) Amorphic c) Hypomorphic d) Heteromorphic e) Homomorphic Answer: c Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 15) Allele b denotes blue eyes in humans and allele B for brown eyes is completely dominant over b. Alele b is partially dominant over bg (green eyes) and bh (hazel eyes). Allele bh is completely dominant over bg (green eyes) What is the dominance hierarchy for this trait? a) B>b>bg>bh b) B>b>bh>bg c) b>B>bg>bh d) B>bg>bh>b e) Bg>bh>b>B Answer: b Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy

16) The test to determine whether similar phenotypic mutations are alleles of the same gene based on the phenotypic effect of combining the mutations in the same individual is known as the: a) Bradford assay b) Complementation test c) Live/Dead assay d) ELISA e) None of these Answer: b Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 17) Mutations that alter some aspect of morphology are known as: a) Lethal mutations b) Sterile mutations c) Visible mutations d) Non-visible mutations e) Polymorphic mutations Answer: c Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy

18) Mutations that limit reproduction are known as: a) Lethal mutations b) Sterile mutations c) Visible mutations d) Non-visible mutations e) Polymorphic mutations Answer: b Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 19) If a mutation occurs and it impairs a vital function of an organism the mutation is known as a: a) Lethal mutation b) Sterile mutation c) Visible mutation d) Non-visible mutation e) Polymorphic mutation Answer: a Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy

20) In mice, a cross between AYA+ heterozygotes produces two kinds of progeny, yellow (AYA+) and gray-brown (A+A+), in a ratio of 2:1. Why is there an altered ratio of individuals produced? a) AYAY individuals die during embryonic development due to the presence of a lethal mutation b) AYA+ individuals are physiologically more able to adapt to the birth process c) A+A+ individuals are the mutant phenotype and occur more frequently in nature d) AYA+ individuals are the mutant phenotype and occur more frequently in nature e) None of these explains why there would be an altered ratio in this case Answer: a Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 21) Which of the following is responsible for the discovery of the “one gene=one polypeptide” theory, which says that one gene is responsible for the formation of one polypeptide? a) Darwin b) Mendel c) Bateson d) Beadle and Tatum e) Schleiden and Schwann Answer: d Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 22) A mutation that actually interfere with the function of the wild-type allele by specifying polypeptides that inhibit, antagonize, or limit the activity of the wild-type polypeptide is known as: a) Loss of function mutation b) Partial loss of function mutation c) Recessive mutation d) Gain of function mutation e) Dominant negative mutation Answer: e Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy 23) Most, but not all, loss of function mutations can be attributed to a: a) Dominant allele b) Recessive allele c) Codominant allele d) Partially dominant allele e) Incompletely dominant allele Answer: b Section: 4.1 Allelic Variation and Gene Function Difficulty: Easy

24) Polypeptides are macromolecules comprised of what monomer subunit? a) Nucleic acids b) Amino acids c) Micromolecules d) Enzymes e) Proteins Answer: b Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy 25) The work of Beadle and Tatum not apply to every gene. Why not? a) Genes never encode polypeptide sequences b) Some genes encode RNA sequences that do not lead to the formation of polypeptides c) Some genes encode other DNA sequences that do not lead to the formation of polypeptides d) All of these are correct e) None of these Answer: b Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy 26) Which of the following is a true statement regarding gene activity? a) Genes do not act in isolation b) Genes act in the context of their environment c) Genes act in concert with other genes d) One gene can influence multiple traits e) All of these are true.

Answer: Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy 27) Which of the following can control gene activity? a) Temperatue b) Biological factors such as an individual's sex c) Diet d) All of these e) None of these Answer: e Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy

28) When individuals do not show a trait even though they have the appropriate genotype, the trait is said to exhibit: a) Codominance b) Incomplete dominance c) Incomplete penetrance d) Incomplete expressivity e) Complete penetrance Answer: c Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy 29) The allele for polydactyly is dominant, but most individuals do not exhibit the polydactyly trait. Which of the following explains why this occurs in nature? a) The allele for polydactyly is completely penetrant b) The allele for polydactyly is never expressed c) The allele for polydactyly is incompletely penetrant d) The allele for polydactyly is variably expressive e) None of these is an adequate explanantion Answer: c Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy 30) In Drosophila that exhibit the lobe eye mutation, some heterozygous flies have tiny compound eyes, whereas others have large, lobulated eyes and between these extremes, there is a full range of phenotypes. This mutation is said to have: a) Incomplete penetrance b) Complete penetrance c) Complete expressivity d) Variable expressivity e) None of these Answer: d Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy 31) Pattern baldness is an example of a genetic trait, often expressed differently in males and females. It is considered a ___________________ trait. a) Sex-linked b) Sex-influenced c) Dominant d) Recessive e) Penetrant Answer: b Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy

32) The work of Bateson and Punnett, who studied the combs of chickens, demonstrated that: a) Two independently assorting genes can affect a trait. b) Two linked genes can affect a trait c) Two independently assorting genes cannot affect a trait d) Two linked genes cannot affect a trait e) None of these Answer: a Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy 33) When two or more genes influence a trait, an allele of one of them may have an overriding effect on the phenotype. When an allele has such an overriding effect, it is said to be: a) Dominant b) Recessive c) Penetrant d) Epistatic e) Expressive Answer: d Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy 34) Summer squash color is determined by the interaction of more than one gene. The presence of CC or Cc allele combinations produces a squash that is white in color, and the C allele is epistatic to the G allele. The presence of GG or Gg produces a squash that is yellow in color, and ccgg produces a squash that is green. After two heterozygous squash are crossed one of the offspring is CcGg. What is the color of the offspring squash? a) White b) Yellow c) Green d) It is impossible to determine based on the information given e) None of these Answer: a Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Medium 35) Bateson and Punnett mated purple sweetpeas with white sweetpeas. F1 hybrids were all purple in color. When the hybrids were mated a ratio of 9 purple : 7 white were observed in the F2 generation. Which of the following could best explain this observation? a) Incomplete penetrance b) Lethal alleles c) Epistasis d) Codominance e) Incomplete dominance Answer: c Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy

36) In maize, overall kernel color is controlled by both the endosperm and the aleurone colors.

The Y allele produces a yellow endosperm while the homozygous recessive (y/y) state produces a white endosperm. Endosperm color is masked by the presence of a colored aleurone. The R and Pr genes control aleurone color. The combination of Pr and R alleles lead to a purple aleurone.

The combination of pr and R alleles lead to a red aleurone. A homozygous recessive (r/r) produces a colorless aleurone, allowing the endosperm color to be expressed. Which allele is epistatic? a) Y b) R c)Pr d) Y and R e) R and Pr Answer: b Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Medium 37) When one gene influences many phenotypic expressions it is said to be: a) Incompletely penetrant b) Variably penetrant c) Pleiotropic d) Epistatic e) Polygenice Answer: c Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy 38) The gene that controls Marfan's Syndrome causes abnormal heart defects, the presence of long limbs, the presence of increased height, and many other physical characteristics. It could be said that the gene that controls Marfan's Syndrome is: a) Expressive b) Penetratn c) Pleiotropic d) Polygenic e) Epistatic Answer: c Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Easy

39) Geneticists use the _______________________ to analyze the effects of matings between relatives. a) Expressivity coefficient b) Inbreeding coefficient c) Mating frequency d) Chi-square value e) Student T-test Answer: b Section: 4.3 Inbreeding: Another Look at Pedigrees Difficulty: Easy 40) Which of the following is a disadvantage for creating inbred animal lines?

1. Increased genetic purity 2. Increased inbreeding depression 3. Increased rate of expression of harmful recessive alleles a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: e Section: 4.3 Inbreeding: Another Look at Pedigrees Difficulty: Easy 41) The inbreeding coefficient can be defined as the: a) Probability that two gene copies in an individual are identical by descent from a common ancestor b) Probability that two gene copies in an individual are different by descent from a common ancestor c) Probability that two gene copies in an individual are identical by descent from two different ancestors d) All of these e) None of these Answer: a Section: 4.3 Inbreeding: Another Look at Pedigrees Difficulty: Easy

42) The inbreeding coefficient can be used to measure: a) The decline in a complex phenotype b) The increased frequency of recessive disorders among the offspring of consanguineous matings c) The closeness of genetic relationships d) All of these e) None of these Answer: d Section: 4.3 Inbreeding: Another Look at Pedigrees Difficulty: Easy 43) The fraction of genes that two individuals share due to common ancestry is known as: a) The inbreeding coefficient b) The coefficient of relationship c) The relative constant d) All of these e) None of these Answer: b Section: 4.3 Inbreeding: Another Look at Pedigrees Difficulty: Easy 44) The inbreeding coefficient is represented by which of the following: a) (1/2)n b) n(1/2) c) (1/2)n * 2 d) (1/2) + n e) n – (1/2) Answer: a Section: 4.3 Inbreeding: Another Look at Pedigrees Difficulty: Easy 45) The coefficient of relationship is represented by which of the following: a) (1/2)n b) n(1/2) c) (1/2)n * 2 d) (1/2) + n e) n – (1/2) Answer: c Section: 4.3 Inbreeding: Another Look at Pedigrees Difficulty: Easy

Question Type: Essay 46) Summer squash color is determined by the interaction of more than one gene. The presence of CC or Cc allele combinations produces a squash that is white in color, and the C allele is epistatic to the G allele. The presence of GG or Gg produces a squash that is yellow in color, and ccgg produces a squash that is green. After two fully heterozygous squash are crossed give the phenotypes, and frequency of occurrence, of the F2 offspring.

Answer: White = 12/16 ; Yellow = 3/16 ; Green = 1/16 Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Hard 47) In maize, overall kernel color is controlled by both the endosperm and the aleurone colors.

The endosperm color is yellow when the allele Y is present. In the homozygous recessive (y/y) condition the endosperm is white. The Y alleles are masked by the presence of a colored aleurone. The R and Pr genes control the aleurone color. The allele Pr interacts with the R allele to produce a purple color, whereas in the recessive condition (pr/pr), the gene interacts with the R allele to produce a red color. If the R allele is present in the homozygous recessive state (r/r) then the aleurone will be colorless, allowing the endosperm color to be expressed.

Two fully heterozygous individuals are crossed. Give the phenotypes, and frequency of occurrence, of the F2 offspring.

Answer: Yellow kernel color = 12/64, White kernel color = 4/64, Red kernel color = 12/64, Purple kernel color =36/64 Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Hard 48) In a mating between individuals with the genotypes IBIO X IOIO, what percentage of the offspring are expected to have the O blood type?

Answer: 50% Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Medium 49) The trait of wavy flowers in petunias is determined by the genetic condition PP'. Plants with flat flowers are PP, while plants with curly flowers are P'P'. A cross is made between two plants each with wavy flowers. If they produce 200 seedlings, what phenotypes and in what numbers would you expect? What is the term for this allelic relationship?

Answer: Curly Flowers = 50 seedlings, Flat Flowers = 50 seedlings, Wavy Flowers= 100 seedlings; Incomplete dominance Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Medium

50) On a national talk show a woman was claiming that a man was the father of her child. She had blood type AB. The child had type B. The suspected father had type O. The suspected father maintained the claim that since he had type O blood he could not be the father. You have been called in as an expert witness. Explain why or why not the man could be the father of the child.

Answer: The man could be the father of the child, because his IOIO genotype can only contribute recessive IO alleles and when crossed with the mothers IAIB genotype would produce children of with either Type A or Type B blood.

Section: 4.2 Gene Action: From Genotype to Phenotype Difficulty: Medium

Chapter 05 Test Bank Question Type: Multiple Choice 1) Which of the following discovered chromosomes in the second half of the nineteenth century? a) Thomas Morgan b) Calvin Bridges c) W. Waldeyer d) Gregor Mendel e) Edmund Beecher Wilson Answer: c Section: 5.1 Chromosomes Difficulty: Easy 2) A diffuse network of thin and loosely coiled chromosomes is referred to as: a) Chromosomes b) Cylinders c) Chromatin d) Chromomeres e) Centromeres Answer: c Section: 5.1 Chromosomes Difficulty: Easy 3) Lightly stained regions of chromatin and darkly stained regions of chromatin are respectively known as: a) Heterochromatin and euchromatin b) Euchromatin and heterochromatin c) Haploid and diploid d) Eukaryotic and prokaryotic e) Chromosome and nucleus Answer: b Section: 5.1 Chromosomes Difficulty: Easy 4) An organism’s standard chromosome number (n) is known as the ________ number. a) Haploid b) Diploid c) Aneuploid d) Polyploid e) Tetraploid Answer: a Section: 5.1 Chromosomes Difficulty: Easy

5) Cells that contain two of each of the chromosomes in a set are known as: a) Haploid b) Diploid c) Aneuploid d) Polyploid e) Monoploid Answer: b Section: 5.1 Chromosomes Difficulty: Easy 6) Which of the following is a possible offspring in a mating between grasshoppers? a) XO male b) XX female c) XO female d) XO male and XX female e) XO male and XO female Answer: d Section: 5.1 Chromosomes Difficulty: Easy 7) In some animal species, like the grasshopper, sex is determined by: a) The presence of the Y chromosome b) The number of X chromosomes in relation to the number of sets of autosomes c) The absence of the Y chromosome d) The shorter nature of the Y chromosome e) None of these Answer: b Section: 5.1 Chromosomes Difficulty: Easy 8) In some animal species, like humans, sex is determined by: a) The presence of the Y chromosome b) The number of X chromosomes in relation to the number of sets of autosomes c) The size of the Y chromosome d) None of these e) All of these Answer: a Section: 5.1 Chromosomes Difficulty: Easy

9) Why is the Y chromosome able to pair with the X chromosome during meiosis? a) The Y chromosome has the same centromere placement as the X chromosome b) The Y chromosome is the same length as the X chromosome c) The X and Y chromosomes share small gene segments that allow them to act like homologues during meiosis d) None of these e) All of these Answer: c Section: 5.1 Chromosomes Difficulty: Easy 10) X and Y chromosomes are known as: a) Autosomes b) Somosomes c) Sex chromosomes d) Non-sex chromosomes e) Somatosomes Answer: c Section: 5.1 Chromosomes Difficulty: Easy 11) All chromosomes except X and Y are known as: a) Autosomes b) Somosomes c) Sex chromosomes d) Somatosomes e) Tetrasomes Answer: a Section: 5.1 Chromosomes Difficulty: Easy 12) In a certain species, a somatic cell has 48 chromosomes. How many chromosomes are in a gamete? a) 12 b) 24 c) 96 d) 144 e) 200 Answer: b Section: 5.1 Chromosomes Difficulty: Medium

13) The haploid number for a cell in a specific organism is 12. What is the number of chromosomes in a somatic cell in this organism? a) 12 b) 24 c) 36 d) 48 e) 96 Answer: b Section: 5.1 Chromosomes Difficulty: Medium 14) A cell is stained and viewed under a microscope and no chromosomes are visible. Which point in the cell cycle is the cell currently in? a) Interphase b) Prophase c) Metaphase d) Anaphase e) Telophase Answer: a Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Medium 15) Which organism that was the first to be used to relate the behavior of chromosomes during meiosis to Mendel's principles of Segregation and Independent Assortment? a) Garden peas b) Summer squash c) Drosophila d) Chickens e) Sweet peas Answer: c Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Easy 16) Which of the following is required to unambiguously link a gene to a chromosome? a) The gene must be defined by a mutant allele b) The chromosome must be morphologically distinguishable c) The pattern of gene transmission must reflect the chromosome's behavior during reproduction. d) All of these e) None of these Answer: d Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Medium

17) Which individual is credited with linking the gene for eye color in Drosophila to the X chromosome? a) Thomas Hunt Morgan b) Calvin Bridges c) Gregor Mendel d) Sutton and Boveri e) None of these Answer: a Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Easy 18) A (w+/ w+) red-eyed Drosophila female is crossed with a white-eyed male. Assuming the trait for eye color is sex-linked, what are the possible phenotypes of the progeny? a) All red-eyed individuals b) Red- and white-eyed females and males c) Only red-eyed females and white-eyed males d) Both red- and white-eyed males and only white-eyed females e) None of these Answer: a Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Medium 19) An organism that has only one copy of a gene is called a: a) Heterozygote b) Homozygote c) Hemizygote d) Zygote e) None of these Answer: c Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Easy 20) A (w+/ w) red-eyed Drosophila female is crossed with a white-eyed male. Assuming the trait for eye color is sex-linked, what are the possible phenotypes of the progeny? a) All red-eyed individuals b) Red- and white-eyed females and males c) Red-eyed females and white-eyed males d) Both red- and white-eyed males and only white-eyed females e) None of these Answer: b Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Medium

21) A white-eyed Drosophila female is crossed with a red-eyed male. Assuming the trait for eye color is sex-linked, what are the possible phenotypes of the progeny? a) All red-eyed individuals b) Red- and white-eyed females and males c) Red-eyed females and white-eyed males d) Red- and white-eyed males and only-white eyed females e) None of these Answer: c Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Medium 22) The view that all genes are located on chromosomes and that Mendel's principles can be explained by the transmissional properties of chromosomes during reproduction is known as: a) The Cell Theory b) The Chromosome Theory of Heredity c) The Sex-Linked Inheritance Theory d) All of these e) None of these Answer: b Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Easy 23) When chromosomes fail to separate during meiosis, thereby producing an egg with two X chromosomes or an egg with no X chromosome at all, this event is referred to as: a) Nondisjunction b) Disjunction c) Polyploidy d) Polysomy e) None of these Answer: a Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Easy 24) A Drosophila is determined to have the genotype XO. What can you determine about this individual based upon this information? a) The individual is male b) The individual is sterile c) The individual is male and sterile d) The individual is female and sterile Answer: d Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Easy

25) In Drosophila, Xw denotes a white eyed allele and X+ denotes the red-eyed allele. An XwXwY female is crossed with an X+Y male. What are the possible progeny genotypes? a) XwXwX+, XwY b) XwXwX+, XwXwY, XwY, X+O c) XwXw, X+Y, XwO d) Y0, XwXwX+, XwXwY, X e) None of these Answer: e Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Easy 26) Down Syndrome occurs when there is an extra copy of chromosome 21. Such individuals therefore have 47 chromosomes. If a female with Down Syndrome mates with a normal male and normal disjunction occurs, what percentage of offspring would be expected to have Down Syndrome? a) 25% b) 50% c) 75% d) 100% e) 0% Answer: b Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Medium 27) Which of Mendel's principles is based on the separation of homologous chromosomes during the anaphase of the first meiotic division? a) Dominance/Recessiveness b) Segregation c) Independent Assortment d) All of these e) None of these Answer: b Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Easy 28) Which of Mendel's principles is a statement about the random alignment of different pairs of chromosomes at metaphase during meiosis? a) Dominance/Recessiveness b) Segregation c) Independent Assortment d) All of these e) None of these Answer: c Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Easy

29) Why is it easier to identify a recessive sex-linked trait in human beings than a recessive autosomal trait? a) A male needs only to inherit one recessive allele to express an X-linked trait b) A male needs to inherit two recessive alleles to express an X-linked trait c) A female only needs to inherit one recessive allele to express an X-linked trait d) All of these e) None of these Answer: a Section: 5.3 Sex-Linked Genes in Humans Difficulty: Medium 30) Which of the following is an X-linked disorder in human beings? a) Hemophilia b) Color blindness c) HIV d) Hemophilia and Color blindness e) All of these Answer: d Section: 5.3 Sex-Linked Genes in Humans Difficulty: Easy 31) A color blind male mates with woman who has normal vision. The woman has no history of colorblindness in her family. What percentage of their sons are expected to be colorblind? a) 25% b) 50% c) 75% d) 100% e) 0% Answer: e Section: 5.3 Sex-Linked Genes in Humans Difficulty: Medium 32) A(n)________________ gene has a locus on both the X-and Y-chromosome a) Autosomal b) Psedoautosomal c) X-linked d) Y-linked e) None of these Answer: b Section: 5.3 Sex-Linked Genes in Humans Difficulty: Easy

33) In humans the Y chromosome carries _________ genes than the X chromosome. a) Fewer b) More c) Similar d) All of these e) None of these Answer: a Section: 5.3 Sex-Linked Genes in Humans Difficulty: Easy 34) Sexual dimorphism can be determined by: a) The environment b) Temperature c) Genetic factors d) All of these e) None of these Answer: d Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Easy 35) In humans an individual who is genotyped as XO is a: a) Male b) Female c) Hermaphrodite d) None of these e) All of these Answer: b Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Easy 36) Which of the following is the part of the Y chromosome that is critical for normal male development? a) SRY b) Pseudoautosomal region c) Centromere d) All of these e) None of these Answer: a Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Easy

37) In a human, what is the best explanation for being XX and phenotypically male? a) One of the X's is incomplete resembling a chromosome. b) A small piece of the Y-chromosome, containing the SRY region, is inserted on the Xchromosome. c) Both the X's have a mutation deleting the "female" forming genes. d) There is a fragment of the Y-chromosome inserted on an autosomal chromosome. e) The SRY gene has mutated.

Answer: b Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Medium 38) The sex-linked disease called testicular feminization, if transferred from mother to son, forms: a) males (XY) that are phenotypically female and sterile. b) females (XY) that are sterile. c) males (XY) that are sterile. d) females (XY) that are phenotypically male and are sterile. e) fertile males (XY) Answer: a Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Medium 39) In Drosophila, the Y chromosome is not required for male sex-determination. However, it is required for : a) Male sterility b) Male fertility c) Female sterility d) Female fertility e) None of these Answer: b Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Easy 40) An individual with two kinds of gametes (i.e. X bearing and Y bearing) is referred to as: a) Homogametic b) Heterogametic c) Heterozygous d) None of these e) All of these Answer: b Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Easy

41) Which of the following is a mechanism that can compensate for an abnormal number of sex chromosomes in an individual?

1. Each X-linked gene could work twice as hard in males as it does in females 2. One copy of each X-linked gene could be inactivated in females 3. Each X-linked gene could work half as hard in females as it does in males a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: e Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Medium 42) An increase in the activity of X-linked genes in males, as is observed in Drosophila, is known as: a) Hypoactivation b) Hyperactivation c) Mosaicisn d) None of these e) All of these Answer: b Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Easy 43) In mammals, dosage compensation for X-linked genes is achieved by: a) Hyperactivation of X-linked genes b) Down regulation of X-linked genes c) Inactivation of one X-chromosome during early development d) All of these e) None of these Answer: d Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Easy 44) Female mammals that contain two types of cell lineages in which the paternal X chromosome is inactivated in some cells and the maternal X chromosome is inactivated in others are known as: a) Genetic twins b) Genetic mosaics c) Genetic anomalies d) None of these e) All of these Answer: b Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Easy

45) An X chromosome that has been inactivated becomes a(an)___________ in female mammals. a) Barr body b) Oncogene c) Bacteriophage d) All of these e) None of these Answer: a Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Easy

Question Type: Essay

46) A researcher crosses a white-eyed female Drosophila with a red-eyed male. Knowing that the gene for eye color is sex-linked, he expects to observe only red-eyed females and white-eyed males in the progeny. However, he observes that while most of the progeny are as expected, a small percentage of the females have white eyes. Explain what may have occurred to cause this result.

Answer: There may have been nondisjunction in either anaphase1 or anaphase2 of meiosis in the egg cell which would lead to an egg cell with an XwXw combination. During fertilization with a normal sperm cell (Xw+Y) from the red=eyed father, a somatic cell with the genotype XwXwY would result, thereby producing a white-eyed female offspring where there should be none. It should be noted that Drosophila follow the sex determination principles like that of the grasshopper. Sex is determined based upon the number of X chromosomes in relation to the number of sets of autosomes, not the presence of a Y chromosome.

Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Hard 47) A researcher crosses a white-eyed female Drosophila with a red-eyed male. Knowing that the gene for eye color is sex-linked, he expects to observe only red-eyed females and white=eyed males in the progeny. However, he observes that while most of the progeny are as expected, a small percentage of the males have red eyes. Explain what may have occurred to cause this result.

Answer: There may have been nondisjunction in either anaphase I or anaphase II of meiosis in the egg cell which would lead to an egg cell with an XwXw combination and an egg cell with no X chromosome. During fertilization with a normal sperm cell (Xw+Y) from the red-eyed father, a somatic cell with the genotype Xw+O would result, thereby producing a red-eyed male offspring where there should be none. **Please note that Drosophila follow the sex-determination principles like that of the grasshopper. Sex is determined based upon the number of X chromosomes in relation to the number of sets of autosomes, not the presence of a Y chromosome.

Section: 5.2 The Chromosome Theory of Inheritance Difficulty: Hard

48) A man with hemophilia mates with a woman who is normal but whose father was a hemophiliac, mother was normal, and whose maternal grandmother was a hemophiliac. What percentage of their sons could be hemophiliacs?

Answer: 50% of the sons could be hemophiliac Section: 5.3 Sex-Linked Genes in Humans Difficulty: Medium 49) A woman, with normal vision whose father was colorblind mates with a man who has normal vision. What are the chances that she will have a son who is colorblind?

Answer: ½ * ¼ = 1/8 chance that she will have a son who is also colorblind.

Section: 5.3 Sex-Linked Genes in Humans Difficulty: Medium 50) Assume that coat color in cats is an X-linked trait. Briefly explain why only females can exhibit a combination of coat colors (i.e., Calico), whereas normal XY males can only exhibit one.

Answer: Females will be mosaics with one X chromosome being inactivated in each cell.

Therefore depending on which X chromosome is inactivated, they can express multiple coat colors, whereas males, being hemizygous, will only express one. Males will only express multiple coat colors if they are an XXY male.

Section: 5.4 Sex Chromosomes and Sex Determination Difficulty: Medium

Chapter 06 Test Bank Question Type: Multiple Choice 1) The analysis of stained chromosomes is the main activity of the discipline called: a) Cytology b) Cytogenetics c) Genetics d) Embryology e) Neonatology Answer: b Section: 6.1 Cytological Techniques Difficulty: Easy 2) Which type of cell is most often used by researchers? a) Dividing cells b) Non-dividing cells c) Apoptosing cells d) None of these e) All of these Answer: a Section: 6.1 Cytological Techniques Difficulty: Easy 3) Which of the following is the most commonly used differential stain for chromosome analysis? a) Giesma b) Quinacrine c) Crystal Violet d) Giema and Quinacrine e) All of these Answer: d Section: 6.1 Cytological Techniques Difficulty: Easy 4) Which technique creates colorful chromosome images by treating chromosome spreads with fluorescently labeled DNA fragments that have been isolated and characterized in the laboratory? a) Giemsa staining b) Quinacrine staining c) Gram staining d) Chromosome painting e) Classical karyotyping Answer: d Section: 6.1 Cytological Techniques Difficulty: Easy

5) A DNA fragment that interacts with the complementary sequence of DNA on the chromosome is known as a(an): a) Probe b) Fingerprint c) Fluorescent dye d) Probe and fingerprint e) All of these Answer: a Section: 6.1 Cytological Techniques Difficulty: Easy 6) By labeling chromosomes with various DNA probes it is possible to: 1. Locate individual gene sequences on a chromosome 2. Compare the similarities between human DNA and another mammal’s DNA 3. Observe any abnormalities in the chromosome structure a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: e Section: 6.1 Cytological Techniques Difficulty: Medium 7) Human diploid cells contain _____ chromosomes. a) 12 b) 24 c) 46 d) 92 e) 23 Answer: c Section: 6.1 Cytological Techniques Difficulty: Easy 8) A pictorial chart of chromosomes arranged from largest to smallest is known as a(an): a) Karyotype b) Genetic analysis c) Pedigree chart d) Amniocentesis e) Chorionic villus sample Answer: a Section: 6.1 Cytological Techniques Difficulty: Easy

9) The human diploid cell has ______ pairs of chromosomes. a) 12 b) 23 c) 24 d) 46 e) 48 Answer: b Section: 6.1 Cytological Techniques Difficulty: Easy 10) The long arm of the chromosome is designated: a) P b) Q c) L d) M e) N Answer: b Section: 6.1 Cytological Techniques Difficulty: Easy 11) The short arm of the chromosome is designated: a) P b) Q c) L d) M e) N Answer: a Section: 6.1 Cytological Techniques Difficulty: Easy 12) The pattern of bands within a chromosome is termed: a) Karyotype b) Chromogram c) Ideogram d) Sonogram e) None of these Answer: c Section: 6.1 Cytological Techniques Difficulty: Easy

13) If a cytogeneticist refers to the 7p region of a chromosome, he/she is referring to the: a) Short arm on chromosome 7 b) Long arm on chromosome 7 c) 7th band on chromosome P d) P band on chromosome 7 e) 7th band on the short arm of chromosome 1 Answer: a Section: 6.1 Cytological Techniques Difficulty: Easy 14) Which of the following can cause a phenotypic change in an organism? a) Too many chromosomes b) Too few chromosomes c) Changes in part of a chromosome d) Too many chromosomes and too few chromosomes e) All of these Answer: e Section: 6.2 Polyploidy Difficulty: Easy 15) An organism that carries extra sets of chromosomes is termed: a) Aneuploid b) Euploid c) Polyploid d) Diploid e) Multiploid Answer: c Section: 6.2 Polyploidy Difficulty: Easy 16) Organisms in which one chromosome is either under- or overrepresented are termed: a) Aneuploid b) Euploid c) Polyploid d) Diploid e) Haploid Answer: a Section: 6.3 Aneuploidy Difficulty: Easy

17) Which of the following terms implies a genetic imbalance? a) Aneuploid b) Polyploid c) Diploid d) Aneuploid and Polyploid e) All of these Answer: a Section: 6.3 Aneuploidy Difficulty: Easy 18) Which of the following is an effect of polyploidy? a) Increased cell size b) Increased organism size c) Organisms that are more robust d) Increased cell size and Organisms that are more robust e) All of these Answer: e Section: 6.3 Aneuploidy Difficulty: Easy 19) Which of the following crop plants are polyploids? a) Wheat b) Coffee c) Strawberries d) All of these e) None of these Answer: d Section: 6.3 Aneuploidy Difficulty: Easy 20) Why are many polyploids sterile? a) Extra sets of chromosomes segregate irregularly in meiosis leading to aneuploidic gametes b) Extra sets of chromosomes segregate regularly in meiosis leading to aneuploidic gametes c) Extra sets of chromosomes segregate irregularly in mitosis leading to aneuploidic gametes d) All of these e) None of these Answer: a Section: 6.3 Aneuploidy Difficulty: Easy

21) In a polyploidy organism, sometimes three homologues synapse and partially pair with each of the others forming a(an): a) Univalent b) Bivalent c) Trivalent d) Trisomic e) Hectovalent Answer: c Section: 6.3 Aneuploidy Difficulty: Easy 22) Which of the following is a method of asexually propagating sterile polyploids? a) Apomixis b) Cultivation from cuttings c) Bulbs d) Grafts e) All of these Answer: e Section: 6.3 Aneuploidy Difficulty: Easy 23) Which of the following would be most likely to be sterile? a) Diploid organisms b) Triploid organisms c) Tetraploid organisms d) All of these e) None of these Answer: b Section: 6.3 Aneuploidy Difficulty: Easy 24) A fertile tetraploid would best be characterized as having a(an): a) Two distinct sets of chromosomes with each set being duplicated b) One distinct set of chromosomes that have been duplicated c) Three distinct sets of chromosomes with each set being duplicated d) Four distinct sets of chromosomes with each set being duplicated e) None of these Answer: a Section: 6.3 Aneuploidy Difficulty: Easy

25) Fertile tetraploids seem to have arisen by chromosome duplication in a hybrid that was produced by a cross of: a) Two different, but related, diploid species b) Two identical diploid species c) Four different bur related haploid species d) Four identical diploid species e) Four identical haploid species Answer: a Section: 6.3 Aneuploidy Difficulty: Medium 26) A hexaploid, such as Triticum aestivum, could best be described as having: a) Three different chromosome sets each of which has been duplicated b) Three different chromosome sets each of which has not been duplicated c) Six different chromosome sets each of which has been duplicated d) Sixteen different chromosome sets each of which has not been duplicated e) Six identical chromosome sets each of which has been duplicated Answer: a Section: 6.3 Aneuploidy Difficulty: Medium 27) Polyploids created from a hybridization between different species are known as: a) Autopolyploids b) Allopolyploids c) Triploids d) Tetraploids e) None of these Answer: b Section: 6.3 Aneuploidy Difficulty: Easy 28) Polyploids created by chromosome duplication within a species are known as: a) Autopolyploids b) Allopolyploids c) Triploids d) Tetraploids e) None of these Answer: a Section: 6.3 Aneuploidy Difficulty: Easy

29) Which of the following is a method that could result in a polyploid organism 1. A cell could go through mitosis without going through cytokinesis.

2. A cell could go through meiosis without properly separating the homologous pairs 3. A cell could go through mitosis while doubling the rounds of cytokinesis a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: d Section: 6.3 Aneuploidy Difficulty: Medium 30) The process that produces polyploid cells in certain cells during development, such as those in the kidney and the liver is known as: a) Aneuploidic interaction b) Endomitosis c) Exomitosis d) Meiosis e) Endomeiosis Answer: b Section: 6.3 Aneuploidy Difficulty: Easy 31) When polyploid formation occurs without the separation of sister chromatids and the duplicated chromosomes pile up next to each other, forming a bundle of strands, the chromosomes are said to be: a) Polyploidy b) Polymitotic c) Polytene d) Aneuploidic e) Tetraploidic Answer: c Section: 6.3 Aneuploidy Difficulty: Easy 32) Which of the following organisms is best known for its polytene chromosomes? a) C. elegans b) Drosophila c) Human kidney cells d) Human liver cells e) None of these Answer: b Section: 6.3 Aneuploidy Difficulty: Easy

33) Which of the following best explains the formation of polytene chromosomes? a) Successive rounds of chromosome replication occur without intervening cell divisions b) Successive rounds of chromosome replication occur with multiple rounds of intervening cell divisions c) Few rounds of chromosome replication with multiple rounds of intervening cell divisions d) All of these e) None of these Answer: b Section: 6.3 Aneuploidy Difficulty: Easy 34) Which of the following is true regarding polytene chromosomes in Drosophila?

1. Homologous polytene chromosomes pair 2. All the centromeres of Drosophila polytene chromosomes congeal into a body called the chromocenter 3. Homologous chromosomes do not pair because they are in somatic cells a) 1 b) 2 c) 3 d) 1 and 2 e) 1 and 3 Answer: d Section: 6.3 Aneuploidy Difficulty: Medium 35) In which phase of the cell cycle is a polytene chromosome most likely to be found? a) Interphase b) Prophase c) Metaphase d) Anaphase e) Telophase Answer: a Section: 6.3 Aneuploidy Difficulty: Easy 36) A numerical change in part of the genome, usually resulting from the loss or gain of a single chromosome is often referred to as: a) Aneuploidy b) Polyploidy c) Triploidy d) Allopolyploidy e) Autopolyploidy Section: 6.3 Aneuploidy Difficulty: Easy

37) Which of the following individuals exhibits aneuploidy?

1. An individual who is missing a chromosome 2. An individual who has gained an extra chromosome 3. An individual who is missing a portion of a chromosome a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 6.3 Aneuploidy Difficulty: Easy 38) Cells with triplicate chromosomes are said to exhibit: a) Down syndrome b) Trisomy c) Autosomy d) Triploidy e) Polyploidy Answer: b Section: 6.3 Aneuploidy Difficulty: Easy 39) An organism in which one chromosome is underrepresented is referred to as a: a) Hypoploid b) Hyperploid c) Polyploid d) Autopolyploid e) Allopolyploid Answer: a Section: 6.3 Aneuploidy Difficulty: Easy 40) An organism in which one chromosome or chromosome segment is overrepresented is referred to as a(n): a) Hypoploid b) Hyperploid c) Polyploid d) Autopolyploid e) Allopolyploid Answer: b Section: 6.3 Aneuploidy Difficulty: Easy

41) Which of the following disorders is not caused by a trisomy in human beings? a) Down Syndrome b) Patau’s Syndrome c) Turner’s Syndrome d) Edwards Syndrome e) Triple-X Syndrome Answer: c Section: 6.3 Aneuploidy Difficulty: Easy 42) The most common cause for trisomy events in human beings is: a) Normal disjunction during meiosis b) Non-disjunction during meiosis c) Normal disjunction during oogenesis d) Normal disjunction during spermatogenesis e) None of these is a cause for trisomy Answer: b Section: 6.3 Aneuploidy Difficulty: Easy 43) When one chromosome is missing in an otherwise diploid individual, this is referred to as: a) Monosomy b) Trisomy c) Triploidy d) Tetraploidy e) Polyploidy Answer: a Section: 6.3 Aneuploidy Difficulty: Easy 44) A missing portion of a chromosome, such as the missing portion of the short arm of chromosome 5, in Cri-du-Chat Syndrome is known as: a) Deletion b) Duplication c) Hyperploidy d) Polyploidy e) Trisomy Answer: a Section: 6.4 Rearrangements of Chromosome Structure Difficulty: Easy

45) The presence of an extra chromosome segment, such as the X chromosome segment in Drosophila that codes for the Bar-eye mutation, is known as: a) Deletion b) Duplication c) Hypoploidy d) Polyploidy e) Trisomy Answer: b Section: 6.4 Rearrangements of Chromosome Structure Difficulty: Easy 46) Chromosomal inversions may result in: 1. The chromosome segment being detached, flipped 180 degrees, and reattached to the existing chromosome portion 2. The order of the inverted segments gene’s being reversed 3. The order of the inverted segments gene’s being replaced with new genes a) 1 b) 2 c) 3 d) 1 and 2 e) 1 and 3 Answer: d Section: 6.4 Rearrangements of Chromosome Structure Difficulty: Medium 47) An individual who possesses one inverted chromosome and one that is not inverted is said to be a(n): a) Inversion heterozygote b) Inversion mutant c) Inversion homozygote d) All of these e) None of these Answer: a Section: 6.4 Rearrangements of Chromosome Structure Difficulty: Easy 48) When a segment from one chromosome is detached and reattached to a non-homologous chromosome this is known as: a) Inversion b) Translocation c) Polyploidy d) Trisomy e) Monosomy Answer: b Section: 6.4 Rearrangements of Chromosome Structure Difficulty: Easy

49) Which of the following is a difference between compound chromosomes and translocations?

1. Compound chromosomes involve fusions of homologous chromosome segments 2. Compound chromosomes involve fusions of non-homologous chromosome segments 3. Translocations involve fusion of non-homologous chromosome segments a) 1 b) 2 c) 3 d) 1 and 2 e) 1 and 3 Answer: e Section: 6.4 Rearrangements of Chromosome Structure Difficulty: Medium

Question Type: Essay 50) Base upon the karyotype in this question, tell what sex the individual is, if the karyotype is normal or abnormal, and if the karyotype shows an abnormality, explain what the abnormality is and what it means to the individual.

Answer: The karyotype shows a female with Down Syndrome. This means that the individual has trisomy 21 and a total of 47 chromosomes. People with Down syndrome are typically short in stature and loose-jointed, particularly in the ankles; they have broad skulls, wide nostrils, large tongues with a distinctive furrowing, and stubby hands with a crease on the palm. Impaired mental abilities require that they be given special education and care. The life span of people with Down syndrome is much shorter than that of other people. Down syndrome individuals also almost invariably develop Alzheimer's disease, a form of dementia that is fairly common among the elderly. However, people with Down syndrome develop this disease in their fourth or fifth decade of life, much sooner than other people.

Section: 6.1 Cytological Techniques Difficulty: Hard

51) Briefly explain the cause of Down Syndrome and why the incidence rate increases with maternal age.

Answer: Trisomy 21 can be caused by chromosome nondisjunction in one of the meiotic cell divisions. The non-disjunction event can occur in either parent, but it seems to be more likely in females. In addition, the frequency of non-disjunction increases with maternal age. This increased risk is due to factors that adversely affect meiotic chromosome behavior as a woman ages. In human females, meiosis begins in the fetus, but it is not completed until after the egg is fertilized. During the long time prior to fertilization, the meiotic cells are arrested in the prophase of the first division. In this suspended state, the chromosomes may become unpaired. The longer the time in prophase, the greater the chance for unpairing and subsequent chromosome nondisjunction. Older females are therefore more likely than younger females to produce aneuploid eggs.

Section: 6.3 Aneuploidy Difficulty: Medium 52) Briefly explain how a pericentric inversion is different from a paracentric inversion and what effect each would have on a chromosome.

Answer: Pericentric inversions include the centromere, whereas paracentric inversions do not.

The consequence is that a pericentric inversion may change the relative lengths of the two arms of the chromosome, whereas a paracentric inversion has no such effect Section: 6.4 Rearrangements of Chromosome Structure Difficulty: Medium 53) Explain how chromosomes that have undergone reciprocal translocation most often pair during meiosis Answer: During meiosis, these translocated chromosomes would be expected to pair with their untranslocated homologues in a cruciform, or crosslike, pattern (Figure 6.23b). The two translocated chromosomes face each other opposite the center of the cross, and the two untranslocated chromosomes do likewise; to maximize pairing, the translocated and untranslocated chromosomes alternate with each other, forming the arms of the cross.

Section: 6.4 Rearrangements of Chromosome Structure Difficulty: Medium 54) Why does cruciform pairing often lead to non-disjunction events causing the formation of aneuploidy gametes?

Answer: Because cruciform pairing involves four centromeres, which may or may not be coordinately distributed to opposite poles in the first meiotic division, chromosome disjunction in translocation heterozygotes is a somewhat uncertain process, prone to produce aneuploid gametes. Altogether there are three possible disjunctional events, illustrated in Figure 6.24 categorized as either adjacent disjunction or alternate disjunction.

Section: 6.4 Rearrangements of Chromosome Structure Difficulty: Hard

Chapter 07 Test Bank Question Type: Multiple Choice 1) Which of the following most accurately represents how Sturtevant created the first chromosome map? a) He used a microscope to view the genes b) He used a set of microcalipers to measure the distance between the genes on the chromosome c) He used the data he retrieved from his experimental crosses to determine the distance between the genes on the chromosome d) All of these e) None of these Answer: c Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Easy 2) Which of the following is a true statement? a) Genes that are on the same chromosome should be inherited together b) Genes that are on the same chromosome proceed through meiosis together c) Genes that are on different chromosomes are not linked d) All of these are true e) None of these are true Answer: d Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Easy 3) The phenomenon in which genes on the same chromosome are separated from each other during meiosis and new combinations of genes are formed is known as: a) Linkage b) Synapsis c) Recombination d) Nondisjunction e) Disjunction Answer: c Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Easy 4) Recombination is commonly the result of a process known as__________ which occurs in Prophase I of Meiosis. a) Crossing over b) Linkage c) Disjunction d) Non-disjunction e) Formation of the mitotic spindles Answer: a Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Easy

5) The frequency that is often used to measure the degree of linkage between genes is known as the: a) Linkage frequency b) Recombination frequency c) Hardy-Weinberg frequency d) Genetic frequency e) Meiosis frequency Answer: b Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Easy 6) Which of the following statement is false regarding linked genes? a) The more closely the genes are linked the less often they recombine b) The recombination frequency for any set of genes often exceeds 50% c) Recombination occurs most frequently between genes that are not closely linked d) All of these are true statements regarding linked genes e) All of these are false statements regarding linked genes Answer: b Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Medium 7) A frequency of recombination that is less than 50% implies: a) The genes are linked on the same chromosome b) The genes are linked on different chromosomes c) The genes are not linked and are on different chromosomes d) The genes assort independently e) The genes are not linked and are on the same chromosome Answer: a Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Easy 8) Which of the following is a true statement about the linkage phase diagram R l / r L 1. The slash (/) separates alleles inherited from different parents 2. The alleles on the left and right of the slash indicate genotype on different homologous chromosomes 3. The genotype has the repulsion linkage phase a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Medium

9) During the crossing over process: a) Genetic information is physically exchanged b) Tetrads are formed c) Only two chromatids cross over at any one point d) All of these are true statements e) None of these are true statements Answer: d Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Easy 10) Typically, each crossover event produces how many recombinants? a) 1 b) 2 c) 3 d) 4 e) 5 Answer: b Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Easy 11) Why doesn’t an exchange between sister chromatids result in the production of a recombinant? a) Sister chromatids are genetically identical and therefore an exchange would not show any change in phenotype b) Sister chromatids are not genetically identical and therefore cannot exchange information c) Sister chromatids are genetically identical and therefore cannot exchange information d) Sister chromatids are not genetically identical and therefore an exchange would not show any change in phenotype e) None of these explains this Answer: a Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Medium 12) Which of the following performed experiments in maize that gave the evidence that recombination was caused by a physical exchange of genetic information? a) Sutton and Boveri b) McClintock and Creighton c) Watson and Crick d) Bateson and Boveri e) Mendel and Morgan Answer: b Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Easy

13) Which of the following is a false statement regarding chiasmata and the time for crossing over? a) The chiasmata are most visible and can be counted during late prophase I of meiosis b) The chiasmata are the points where the physical exchange of genetic information occurs c) Crossing over occurs in early to mid prophase I before chiasmata can be viewed d) Crossing over occurs in late prophase at the time chiasmata can be viewed e) The number of chiasmata is roughly proportional to the chromosome length Answer: d Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Medium 14) Linked genes on a chromosome can be mapped by: a) Determining how often their alleles recombine b) Determining how many individuals show the wild type phenotype c) Determining the physical structure of the chromosome d) Performing a karyotype e) None of these Answer: a Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Easy 15) Which of the following can be used to estimate the number of crossovers that occur during meiosis? a) Count chiasmata b) Count recombinant chromosomes c) Count centromeres d) Count chiasmata and count recombinant chromosomes e) Count recombinant chromosomes and count centromeres Answer: d Section: 7.1 Linkage, Recombination, and Crossing Over Difficulty: Medium 16) The distance between two points on the genetic map of a chromosome is equal to: a) The average number of crossovers between them b) The exact number of crossovers between them as measured in one cell c) The estimated number of crossovers between them as measured in one cell d) None of these e) All of these Answer: a Section: 7.2 Chromosome Mapping Difficulty: Easy

17) The genetic map distance between two loci is roughly equal to a) The estimated number of crossovers that occur between two points b) The recombination frequency written as a percentage c) The non-recombination frequency written as a percentage d) The number of parental phenoytpes seen in the progeny e) None of these Answer: b Section: 7.2 Chromosome Mapping Difficulty: Easy 18) Wild-type Drosophila females were mated to males homozygous for two autosomal mutations—vestigial (vg), which produces short wings, and black (b), which produces a black body. All the F1 flies had long wings and gray bodies; thus, the wild-type alleles (vg+ and b+) are dominant. The F1 females were then testcrossed to vestigial, black males, and the F2 progeny were sorted by phenotype and counted. Simple analysis indicates that, on average, 18 out of 100 chromosomes recovered from meiosis had a crossover between vg and b. Thus, vg and b are how far apart? a) 82 centimorgans b) 82 morgans c) 18 centimorgans d) 18 morgans e) 100 centimorgans Answer: c Section: 7.2 Chromosome Mapping Difficulty: Medium 19) P1 blue-flowered, short-stalked plants (++++) and white-flowered, long-stalked plant (ffss).

The resulting F1 offspring (++fs) are also crossed to produce the following F2 progeny: 400 blue, short 400 white, long 100 blue, long 100 white, short What is the recombination frequency, as shown in the F2 generation? a) 0.25 b) .20 c) .08 d) .10 e) .05 Answer: b Section: 7.2 Chromosome Mapping Difficulty: Hard

20) You sample a number of cells that have undergone meiosis looking for the production of recombinants between the genes ec, vg, and s. You find 10 cells with no crossovers between ec and s, 30 with 1 crossover between ec and s, 10 with 2 crossovers between ec and s, and zero with three or more crossovers. What is the map distance between the genes ec and s? a) 0.2 cM b) 0.6 cM c) 0.8 cM d) 1.0 cM e) 1.2 cM Answer: d Section: 7.2 Chromosome Mapping Difficulty: Hard 21) A mating between a true breeding red, long-stamen plant with a true breeding, white, shortstamen plant yields an F1. This F1 is self-fertilized to produce an F2 that has 300 red long-stamen plants and 100 white short-stamen plants. What can you conclude?

1. The genes are tightly linked 2. The genes are not tightly linked 3. There were no observable crossover events between the genes a) 1 b) 2 c) 3 d) 1 and 3 e) 2 and 3 Answer: d Section: 7.2 Chromosome Mapping Difficulty: Medium 22) A coefficient of coincidence that is equal to 1 implies: a) That the crossover events are independent b) That the crossover events are not independent c) That one crossover event interferes with another crossover event d) That one crossover event catalyzes another crossover event e) None of these Answer: a Section: 7.2 Chromosome Mapping Difficulty: Easy

23) In a data set to determine crossover frequency between the sc, ec, and cv genes in Drosophila, the crossover frequency between ec and sc = 0.15, and ec and cv = .25. The number of double crossover flies was 15 out of 500 flies. Assuming sc, ec, cv is the respective order of the genes, what can we conclude from this result? a) The coefficient of coincidence is near zero. b) The expected number of crossovers and the observed numbers are different. c) The genes are tightly linked. d) The coefficient of coincidence is near one. e) There is a large amount of interference between these genes.

Answer: d Section: 7.2 Chromosome Mapping Difficulty: Medium 24) Which of the following is a true statement regarding the use of recombination frequencies to estimate genetic distance? a) It is most accurate when the genes are far apart from one another b) It is not accurate when the genes are far apart from one another c) It is not accurate when the genes are close together d) All of these are true e) None of these is true Answer: b Section: 7.2 Chromosome Mapping Difficulty: Easy 25) An average of one chiasma during meiosis is equal to how many centiMorgans of genetic distance? a) 10 b) 20 c) 30 d) 40 e) 50 Answer: e Section: 7.2 Chromosome Mapping Difficulty: Easy 26) In a three point test cross there are two gb bt cq progeny and four + bt + progeny out of a hundred progeny, which are the two rarest groups. What can you conclude? a) The order of the genes is gb cq bt. b) The 6 progeny are the result of a double crossover. c) The genes are tightly linked. d) The order of the genes cannot be determined. e) The order of the genes is bt gb cq.

Answer: b Section: 7.2 Chromosome Mapping Difficulty: Medium

27) True genetic distance may be much greater than the observed recombination frequency.

Which is not an explanation of this phenomenon? a) The genes are far apart. b) Two crossover events may not produce recombinant chromosomes. c) Four crossover events may not produce recombinant chromosomes. d) Interference with 2 genes that are tightly linked affects the recombination frequency. e) A double crossover contributes to the average number of exchanges on a chromosome.

Answer: d Section: 7.2 Chromosome Mapping Difficulty: Medium 28) Mapping positions of genes that can be tied to locations on the cytological map of a chromosome is known as: a) Cytological analysis b) Cytogenetic mapping c) Recombination mapping d) Genetic fingerprinting e) All of these Answer: b Section: 7.3 Cytogenetic Mapping Difficulty: Easy 29) The basic principle in deletion mapping is that: a) A deletion that uncovers a recessive mutation must lack a wild-type copy of the mutant gene b) A deletion that covers a recessive mutation must lack a wild-type copy of the mutant gene c) A deletion that uncovers a recessive mutation must contain a wild type copy of the mutant gene d) A deletion that covers a recessive mutation must contain the recessive mutation of the gene e) All of these Answer: a Section: 7.3 Cytogenetic Mapping Difficulty: Easy 30) The basic principle in duplication mapping is that: a) A duplication that covers a recessive mutation must contain a wild-type copy of the mutant gene b) A duplication that uncovers a recessive mutation must contain a wild type copy of the mutant gene c) A duplication that covers a recessive mutation must contain the recessive mutation of the gene d) A duplication that uncovers a recessive mutation must lack a wild type copy of the mutant gene e) A duplication that covers the recessive mutation must lack a wild type copy of the mutant gene Answer: a Section: 7.3 Cytogenetic Mapping Difficulty: Easy

31) Which of the following is a true statement regarding the frequency of crossing over in chromosomes?

1. Crossing over is less likely to occur near the ends of a chromosome 2. Crossing over is less likely to occur near the centromere 3. Crossing over is more likely to occur near the ends of the chromosome a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: d Section: 7.3 Cytogenetic Mapping Difficulty: Easy 32) Which of the following is a true statement regarding genetic maps and cytological maps? a) The genetic distance and physical distance on cytological maps are not always uniform b) Genetic maps and cytological maps are colinear c) Genetic distances are not proportional to cytological distances d) All of these are true e) None of these are true Answer: d Section: 7.3 Cytogenetic Mapping Difficulty: Easy 33) In mapping a homozygous recessive mutant with a homozygous deletion in flies: a) The mutant phenotype will not appear in offspring if the allele lies within the deletion. b) The wild-type phenotype will always appear if the gene lies within the deletion. c) The mutant phenotype will always appear in progeny if the gene lies within the deletion. d) All of these e) None of these Answer: d Section: 7.3 Cytogenetic Mapping Difficulty: Medium 34) In order to predict inheritance patterns for human traits, it is necessary to collect data from: a) Karyotypes b) Pedigrees c) Amniocentesis d) Medical history e) All of these Answer: b Section: 7.3 Cytogenetic Mapping Difficulty: Medium

35) Some inversion chromosomes are called ____________ because they allow a mutant chromosome to be kept in the heterozygous condition over the inversion. a) Recombiners b) Balancers c) Unbalancers d) Shufflers e) None of these Answer: b Section: 7.3 Cytogenetic Mapping Difficulty: Medium

36) For the pedigree shown here, nail-patella syndrome was found to be linked to which blood type? a) A b) B c) O d) AB e) none of the above Answer: b Section: 7.4 Linkage Analysis in Humans Difficulty: Medium 37) What is the percent recombination in the offspring of individuals II-1 and II-2? a) 0/10 b) 2/10 c) 3/10 d) 4/10 e) 7/10 Answer: c Section: 7.4 Linkage Analysis in Humans Difficulty: Medium

38) Which of the following is a true statement regarding inversions and recombination in chromosomes? a) Inversion typically suppresses recombination b) The inhibition of crossing over that occurs near the breakpoints of the inversion is compounded by the selective loss of chromosomes that have undergone crossing over within the inverted region. c) Crossing over is usually inhibited near the breakpoints of a rearrangement in heterozygous condition d) All of these are true e) None of these are true Answer: d Section: 7.5 Recombination and Evolution Difficulty: Medium 39) Which of the following is true regarding recombination and evolution?

1. Recombination can bring favorable mutations together 2. Chromosome rearrangements can suppress recombination 3. Recombination is under genetic control a) 1 b) 2 c) 3 d) Both 1 and 2 e) All of these Answer: e Section: 7.5 Recombination and Evolution Difficulty: Easy 40) An individual with two a+b+c+d+e+ chromosomes is crossed with an individual with two a-dc-b-e- chromosomes. What will happen in the F1 during meiosis? a) Acentric fragments cannot be formed. b) Acentric fragments cannot be formed. c) The F1 will form many viable gametes. d) Recombination will be suppressed between the sister chromatids. e) All of these Answer: d Section: 7.5 Recombination and Evolution Difficulty: Medium Question Type: Essay 41) 150 out of 1000 chromosomes are recombinant between two genes. How far apart are they located on a genetic map?

Answer: 15 map units or centiMorgans Section: 7.2 Chromosome Mapping Difficulty: Medium

42) How would one estimate the genetic map distance between two genes on the same chromosome using the number of crossing over events in a three-point testcross?

Answer: In a three factor cross, the total number of noncrossover, single, and double crossover chromosomes is determined and the frequency of single and double crossover chromosomes is added and then converted to a percentage which then provides the map distance in cM between the genes Section: 7.2 Chromosome Mapping Difficulty: Medium 43) In maize, three recessive genes (z, xt, and cm) are linked on chromosome 3. A homozygous plant for the recessives is crossed with a wild-type plant. The F1 is crossed to get an F2 generation with the following results: z xt cm : 44 z + cm : 460 + xt cm : 39 z xt +: 164 + xt +: 474 z + +: 35 + + cm: 158 + + +: 40 What is the order of the three genes on chromosome 3?

Answer: z cm xt Section: 7.2 Chromosome Mapping Difficulty: Hard 44) In maize, three recessive genes (z, xt, and cm) are linked on chromosome 3. A homozygous plant for the recessives is crossed with a wild-type plant. The F1 is crossed to get an F2 generation with the following results: z xt cm : 44 z + cm : 460 + xt cm : 39 z xt +: 164 + xt +: 474 z + +: 35 + + cm: 158 + + +: 40 What is the map distance between z and cm ?

Answer: 28 map units or centiMorgans Section: 7.2 Chromosome Mapping Difficulty: Hard 45) In maize, three recessive genes (z, xt, and cm) are linked on chromosome 3. A homozygous plant for the recessives is crossed with a wild-type plant. The F1 is crossed to get an F2 generation with the following results: z xt cm : 44 z + cm : 460 + xt cm : 39 z xt +: 164 + xt +: 474 z + +: 35 + + cm: 158 + + +: 40 Answer: 11 map units or centiMorgans Section: 7.2 Chromosome Mapping Difficulty: Hard

Chapter 08 Test Bank Question Type: Multiple Choice 1) Which of the following is an advantage of using bacteria for research over an organism like Drosophila?

1. Bacteria are small and reproduce quickly 2. Bacteria can grow on artificial biochemically defined media 3. Bacteria have a simple structure and physiology a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: e Section: 8.1 Viruses and Bacteria in Genetics Difficulty: Easy 2) Which of the following statements is false with regards to bacteria as a model organism for genetic study? a) Genetic variability is difficult to detect. b) Bacteria are ideal for studying fundamental biological processes because they are simple in structure. c) Bacteria can grow on artificial media that can be altered by the researcher. d) Bacteria can reproduce more quickly than Drosophila e) All of these are false Answer: a Section: 8.1 Viruses and Bacteria in Genetics Difficulty: Easy 3) A virus that infects a bacterial cell is known as a: a) Bacteriolysin b) Bacteriophage c) Bacteria virus d) Baculovirus e) None of these Answer: b Section: 8.2 The Genetics of Viruses Difficulty: Easy

4) Which of the following statements is true with regard to viruses?

1. Viruses can only reproduce by infecting living cells 2. Viruses can be considered both living and non-living depending on the state they are in 3. Viruses that infect bacterial cells are known as bacteriophages a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 8.2 The Genetics of Viruses Difficulty: Easy 5) Two of the most important bacteriophages that have played a role in the study of genetics are: a) E. coli and T4 b) T4 and TMV c) T4 and Lambda d) Lambda and E. coli e) Lambda and TMV Answer: c Section: 8.2 The Genetics of Viruses Difficulty: Easy 6) Which of the following is a true statement about the phage T4?

1. It is a temperate phage 2. It is a virulent phage 3. It can infect the bacterium E. coli a) 1 b) 2 c) 3 d) 1 and 3 e) 2 and 3 Answer: e Section: 8.2 The Genetics of Viruses Difficulty: Medium

7) Which of the following is a true statement about the phage Lambda?

1. It is a temperate phage 2. It is a virulent phage 3. It can infect the bacterium E. coli a) 1 b) 2 c) 3 d) 1 and 3 e) 2 and 3 Answer: d Section: 8.2 The Genetics of Viruses Difficulty: Medium 8) Temperate and virulent phages differ in that: a) Virulent phages kill the host cell and temperate phages never kill the host cell b) Virulent phages never kill the host cell and temperate phages always kill the host cell c) Virulent phages kill the host cell and temperate phages can kill the host cell or live in a symbiotic state with the host cell d) Temperate phages kill the host cell and virulent phages can kill the host cell or live in a symbiotic state with the host cell e) All of these are true Answer: c Section: 8.2 The Genetics of Viruses Difficulty: Easy 9) The biological purpose of T4 lysozyme during phage propagation is to: a) Open the phage head to release viral DNA into the host cell. b) Disrupt the nuclear membrane of the host cell. c) Lyse host bacteria to release newly formed phage. d) Assemble the head and tail fibers of new phages. e) Shut off transcription and translation of bacterial genes in the cell.

Answer: c Section: 8.2 The Genetics of Viruses Difficulty: Easy 10) T4 contains a modified DNA base. Which base is modified and for what purpose? a) Cytosine; to protect phage DNA from phage encoded DNAse b) Thymidine; to protect phage DNA from phage encoded DNAse c) Cytosine; to increase transcription from phage promoters d) Thymidine; to increase transcription from phage promoters e) Adenine; to help rupture the bacterial plasma membrane Answer: a Section: 8.2 The Genetics of Viruses Difficulty: Easy

11) The modified nitrogenous base found in the T4 phage DNA also contains an attached glucose derived residue that is used for what purpose? a) To increase transcription from phage promoters b) To protect phage from phage encoded nucleases c) To protect phage from host encoded restriction enzymes. d) To indicate a site for head-full packaging e) All of the above Answer: b Section: 8.2 The Genetics of Viruses Difficulty: Easy 12) Which of the following is a true statement regarding the bacteriophage lambda?

1. It is smaller than T4 2. It can undergo a lytic or lysogenic life cycle 3. It is comprised of a double stranded linear piece of DNA encased in a protein coat a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: e Section: 8.2 The Genetics of Viruses Difficulty: Medium 13) In a lysogenic pathway, the viral DNA is: 1. Inserted into the host chromosome 2. Replicated with the host chromosome 3. Integrated by site specific recombination a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 8.2 The Genetics of Viruses Difficulty: Medium 14) When bacteriophage lambda integrates into the bacterial chromosome, it does so at: a) An att P site in the host chromosome. b) An att B site in the phage chromosome. c) An att B site in the host chromosome using a lambda-encoded integrase d) An att B site in the host chromosome using a bacteria-encoded integrase. e) An att B site in both the host and phage chromosomes Answer: c Section: 8.2 The Genetics of Viruses Difficulty: Medium

15) In its integrated state, the lambda chromosome is known as a: a) Lyophilized phage b) Prophage c) Metaphage d) Lytic phage e) None of these Answer: b Section: 8.2 The Genetics of Viruses Difficulty: Easy 16) The lambda phage can cause lysis in a bacterium when: 1. The lamda DNA is spontaneously excised from the host chromosome 2. The lambda DNA is induced to excise from the host chromosome 3. The lambda DNA excises from the host chromosome and moves back into the phage a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: d Section: 8.2 The Genetics of Viruses Difficulty: Medium 17) Which of the following is required for the lambda phage to cause lysis in a bacterial cell? a) Precise excision of phage DNA from the host chromosome b) Lambda integrase c) Lambda excisase d) All of these e) None of these Answer: d Section: 8.2 The Genetics of Viruses Difficulty: Easy 18) Which of the following is required site-specific recombination between the circular lambda DNA and the circular E. coli chromosome? a) Precise excision of phage DNA from the host chromosome b) Lambda integrase c) Lambda excisase d) All of these e) None of these Answer: b Section: 8.2 The Genetics of Viruses Difficulty: Easy

19) Gene transfer in bacteria is: a) Unidirectional – from donor cells to recipient cells b) Unidirectional – from recipient cells to donor cells c) Bidirectional d) Mutidirectional e) None of these Answer: a Section: 8.3 The Genetics of Bacteria Difficulty: Easy 20) What would you expect to observe if three crossovers occur in a bacterial chromosome? a) The segment will be unable to insert into the bacterial chromosome. b) The fragment to be inserted will be degraded. c) The fragment will insert, producing an unstable linear chromosome that will be degraded. d) The fragment will insert and become integrated into the E. coli chromosome. e) Both a and b.

Answer: c Section: 8.3 The Genetics of Bacteria Difficulty: Medium 20) Wild type bacteria that can synthesize all necessary metabolites are known as: a) Prototrophs b) Auxotrophs c) Phototrophs d) Heterotrophs e) Chemoheterotrophs Answer: a Section: 8.3 The Genetics of Bacteria Difficulty: Easy 22) Bacteria that cannot synthesize all the necessary amino acids are known as: a) Prototrophs b) Auxotrophs c) Phototrophs d) Heterotrophs e) Chemotrophs Answer: b Section: 8.3 The Genetics of Bacteria Difficulty: Easy

23) Which of the following can be mutated in bacteria?

1. Colony morphology 2. Cell morphology 3. Ability to use an energy source a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy 24) A mutant of E. coli is designated Strr and arg-. Which of the following best describes this genotype? a) antibiotic resistant, prototrophic b) antibiotic resistant, auxotrophic c) antibiotic sensitive, prototrophic d) antibiotic sensitive, auxotrophic e) antibiotic resistant, carbon-source mutant Answer: b Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Medium 25) Plasmids are: 1. Autonomously replicating 2. Circular 3. Extrachromosomal DNA a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: e Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy 26) Which of the following is true regarding bacteria? a) Bacteria reproduce by simple fission b) Bacteria are monoploid and multinucleate c) Bacteria do not undergo mitosis d) Bacteria do not undergo meiosis e) All of these are true Answer: e Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Medium

27) With rare exceptions, after gene transfer occurs, the recipient cells become: a) Full diploids b) Partial diploids c) Full haploids d) Partial haploids e) Triploids Answer: b Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy 28) Recombination between bacteria occurs between: a) The main chromosome of the donor and a fragment of the recipient. b) The main chromosome of the recipient and a fragment of the donor. c) The main chromosome of the donor and the main chromosome of the recipient. d) The main chromosome of the recipient and the main chromosome of the donor. e) None of the above Answer: b Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy 29) Which process relies on phage as a vehicle of DNA transport? a) Conjugation b) Transformation c) Transduction d) Conjugation and Transformation e) Transformation and Transduction Answer: c Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy 30) Which process requires cell-to-cell contact for DNA transfer? a) Conjugation b) Transformation c) Transduction d) Conjugation and Transformation e) Transformation and Transduction Answer: a Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy

31) While conducting research you notice that when DNAse is added to the medium, genetic exchange no longer occurs in your bacteria. What can you conclude about the method of genetic exchange that was occurring? a) Transduction was taking place before the addition of DNAse b) Transformation was taking place before the addition of DNAse c) Conjugation was taking place before the addition of DNAse d) Binary Fission was taking place before the addition of DNAse e) There is not enough information provided to determine what method of genetic exchange was occurring prior to the addition of DNAse Answer: b Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Medium 32) Cells that are capable of DNA uptake are known as: a) Capable b) Competent c) Conjugative d) Universal recipient e) None of these Answer: b Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy 33) During B. subtilis transformation, which of the following statements is incorrect? a) Large, double-stranded DNA fragments are bound to the cell surface b) Bound DNA is attached to specific receptor sites c) DNA moves across the membrane in an energy independent manner d) One strand of the double-stranded DNA is hydrolyzed during uptake by a bound DNAse e) One strand of DNA is protected from degradation by a coating of single stranded binding proteins and the RecA protein Answer: c Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy 34) A recombinant double helix that has one allele on one strand and the other allele on the second strand is known as a: a) Heteroduplex b) Homoduplex c) Heterotriplex d) Homotriplex e) Diploid Answer: a Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy

35) In F factor mating, which of the following statements is incorrect? a) F+ act as donors. b) Genes are carried on the F factor encoding for cell-cell contact and the formation of sex pili. c) Recombination can occur between IS in the F factor and chromosome to generate F' cells. d) F factor is regarded as an episome. e) F factor is transferred as a single strand.

Answer: c Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Medium 36) Why is the F factor considered to be an episome?

1. The F factor can exist in an autonomous or integrated state 2. The F factor can not be fully transferred between bacteria 3. The F factor does not exist in an autonomous state a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: a Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Medium 37) A cell that carries an integrated F factor is best known as a/an: a) Episome b) Hfr cell c) F' cell d) F+ cell e) None of the above Answer: b Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy 38) The ability of episomes to insert themselves into chromosomes depends on the presence of short DNA sequences called a) RecA b) attP c) ComEA d) insertion sequences (or IS elements) e) OriC Answer: d Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy

39) Hfr mapping a) is carried out using interrupted mating experiments. b) maps gene distance as a unit of base pairs rather than a crossover frequency. c) Requires F+ to integrate into a multiple positions to map chromosomal genes. d) Only maps genes that are more than 4 minutes apart. e) None of the above Answer: a Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy 40) Which form of replication mediates the transfer of chromosomes from Hfr cells to F- cells? a) Rolling circle replication b) Bidirectional replication c) Semi-conservative replication d) Conservative replication e) Dispersive replication Answer: a Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy 41) Which of the following statements is not correct about specialized transduction? a) The phage chromosome carries only specific portions of the bacterial genome. b) It requires a temperate phage for DNA transfer because it is capable of integrating into the host chromosome. c) It requires precise excision of the integrated prophage to transfer chromosomal markers. d) Its particles carry both bacterial and viral DNA in the same phage head. e) All of these are correct Answer: c Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Medium 42) A stable transductant is formed when a) There is a double crossover between a specialized transducing phage and the bacterial chromosome, resulting in an exchange of the shared bacterial marker. b) A phage integrates into the chromosome. c) A phage integrates into the chromosome and cannot excise due the lack of helper phage. d) Improper excision removes genes involved with lytic growth. e) None of these Answer: a Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Medium

43) A Hfr strain is produced by: a) The excision of a lambda phage b) The integration of the F factor into a bacterial chromosome c) The integration of lambda phage DNA into a bacterial chromosome d) All of these e) None of these Answer: b Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy 44) Modified F factors that also carry extra bacterial genes are known as: a) F + b) Hfr c) Fd) F' e) None of the above Answer: d Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy 45) Transfer of F' factors into recipient cells is a process known as: a) Conjugation b) Sexjugation c) Transduction d) Sexduction e) Transformation Answer: d Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Easy Question Type: Essay 46) Compare the procedure for mapping bacteriophage genes to mapping the genes in an organism such as Drosophila.

Answer: Because viruses have a single chromosome that does not go through meiosis, the mapping procedure is somewhat different from that used for an organism like Drosophila.

Crosses are performed by simultaneously infecting host bacteria with two different types of phage and then screening the progeny phage for recombinant genotypes. Map distances, in centiMorgans, are then calculated as the average number of crossovers that have occurred between genetic markers. For short distances, map distances are approximately equal to the percentage of recombinant chromosomes among the progeny.

Section: 8.2 The Genetics of Viruses Difficulty: Medium

47) Briefly compare the three parasexual processes of genetic exchange that occurs in bacteria.

Answer: Transformation involves the uptake of free DNA molecules released from one bacterium (the donor cell) by another bacterium (the recipient cell). It does not require cell to cell contact and is sensitive to the addition of DNAse. Conjugation involves the direct transfer of DNA from a donor cell to a recipient cell, via a plasmid. It does require cell to cell contact and is not sensitive to the addition of DNAse. Transduction occurs when bacterial genes are carried from a donor cell to a recipient cell by a bacteriophage. It does not require cell to cell contact and is not sensitive to the addition of DNAse. Section: 8.3 The Genetics of Bacteria Difficulty: Medium 48) Explain how transformation takes place in a bacterium such as B. subtilis.

Answer: ComEA and ComG proteins bind double-stranded DNA to the surfaces of competent cells. As the bound DNA is pulled into the cell by the ComFA DNA translocase (an enzyme that moves or "translocates" DNA), one strand of DNA is degraded by a deoxyribonuclease (an enzyme that degrades DNA), and the other strand is protected from degradation by a coating of single-stranded DNA-binding protein and RecA protein (a protein required for recombination).

With the aid of RecA and other proteins that mediate recombination, the single strand of transforming DNA invades the chromosome of the recipient cell, pairing with the complementary strand of DNA and replacing the equivalent strand. The replaced recipient strand is then degraded. If the donor and recipient cells carry different alleles of a gene, the resulting recombinant double helix will have one allele in one strand and the other allele in the second strand.

Section: 8.3 The Genetics of Bacteria Difficulty: Medium 49) Four Hfr strains of E. coli transferred a series of genes in the order of left to right as follows Hfr1: BAKJI Hfr3: ABC Hfr2: EFGHI Hfr4: FEDCB Construct a single genetic map, indicating the relative site of the F factor insertion, and the direction of transfer.

Answer: > < > < |A-B-|-C-D-E-|-F-|-G-H-I-J-K Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Medium

50) Using two strains of T4, with genotypes xyz and x+y+z+, the results of recombination are shown below. Are x, y, and z linked and what is the order of the genes?

Class Frequency xyz 0.30 x+y+z+ 0.31 x y+z+ 0.05 x+y z 0.05 x y z+ 0.12 x+y+z 0.14 x y+z 0.02 Answer: They are linked and the order places y in the middle (xyz or zyx) Section: 8.4 Mechanism of Genetic Exchange in Bacteria Difficulty: Medium

Chapter 09 Test Bank Question Type: Multiple Choice 1) Why were some geneticists reluctant to accept the idea of nucleic acids as genetic material even after the discovery of the structure of DNA in 1953? a) Proteins were larger than nucleic acids b) Nucleic acids exhibit less structural variability than proteins c) It was proven in 1871 that proteins were the source of genetic material d) All of these e) None of these Answer: b Section: 9.1 Proof That Genetic Information is Stored in DNA Difficulty: Easy 2) Which large organic molecules are essential chromosome components? a) Lipids and proteins b) Proteins and nucleic acids c) Nucleic acids and polysaccharides d) Proteins and polysaccharides Answer: b Section: 9.1 Proof That Genetic Information is Stored in DNA Difficulty: Easy 3) Before direct proof of DNA being the genetic material, indirect evidence suggesting DNA as the appropriate macromolecule, included a) Gametes have 1/2 the amount of DNA as somatic cells b) Molecular composition of DNA is the same in different cells of an organism, whereas RNA and protein contents vary c) A correlation between DNA content and the number of chromosome sets per cell d) DNA is more stable than RNA or protein e) All of these Answer: e Section: 9.1 Proof That Genetic Information is Stored in DNA Difficulty: Medium 4) In Sia and Dawson's 1931 experiment, a) Mice were required to demonstrate the transforming principle b) Used serum to precipitate IIIS cells from a mixture of heat-killed IIIS and living IIR cells c) They showed that mice play no direct role in the transforming principle d) Heat-killed IIR cells mixed with living IIIS cells gave rise to IIR colonies e) None of these Answer: c Section: 9.1 Proof That Genetic Information is Stored in DNA Difficulty: Easy

5) After Griffiths 1928 experiments leading to the discovery of the transforming principle, subsequent experimentation demonstrated that: a) The transforming principle could not be carried out in vitro b) The conversion of Type IIIR to Type IIIS was due to mutation c) The process of heat-killing cells was not entirely effective. d) DNAse treatment eliminates all transforming activity e) RNAse treatment eliminates all transforming activity Answer: d Section: 9.1 Proof That Genetic Information is Stored in DNA Difficulty: Easy 6) The first direct evidence indicating that DNA, rather than RNA or protein, is the genetic material in bacteria was gathered by: a) Griffith b) Avery, McLeod, and McCarty c) Hershey and Chase d) Watson and Crick e) Sia and Dawson Answer: b Section: 9.1 Proof That Genetic Information is Stored in DNA Difficulty: Easy 7) Whose experiments provided direct evidence indicating that DNA, rather than RNA or protein, is the genetic material in bacteriophages? a) Griffith b) Avery, McLeod, and McCarty c) Hershey and Chase d) Watson and Crick e) Sia and Dawson Answer: c Section: 9.1 Proof That Genetic Information is Stored in DNA Difficulty: Easy 8) Which of the following materials was used by Avery, McLeod, and McCarty in their experimentation to identify the transforming agent? a) DNAse b) RNAse c) Protease d) RNAse and Protease e) All of these Answer: e Section: 9.1 Proof That Genetic Information is Stored in DNA Difficulty: Easy

9) Hershey and Chase used radiolabeled T2 phages to identify which component of the phage entered E. coli during infection. Which of the following correctly pairs the radiolabel with the appropriate phage component? a) 32P DNA, 35S Protein b) 32P Protein, 35S RNA c) 32P RNA, 35S DNA d) 32P Protein, 35S DNA e) 32P Protein, 35S Protein Answer: a Section: 9.1 Proof That Genetic Information is Stored in DNA Difficulty: Medium 10) The work carried out by Hershey and Chase had one flaw: a high degree of 35S was found inside the cells. How was this flaw later addressed? a) Protease treatment b) Shearing for longer periods of time c) Transfection of protoplasts with pure phage DNA d) Using alternative isotopes. e) This flaw was never addressed Answer: c Section: 9.1 Proof That Genetic Information is Stored in DNA Difficulty: Easy 11) Tobacco mosaic virus (TMV) is composed of a) protein b) RNA c) DNA d) Both a and b e) All of the above Answer: d Section: 9.1 Proof That Genetic Information is Stored in DNA Difficulty: Easy 12) RNA was first identified as the genetic material of the tobacco mosaic virus (TMV) through an experiment in which: a) Protein and nucleic acid were radioactively labeled b) Protein, RNA, and DNA were digested using various enzymes c) The protein coats and RNA from two distinctive TMV strains were exchanged d) RNA underwent CsCl density-dependent centrifugation e) None of these Answer: c Section: 9.1 Proof That Genetic Information is Stored in DNA Difficulty: Easy

13) Which of the following is true regarding the structure of DNA? a) Double stranded molecule b) Single stranded molecule c) Composed of nucleotides d) Double stranded molecule and Composed of nucleotides e) Single stranded molecule and Composed of nucleotides Answer: d Section: 9.2 The Structures of DNA and RNA Difficulty: Easy 14) Which of the following is true regarding the structure of RNA? a) RNA is a double stranded molecule. b) RNA is a single stranded molecule. c) RNA is composed of nucleotides. d) Both a and c. e) Both b and c.

Answer: e Section: 9.2 The Structures of DNA and RNA Difficulty: Easy 15) Which of the following statements concerning DNA structure is incorrect? a) The two strands of a double helix are complementary and anti-parallel b) he structure was first predicted by Watson and Crick in 1953 to be arranged as a double-stranded helix. c) The final structure disproved the work leading to Chargaff's rule. d) X-ray diffraction patterns indicated a double-stranded structure with repeating substructure every 0.34 nm e) None of these Answer: c Section: 9.2 The Structures of DNA and RNA Difficulty: Medium 16) Which of the following is not accurate according to Chargaff's rules? a) A = T b) C = G c) A+T = C+ G d) A+G = C + T e) Purines = Pyrimidines Answer: d Section: 9.2 The Structures of DNA and RNA Difficulty: Easy

17) Which type of chemical bond attaches deoxyribose molecules together to form the backbone of a DNA molecule? a) Hydrogen b) Phosphodiester c) Peptide d) Ionic e) Covalent Answer: b Section: 9.2 The Structures of DNA and RNA Difficulty: Easy 18) Which type of chemical bond connects nitrogenous bases to each other in the center of a DNA molecule? a) Hydrogen b) Phosphodiester c) Peptide d) Ionic e) Covalent Answer: a Section: 9.2 The Structures of DNA and RNA Difficulty: Easy 19) Which of the following statements concerning the various forms of DNA is incorrect? a) B-DNA is the conformation under physiological conditions. b) A-DNA occurs in a partially dehydrated environment. c) DNA-RNA duplexes exist in an A-form. d) B and Z are right handed helices, while A is left handed. e) All of these are incorrect Answer: d Section: 9.2 The Structures of DNA and RNA Difficulty: Medium 20) A single strand of a double-stranded helix has the sequence 5'-GTTTAACGTAT-3'. If the helix was in B-form, how many turns (360°) would the helix make? a) 1 b) 1.1 c) 1.5 d) 2 e) 2.1 Answer: b Section: 9.2 The Structures of DNA and RNA Difficulty: Easy

21) DNA extracted from a bacterium was identified as having a guanine content of 14%. What is the percentage of C? a) 14% b) 28% c) 36% d) 78% e) There is not enough information to determine the percentage of C. Answer: a Section: 9.2 The Structures of DNA and RNA Difficulty: Easy 22) DNA extracted from a cheek cell has an adenine content of 23% A. What is the percentage of C? a) 23% b) 27% c) 54% d) 77% e) There is not enough information to determine the percentage of C. Answer: b Section: 9.2 The Structures of DNA and RNA Difficulty: Medium 23) If you measure 23% U, what is the percentage of C in a given RNA molecule? a) 23% b) 27% c) 54% d) 77% e) There is not enough information to determine the percentage of C. Answer: e Section: 9.2 The Structures of DNA and RNA Difficulty: Medium 24) What kind of data on DNA structure was gathered by Rosalind Franklin and Maurice Wilkins? a) 3D model b) Base composition data c) X-ray diffraction pattern data d) All of these e) None of these Answer: c Section: 9.2 The Structures of DNA and RNA Difficulty: Easy

25) In aqueous solutions with low concentrations of salt, the DNA takes on the ______conformation. a) A-DNA b) B-DNA c) Z-DNA d) A-DNA or B-DNA e) all of the above Answer: b Section: 9.2 The Structures of DNA and RNA Difficulty: Easy 26) Suppose you have a tube containing DNA, but you have accidentally resuspended your DNA in a buffer with an extremely high salt concentration. The DNA takes on the ______conformation. a) A-DNA b) B-DNA c) Z-DNA d) A-DNA or B-DNA e) all of the above Answer: a Section: 9.2 The Structures of DNA and RNA Difficulty: Easy 27) You are doing X-ray diffraction analysis, and you measure DNA that has 12 base pairs per turn and a diameter of 1.8 nm. This DNA is in the ______conformation. a) A-DNA b) B-DNA c) Z-DNA d) A-DNA or B-DNA e) all of the above Answer: b Section: 9.2 The Structures of DNA and RNA Difficulty: Easy 28) In a certain DNA molecule one strand is cleaved and when the complementary strand is rotated at one end. This configuration is called: a) Relaxation b) Supercoiling c) Splicing d) All of these e) None of these Answer: b Section: 9.2 The Structures of DNA and RNA Difficulty: Easy

29) Underwound DNA is said to exhibit: a) Positive supercoiling b) Negative supercoiling c) Positive relaxation d) Negative relaxation e) None of these Answer: b Section: 9.2 The Structures of DNA and RNA Difficulty: Easy 30) Functional DNA molecules in cells typically exhibit: a) Positive supercoiling b) Negative supercoiling c) Positive relaxation d) Negative relaxation e) None of these Answer: b Section: 9.2 The Structures of DNA and RNA Difficulty: Easy 31) Prokaryotic chromosome structure: a) Is in a loosely condensed state in vivo called the "folded genome" b) Has 50 to 100 domains or loops, each of which is independently negatively supercoiled c) Consists only of DNA and protein d) Is unaffected by RNAse treatment e) Is unorganized as a "naked molecule of DNA" Answer: b Section: 9.3 Chromosome Structure in Viruses and Prokaryotes Difficulty: Easy 32) You have isolated E. coli DNA and have treated it with DNase. What effect will this have on the chromosome? a) It will introduce single-stranded nicks in the DNA. b) It will partially unfold the genome by removing anchors at the loops. c) It will reduce supercoiling. d) Both a and b. d) All of the above.

Answer: d Section: 9.3 Chromosome Structure in Viruses and Prokaryotes Difficulty: Medium

33) You have isolated E. coli DNA and have treated it with RNase. What effect will this have on the chromosome? a) It will introduce single-stranded nicks in the DNA. b) It will partially unfold the genome by removing anchors at the loops. c) It will reduce supercoiling. d) Both a and b. d) All of the above.

Answer: b Section: 9.3 Chromosome Structure in Viruses and Prokaryotes Difficulty: Medium 34) A structure known as _________________ is the functional state of a bacterial chromosome. a) Folded genome b) Haploid genome c) Diploid genome d) Circular genome e) Linear genome Answer: a Section: 9.3 Chromosome Structure in Viruses and Prokaryotes Difficulty: Easy 35) Which class of proteins is associated with DNA in chromatin? a) Histones b) Non-histone chromosomal proteins c) Enzymes d) Histones and non-histone chromosomal proteins e) All of these Answer: d Section: 9.4 Chromosome Structure in Eukaryotes Difficulty: Easy 36) Which amino acids are responsible for the basic charge of histones? a) Glycine and Valine b) Arginine and Lysine c) Leucine and Isoleucine d) Phenylalanine and Tyrosine e) Proline and Methionine Answer: b Section: 9.4 Chromosome Structure in Eukaryotes Difficulty: Easy

37) In the DNA found in sperm, ________ replace histones in DNA packaging. a) nonhistone chromosomal proteins b) solenoids c) enzymes d) protamines e) None of the above Answer: b Section: 9.4 Chromosome Structure in Eukaryotes Difficulty: Easy 38) The nucleosome core is comprised of which histone(s)? a) H1 b) H1 and H2a c) H1, H2a, H3 d) H2a, H3, and H4 e) H1, H2a, H3, and H4 Answer: d Section: 9.4 Chromosome Structure in Eukaryotes Difficulty: Easy 39) Which histone(s) is/are involved in condensing the 11-nm nucleosome fiber into the 30-nm chromatin fiber? a) H1 b) H1 and H2a c) H1, H2a, H3 d) H2a, H3, and H4 e) H1, H2a, H3, and H4 Answer: a Section: 9.4 Chromosome Structure in Eukaryotes Difficulty: Easy 40) Which proteins are involved in condensing the 30-nm chromatin fiber into the tightly packed metaphase chromosomes? a) Histones H1 and H2a b) Nonhistone chromosomal proteins c) Linkers d) Histones H2a, H3, and H4 e) Telomeres Answer: b Section: 9.4 Chromosome Structure in Eukaryotes Difficulty: Easy

41) Three levels of condensation are required to package metaphase chromosomes. In order of complexity, lowest to highest, they are: a) Supercoiling into nucleosomes, chromatin fiber condensation, scaffold formation b) Chromatin fiber condensation, supercoiling into nucleosomes, scaffold formation c) Scaffold formation, supercoiling into nucleosomes, chromatin fiber condensation d) Supercoiling into nucleosomes, scaffold formation, chromatin fiber condensation e) Scaffold formation, chromatin fiber condensation, supercoiling into nucleosomes Answer: a Section: 9.4 Chromosome Structure in Eukaryotes Difficulty: Medium 42) The functions associated with telomeres are to: a) Prevent ribonucleases from degrading the ends of linear RNA primer molecules b) Allow the fusion of broken chromosomal ends c) Facilitate replication of chromosomes without the loss of termini d) Ensure the appropriate segregation of chromosomes e) Provide chromosomal anchorage to spindle-fibers Answer: c Section: 9.5 Special Features of Eukaryotic Chromosomes Difficulty: Easy 43) Satellite DNA is a) Repetitive b) Randomly located throughout the genome c) Prokaryotic d) Expressed e) None of these Answer: a Section: 9.5 Special Features of Eukaryotic Chromosomes Difficulty: Easy 44) Which of the following characteristics or functions is not believed to be exhibited by highly repetitive DNA? a) Structural or organizational roles for the chromosome. b) Transposable elements c) Heterochromatin d) Influencing regions of chromosome pairing e) Junk Answer: b Section: 9.5 Special Features of Eukaryotic Chromosomes Difficulty: Easy

45) Which protein variant of histone H3 is involved in packaging of the heterochromatin near the centromere of eukaryotic DNA? a) Shelterin b) CENP-A c) Linkers d) POT-1 e) TRF1 Answer: b Section: 9.4 Chromosome Structure in Eukaryotes Difficulty: Easy Question Type: Essay 46) Compare the structure of DNA to the structure of RNA. Include at least three differences.

Answer: In DNA, the sugar is 2-deoxyribose (thus the name deoxyribonucleic acid); in RNA, the sugar is ribose (thus ribonucleic acid). Four different bases commonly are found in DNA: adenine (A), guanine (G), thymine (T), and cytosine (C). RNA also usually contains adenine, guanine, and cytosine but has a different base, uracil (U), in place of thymine. Adenine and guanine are double-ring bases called purines; cytosine, thymine, and uracil are single-ring bases called pyrimidines. Both DNA and RNA, therefore, contain four different subunits, or nucleotides: two purine nucleotides and two pyrimidine nucleotides (Figure 9.6). In polynucleotides such as DNA and RNA, these subunits are joined together in long chains (Figure 9.7). RNA usually exists as a single-stranded polymer that is composed of a long sequence of nucleotides. DNA has one additional—and very important—level of organization: it is usually a double-stranded molecule.

Section: 9.2 The Structures of DNA and RNA Difficulty: Medium 47) While discussing the structure of DNA with your classmates you notice that your study partner has drawn a DNA molecule that looks like the following.

3'—ATCGGUTCCAAA—5' 3'—TAACCAAGGTTG—5' What is incorrect about this drawing of the DNA molecule?

Answer: There are several problems with the above drawing. Some of the problems include the strands are not helical; the strands are not anti-parallel; DNA does not contain the base Uracil; also the bonding of the bases is not correct. C does not bind to A and A does not bind to G. Section: 9.2 The Structures of DNA and RNA Difficulty: Medium 48) What were the two major pieces of evidence that led to the deduction of DNA structure by Watson and Crick and who supplied this information?

Answer: Chargaff: Ratio of A-to-T and G-to-C is one, led to idea of basepairing.

Franklin and Wilkins: X-ray diffraction patterns of DNA, double helical nature of DNA. Section: 9.2 The Structures of DNA and RNA Difficulty: Medium

49) What are the three major forms of DNA? Identify a major context in which they are found, and what the requirements are for each.

Answer: B-DNA: in vivo, requires a high degree of hydration.

A-DNA: in RNA:DNA duplexes, in high concentrations of salts or partially dehydrated state.

Z-DNA: unknown, possible role in gene regulation, GC-rich alternating purine and pyrimidine residues.

Section: 9.2 The Structures of DNA and RNA Difficulty: Medium 50) Describe the three levels of condensation required to package the 103 to 105 μm of DNA in a eukaryotic chromosome into a metaphase structure a few microns long.

Answer: The first level of condensation involves packaging DNA as a negative supercoil into nucleosomes, to produce the 11-nm diameter interphase chromatin fi ber. This clearly involves an octamer of histone molecules, two each of histones H2a, H2b, H3, and H4.

2. The second level of condensation involves an additional folding or supercoiling of the 11-nm nucleosome fi ber, to produce the 30-nm chromatin fi ber. Histone H1 is involved in this supercoiling.

3. Finally, nonhistone chromosomal proteins form a scaffold that is involved in condensing the 30-nm chromatin fi ber into the tightly packed metaphase chromosomes. This third level of condensation appears to involve the separation of segments of the giant DNA molecules present in eukaryotic chromosomes into independently supercoiled domains or loops. The mechanism by which this third level of condensation occurs is not known.

Section: 9.4 Chromosome Structure in Eukaryotes Difficulty: Medium

Chapter 10 Test Bank Question Type: Multiple Choice 1) DNA replication occurs in which manner? a) Conservative b) Semi-Conservative c) Dispersive d) Conservative and Semi-Conservative e) Semi-Conservative and Dispersive Answer: b Section: 10.1 Basic Features of DNA Replication In Vivo Difficulty: Easy 2) Which of the following statements is not true regarding DNA replication? a) DNA replication occurs in a semi-conservative manner. b) DNA replication begins at unique initiation points c) DNA replication proceeds bidirectionally from the initiation point. d) DNA replication proceeds in one direction from the initiation point. e) All of these are false Answer: d Section: 10.1 Basic Features of DNA Replication In Vivo Difficulty: Medium 3) The synthesis of new DNA from parental DNA involves which three steps? a) Replication, transcription, and translation b) Replication, gene expression and translation c) Initiation, elongation, and termination d) Initiation, gene expression, and transcription e) Elongation, termination, and gene expression Answer: c Section: 10.1 Basic Features of DNA Replication In Vivo Difficulty: Easy

4) In order for DNA to replicate in a semi-conservative manner, what must occur?

1. The double stranded parent DNA molecule must be separated by breaking the hydrogen bonds between the nitrogenous bases 2. Both parent strands of DNA must serve as a template for the synthesis of the new strands 3. The parent strands of DNA must re-anneal after serving as a template for the synthesis of new DNA. a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: d Section: 10.1 Basic Features of DNA Replication In Vivo Difficulty: Medium 5) Meselson and Stahl are best known for demonstrating that: a) DNA replication occurs in a conservative manner in E. coli b) DNA replication occurs in a semi-conservative manner in E. coli c) DNA replication occurs in a conservative manner in eukaryotes d) DNA replication occurs in a semi-conservative manner in eukaryotes e) None of these Answer: b Section: 10.1 Basic Features of DNA Replication In Vivo Difficulty: Easy 6) The work of Meselson and Stahl: a) Showed that DNA replication semiconservative, which means that the parental double helix is conserved. b) Used the radioisotope 3H-thymidine in conjunction with centrifugation to separate DNA on the basis of size c) Showed that the DNA, after one round of replication, was of intermediate density thereby showing that the DNA molecules contained one parent strand and one new strand. d) All of these e) None of these Answer: c Section: 10.1 Basic Features of DNA Replication In Vivo Difficulty: Easy

7) The idea that eukaryotic chromosomes replicate semiconservatively was supported by Taylor et al, in experiments using 3H-thymidine labeled DNA from ______: a) E. coli, and saw that the DNA was of intermediate density after one round of replication b) Vicia faba, and saw that only one of the chromatids of each pair was radioactive after a second round of c-metaphase. c) Vicia faba, and saw that only one of the chromatids of each pair was radioactive after a single round of c-metaphase. d) E. coli, and saw that the DNA was of light density after one round of replication e) None of these Answer: b Section: 10.1 Basic Features of DNA Replication In Vivo Difficulty: Easy 8) The proof that eukaryotic chromosomal replication was a semi-conservative process: a) Used 3H-thymidine to label E. coli DNA. b) Used 3H-thymidine to label human cell DNA. c) Relied on the assumption that each chromosome in the broad bean consisted of a single DNA molecule. d) Did not require the fact that each chromosome consists of a single strand of DNA. e) None of these Answer: c Section: 10.1 Basic Features of DNA Replication In Vivo Difficulty: Easy 9) Which of the following technologies aided in the visualization of the replication forks in E. coli? a) Autoradiography b) Gel electrophoresis c) SDS-PAGE d) Southern blotting e) HPLC Answer: a Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 10) E. coli chromosomes: a) Replicate in a unidirectional pattern from a single, unique origin, called oriC. b) Adopt a θ-shaped conformation during replication. c) Consist of two Y-shaped replication forks which move in same direction d) Have AT-rich regions at the origin of replication to maintain a tight conformation between strands, this facilitates the formation of the replication bubble. e) None of these Answer: b Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy

11) When John Cairns first visualized the replication of E. coli chromosomes in 1963 what did he conclude?

1. The chromosomes of E. coli are circular structures that exist as θ-shaped intermediates during replication.

2. The unwinding of the two complementary parental strands and their semiconservative replication occur simultaneously or are closely coupled.

3. Semiconservative replication of the E. coli chromosome started at a specific site, and proceeded sequentially and unidirectionally around the circular structure a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 10.2 DNA Replication in Prokaryotes Difficulty: Medium 12) Since the parental double helix must rotate 360° to unwind each gyre of the helix, during the semi-conservative replication of the bacterial chromosome, some kind of “swivel” must exist.

What do geneticists now know that the required swivel is? a) Topoisomerase b) Helicase c) A transient single-strand break produced by the action of topoisomerases d) A transient single-strand break produced by the action of helicases e) A transient single-strand break produced by the action of Ligase Answer: c Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 13) How many origins of replication are typically found in a bacterial or viral chromosome? a) One b) Two c) Three d) Four e) Five Answer: a Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy

14) In the E. coli chromosome the origin of replication, called oriC, is characterized as being rich in: a) A-T base pairs b) A-G base pairs c) C-G base pairs d) C-T base pairs e) G-T base pairs Answer: a Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 15) A replication bubble is: a) A localized region of DNA strand separation b) A localized region of RNA strand separation c) A localized region of DNA strand attachment d) All of these e) None of these Answer: a Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 16) In the yeast Saccharomyces cerevisiae, segments of chromosomal DNA that allow a fragment of circularized DNA to replicate as an independent unit are known as: a) Bidirectionally Replicating Units b) Autonomously Replicating Sequences c) Extrachromosomally Replicating Elements d) Unique Origins of Replication e) None of these Answer: b Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 17) What purpose is served by DNA ligase? a) To break hydrogen bonds in the DNA structure b) To seal single stranded breaks in the DNA double helices c) To seal double stranded breaks in the DNA single helices d) All of these e) None of these Answer: b Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy

18) Schnös and Inman demonstrated that bidirectional replication occurs by: a) visualizing phage replication forks in conjunction with denaturation mapping by electron microscopy b) visualizing phage replication forks using gel electrophoresis c) observing the θ-shaped replicative structure of the lambda chromosome d) All of these e) None of these Answer: a Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 19) Which of the following does not replicate in a bidirectional format? a) Lambda phage b) E. coli c) SV40 d) Coliphage P2 e) All of these replicate bidirectionally Answer: d Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 20) Consider the DNA template to be used for replication: 3'—ATCGGGAAATTCGGA—5' Which of the following choices shows the correctly replicated new strand of DNA? a) 3'---ATCGGGAAATTCGGA—5' b) 5'---ATCGGGAAATTCGGA---3' c) 3'---TAGCCCTTTAAGCCT---5' d) 3'---TAGCCCTTTAAGCCT---5' e) 5'---UAGCCCUUUAAGCCU---3' Answer: c Section: 10.2 DNA Replication in Prokaryotes Difficulty: Medium 21) Consider the replicated strand of DNA: 3'—ATCGGGAAATTCGGA—5' Which of the following choices shows the template strand of DNA used for replication? a) 3'---ATCGGGAAATTCGGA—5' b) 5'---ATCGGGAAATTCGGA---3' c) 3'---TAGCCCTTTAAGCCT---5' d) 3'---TAGCCCTTTAAGCCT---5' e) 5'---UAGCCCUUUAAGCCU---3' Answer: c Section: 10.2 DNA Replication in Prokaryotes Difficulty: Medium

22) Eukaryotic chromosomal replication: 1. Is bidirectional 2. Occurs from multiple origins 3. Has sequence-specific origins (termed autonomously replicating sequences) a) 1 b) 2 c) 3 d) 2 and 3 e) All of these Answer: e Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 23) Which of the following enzymes catalyzes DNA synthesis? a) DNA ligase b) DNA helicase c) DNA polymerase d) DNA gyrase e) None of these Answer: c Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 24) Which of the following must be present in order for DNA polymerase to be active?

1. Mg2+ 2. dNTPs 3. Existing DNA template with a free 3’ end a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: e Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy

25) A researcher is attempting to replicate DNA in vitro. He adds all of the components to the tube except Mg2+. What kind of result can he expect to observe? a) The DNA will replicate at a normal rate b) The DNA will replicate but at a much slower rate c) The DNA will not replicate because DNA polymerase will not be active d) The DNA will not replicate because DNA polymerase was not added e) The DNA will not replicate because the Mg2+ is what catalyzes the DNA synthesis Answer: c Section: 10.2 DNA Replication in Prokaryotes Difficulty: Medium 26) The direction of DNA synthesis is: a) Always 3'→5' b) Always 5'→3' c) Both 5'→3' and 3'→5' d) Neither 5'→3' or 3'→5' e) DNA synthesis is not ordered. It is a random process Answer: b Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 27) A researcher is attempting to replicate DNA in vitro. All necessary components are added, however, the DNA that is used does not have a free 3' hydroxyl. What kind of result is expected? a) The DNA will replicate at a normal rate b) The DNA will replicate but at a much slower rate c) The DNA will not replicate because DNA polymerase will not be active d) The DNA will not replicate because DNA polymerase can not add dNTPs to the existing DNA e) The DNA will not replicate because the free 3'hydroxyl is what catalyzes the DNA synthesis Answer: d Section: 10.2 DNA Replication in Prokaryotes Difficulty: Medium 28) Which of the following is a true function of DNA Polymerase I? 1. Replicase 2. 5’→3’ exonuclease 3. 3’→5’ exonuclease a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: e Section: 10.2 DNA Replication in Prokaryotes Difficulty: Medium

29) Which DNA polymerase molecule acts as the replicase in E. coli? a) DNA Polymerase I b) DNA Polymerase II c) DNA Polymerase III d) DNA Polymerase IV e) DNA Polymerase V Answer: c Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 30) Which of the following characteristics is not a commonality between prokaryotic and eukaryotic DNA polymerases? a) 3'→5' exonuclease activity. b) They consist of a single polypeptide chain. c) 5'→3' polymerase activity. d) They require a free 3'-OH end on a pre-existing primer. e) Prokaryotic and eukaryotic DNA polymerases share all of these characteristics.

Answer: b Section: 10.2 DNA Replication in Prokaryotes Difficulty: Medium 31) Proofreading activities of E. coli DNA polymerases is carried out by: a) Polymerase I by the 3'→5' exonuclease activity b) Polymerase I by the 5'→3' exonuclease activity c) Polymerase III by the 3'→5' exonuclease activity d) Polymerase III by the 5'→3' exonuclease activity e) Polymerase IV by the 5'→3' endonuclease activity Answer: a Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 32) The 3'→5' exonuclease activities of DNA polymerases read nascent strands as they are synthesized, removing any mispaired nucleotides at the 3 termini of primer strands. This action is known as: a) Rereading b) Proofreading c) Editing d) Splicing e) Lysing Answer: b Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy

33) Complementary strands of a DNA double helix have opposite chemical polarity but are synthesized in the same direction (replication fork movement). This is accounted for by: a) Using a 3'→5' polymerase to synthesize the lagging strand. b) Using discontinuous synthesis on the leading strand. c) Using the polymerase activity of DNA ligase. d) Using the 5'→3' polymerase activity of Pol III to synthesize multiple short segments which are later covalently joined. e) None of these Answer: d Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 34) DNA replication requires a free 3'-OH to initiate polymerase activity. This is accomplished by synthesizing: a) A temporary RNA primer removed by Pol I. b) A temporary RNA primer removed by Pol III. c) A permanent DNA primer synthesized by Pol III. d) A permanent DNA primer synthesized by Pol I. e) DNA replication does not require a free 3'-OH Answer: a Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 35) During replication, the double helix undergoes torsional stress during strand separation and polymerase activity. Several proteins are employed to counter this stress and assist strand separation. Which of the following proteins involved with this process is defined incorrectly below? a) SSB: promote the separation of complementary sequence. b) Helicase: unwinding of the parental DNA strands. c) Topoisomerase I: removes and introduces supercoils one at a time. d) Topoisomerase II: removes and introduces supercoils two at a time. e) Gyrase: topoisomerase II in E. coli.

Answer: c Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy

36) Which of the following prokaryotic replication events is incorrect? a) Interaction of pre-priming proteins with oriC, such as DnaA, DnaB, and DnaC. b) Initiation of Okazaki fragments on the lagging strand carried out by the primosome, a protein complex of primase and DNA helicase. c) RNA primers are covalently extended with deoxynucleotides by Pol I d) The replisome contains two Pol III holoenzymes, one for the leading and one for lagging strands. e) Only one of the two Pol III holoenzymes contains a primosome.

Answer: c Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 37) Rolling circle replication: a) Is only found in viruses and bacteria. b) Proceeds bidirectionally. c) Is initiated by an exonucleolytic cleavage event. d) Can only produce single-stranded progeny molecules. e) Generates single-stranded tails longer than length of the parent circular molecule.

Answer: e Section: 10.2 DNA Replication in Prokaryotes Difficulty: Easy 38) One similarity between prokaryotic and eukaryotic DNA replication is: a) Distinct leading and lagging strands b) RNA primer length c) Multiple origins of replication d) Having DNA replication occur during a specific portion of the cell cycle. e) Use of two different polymerases for leading and lagging strand synthesis.

Answer: a Section: 10.3 Unique Aspects of Eukaryotic Chromosome Replication Difficulty: Easy 39) Eukaryotes were found to have multiple replicons per chromosome by: a) 2H thymidine pulse experiments b) 3H thymidine pulse-chase experiments c) 1H thymidine pulse experiments d) 2H thymidine pulse-chase experiments e) None of these Answer: b Section: 10.3 Unique Aspects of Eukaryotic Chromosome Replication Difficulty: Easy

40) In eukaryotes, two DNA polymerases were found to be essential to leading and lagging strand synthesis of the SV40 virus. They are: a) δ, ε b) α, β c) γ, δ d) ε, κ e) θ, λ Answer: a Section: 10.3 Unique Aspects of Eukaryotic Chromosome Replication Difficulty: Easy 41) The RNA containing enzyme that participates in the addition of the telomeric ends of the chromosome is known as: a) Replicase b) Gyrase c) Telomerase d) Ligase e) Polymerase V Answer: c Section: 10.3 Unique Aspects of Eukaryotic Chromosome Replication Difficulty: Easy 42) Which of the following statements about telomerase are true: 1. It extends the 3' end of a linear chromosome one repeat at a time.

2. It uses a built-in DNA to act as a template for synthesis of the telomeric repeat.

3. Without telomerase, linear chromosomes would become progressively shorter. a) 1 b) 2 c) 3 d) 1 and 2 e) 1 and 3 Answer: e Section: 10.3 Unique Aspects of Eukaryotic Chromosome Replication Difficulty: Medium 43) Which subunit of the Pol III enzyme forms a ring-like structure around DNA, acting like PCNA in eukaryotes, in order to increase the enzyme's processivity? a) α b) β c) γ d) δ e) θ Answer: b Section: 10.3 Unique Aspects of Eukaryotic Chromosome Replication Difficulty: Easy

44) Telomere length has not been correlated with: a) Aging b) Sex determination c) Progeria d) Cancer e) All of these Answer: b Section: 10.3 Unique Aspects of Eukaryotic Chromosome Replication Difficulty: Easy 45) Telomerase inhibitors are thought to be one possible treatment for which of the following? a) aging b) progeria c) cancer d) aging and progeria e) progeria and cancer Answer: c Section: 10.3 Unique Aspects of Eukaryotic Chromosome Replication Difficulty: Easy Question Type: Essay 46) Briefly compare the origins of replication in bacterial chromosomes and eukaryotic chromosomes?

Answer: In bacterial and viral chromosomes, there is usually one unique origin per chromosome, and this single origin controls the replication of the entire chromosome. In the large chromosomes of eukaryotes, multiple origins collectively control the replication of the giant DNA molecule present in each chromosome. The multiple origins of replication in eukaryotic chromosomes also appear to be specific DNA sequences. Current evidence indicates that these multiple replication origins in eukaryotic chromosomes also occur at specific sites. Each origin controls the replication of a unit of DNA called a replicon; thus, most prokaryotic chromosomes contain a single replicon, whereas eukaryotic chromosomes usually contain many replicons.

Section: 10.1 Basic Features of DNA Replication In Vivo Difficulty: Medium 47) In what ways are the thermodynamic properties of A:T pairing (as opposed to G:C pairing) utilized in a biological context such as DNA replication? What about in an experimental context?

Answer: A:T pairing has only 2 H-bonds, whereas G:C pairing has 3 H-bonds, thus making AT-rich regions more susceptible to strand separation. In a biological context, AT-rich regions are a part of the oriC in E. coli to aid in strand separation to allow the formation of replication forks (Also, -10 sequence in transcription bubble). In an experimental context, denaturation mapping uses melted AT-rich regions as physical markers when deducing the directionality of DNA replication.

Section: 10.1 Basic Features of DNA Replication In Vivo Difficulty: Medium

48) Pol I of E. coli has three distinct activities. List each activity and define its principle function.

Answer: 5→3 polymerase activity: synthesizes short stretches of nucleotides; DNA replacement of RNA primers.

5→3 exonuclease activity: removal of RNA primers.

3→5 Exonuclease activity: proofreading activity removes unpaired bases from 3' end.

Section: 10.2 DNA Replication in Prokaryotes Difficulty: Medium 49) Briefly explain the mechanism for rolling circle replication.

Answer: As the name implies, rolling-circle replication is a mechanism for replicating circular DNA molecules. The unique aspect of rolling-circle replication is that one parental circular DNA strand remains intact and rolls (thus the name rolling circle) or spins while serving as a template for the synthesis of a new complementary strand. Replication is initiated when a sequence-specific endonuclease cleaves one strand at the origin, producing 3′-OH and 5′-phosphate termini. The 5′ terminus is displaced from the circle as the intact template strand turns about its axis. Covalent extension occurs at the 3′-OH of the cleaved strand. Since the circular template DNA may turn 360° many times, with the synthesis of one complete or unit-length DNA strand during each turn, rolling-circle replication generates single-stranded tails longer than the contour length of the circular chromosome. Rolling-circle replication can produce either single-stranded or double-stranded progeny DNAs. Circular single-stranded progeny molecules are produced by site-specific cleavage of the single-stranded tails at the origins of replication and recircularization of the resulting unit-length molecules. To produce double-stranded progeny molecules, the single-stranded tails are used as templates for the discontinuous synthesis of complementary strands prior to cleavage and circularization. The enzymes involved in rolling-circle replication and the reactions catalyzed by these enzymes are basically the same as those responsible for DNA replication involving θ-type intermediates.

Section: 10.2 DNA Replication in Prokaryotes Difficulty: Medium 50) How are the telomeric regions on the ends of the linear eukaryotic chromosomes preserved during replication?

Answer: Telomerase recognizes the G-rich telomere sequence on the 3′ overhang and extends it 5'→3' one repeat unit at a time. Telomerase does not fill in the gap opposite the 3′ end of the template strand; it simply extends the 3′ end of the template strand. The unique feature of telomerase is that it contains a built-in RNA template. After several telomere repeat units are added by telomerase, DNA polymerase catalyzes the synthesis of the complementary strand.

Section: 10.3 Unique Aspects of Eukaryotic Chromosome Replication Difficulty: Medium

Chapter 11 Test Bank Question Type: Multiple Choice 1) Which process transfers information from DNA to RNA? a) Replication b) Transcription c) Translation d) Splicing e) None of these Answer: b Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy 2) Which process transfers information from RNA molecules to proteins? a) Replication b) Transcription c) Translation d) Splicing e) None of these Answer: c Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy 3) Which of the following best illustrates the central dogma of biology? a) DNA→Protein→RNA b) RNA→DNA→Protein c) DNA→DNA→Protein d) DNA→RNA→Protein e) Protein→RNA→DNA Answer: d Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy 4) Which of the following is a true statement regarding the transfer of information from DNA to RNA. 1. It is always irreversible 2. It occurs during a process known as translation 3. It is sometimes reversible a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: c Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Medium

5) How many strands of DNA are used as a template during the process of transcription? a) One b) Two c) Three d) Four e) None of these Answer: a Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy 6) During transcription the newly formed complementary strand of RNA is known as the: a) Translate b) Transcript c) cRNA d) cDNA e) Genetic Code Answer: b Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy 7) Nucleotide triplets in the gene transcript known as _____________ specify the amino acids which will be added during translation. a) Genes b) Alleles c) Codons d) Anticodons e) tRNA Answer: c Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy 8) Consider the DNA template 3'—AAATTTTAGCCA—5'. When transcribed, which of the following is the correct resulting transcript? a) 5'—TTTAAATCGGT—3' b) 5'—UUUAAAUGCCA—3' c) 5'—UUUAAAUCGGU—3' d) 3'—TTTAAATCGGT—5' e) 3'—AAATTTTAGCCA—5' Answer: c Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy

9) Which part of the eukaryotic cell is the location of transcription? a) Nucleus b) Mitochondria c) Ribosome d) Cytosol e) Lysosome Answer: a Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy 10) Which part of the cell is the location for translation? a) Nucleus b) Mitochondria c) Ribosome d) Cytosol e) Lysosome Answer: c Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy 11) Prokaryotic and eukaryotic RNA's differ in that: a) rRNA molecules are only subunits of the prokaryotic ribosome. b) Eukaryotes have a pre-mRNA that requires splicing to create the functional transcript c) Prokaryotes have non-coding sequences that are removed during RNA-transcript processing d) Eukaryotes have simpler ribosomes, consisting of fewer subunits. e) None of these Answer: b Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Medium 12) The RNA molecules that are intermediaries between DNA and polypeptides are known as: a) tRNA b) mRNA c) rRNA d) sRNA e) pRNA Answer: b Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy

13) Small RNA molecules that function as adaptors between amino acids and the codons in mRNA during translation are known as: a) tRNA b) mRNA c) rRNA d) sRNA e) pRNA Answer: a Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy 14) Which type of RNA are structural components of spliceosomes, the nuclear structures that excise introns from nuclear genes? a) tRNA b) mRNA c) rRNA d) sRNA e) pRNA Answer: d Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy 15) What name is given to the short (20 to 22–nucleotide) single-stranded RNAs that are cleaved from hairpin-shaped precursors and block the expression of complementary mRNAs by either causing their degradation or repressing their translation? a) tRNA b) mRNA c) rRNA d) sRNA e) pRNA Answer: b Section: 11.1 Transfer of Genetic Information: The Central Dogma Difficulty: Easy 16) Which of the following is a way in which RNA synthesis is similar to DNA synthesis? a) Precursors are ribonucleoside triphosphates. b) Only one strand of DNA is used as a template for the synthesis of a complementary RNA chain in any given reaction. c) The basepair sequence of the mRNA is identical to the nontemplate strand of DNA except that U substitutes for T. d) RNA chains can be initiated de novo. e) None of these Answer: c Section: 11.2 The Process of Gene Expression Difficulty: Medium

17) mRNA strands that will specify amino acids in the protein gene product are also known as: a) Coding strands b) Sense strands c) Antisense strands d) Coding strands and Sense strands e) Coding strands and Antisense strands Answer: d Section: 11.2 The Process of Gene Expression Difficulty: Easy 18) Which enzyme catalyzes the addition of new ribonucleotides to the free 3' end of the growing molecule? a) DNA Polymerase b) RNA Polymerase c) Ligase d) DNA Gyrase e) Helicase Answer: b Section: 11.2 The Process of Gene Expression Difficulty: Easy 19) RNA polymerase binds to a specific nucleotide sequence on the template strand known as a/an: a) Operator b) Promoter c) Enhancer d) Silencer e) None of these Answer: b Section: 11.2 The Process of Gene Expression Difficulty: Easy 20) RNA synthesis takes place within a locally unwound segment of DNA, sometimes called the: a) Translation bubble b) Replication bubble c) Transcription bubble d) DNA/RNA hybrid e) Operator bubble Answer: b Section: 11.2 The Process of Gene Expression Difficulty: Easy

21) The initiation of transcription is characterized by: a) The binding of RNA polymerase to the promoter region of the template DNA b) The binding of DNA polymerase to the operator region of the template RNA c) The binding of helicase to the operator region of the template RNA d) The binding of ligase to the promoter region of the template DNA e) None of these Answer: a Section: 11.3 Transcription in Prokaryotes Difficulty: Easy 22) In which direction does RNA synthesis occur? a) 5' → 3' b) 3' → 5' c) 1' →3' d) 5' → 3'and 3'→5' e) RNA synthesis occurs randomly without a specific direction Answer: a Section: 11.3 Transcription in Prokaryotes Difficulty: Easy 23) Which of the following is the correct order of the steps of transcription? a) Initiation, termination, elongation b) Initiation, transcription, elongation c) Initiation, elongation, termination d) Elongation, transcription, termination e) Termination, elongation, initiation Answer: c Section: 11.3 Transcription in Prokaryotes Difficulty: Easy 24) Which of the following are biochemical features of RNA polymerase?

1.

5'→3' polymerase activity.

2.

Recognition of a promoter sequence to initiate transcription with help from factors.

3.

Local unwinding of DNA, called a transcription bubble, provides a template for enzymatic activity. a) 1 b) 2 c) 3 d) 2 and 3 e) All of these Answer: e Section: 11.3 Transcription in Prokaryotes Difficulty: Medium

25) Which of the following E. coli RNA polymerase subunits are responsible only for promoter specific transcription in vitro? a) α b) β c) β' d) σ e) α2ββ′ Answer: d Section: 11.3 Transcription in Prokaryotes Difficulty: Easy 26) The function of the core enzyme of RNA polymerase is: a) To recognize and bind RNA polymerase to the transcription initiation or promoter sites in DNA b) To recognize and bind DNA polymerase to the transcription initiation or promoter sites in RNA c) To catalyze RNA synthesis from DNA templates in vitro d) To catalyze DNA synthesis from DNA templates in vitro e) To catalyze DNA synthesis from RNA templates in vitro Answer: c Section: 11.3 Transcription in Prokaryotes Difficulty: Medium 27) During a research event, the σ subunit is removed from the RNA polymerase enzyme. How will this affect the initiation of transcription? a) Transcription will initiate and proceed normally b) Transcription initiation will be stopped c) Transcription will initiate at random spots along the template d) All of these e) None of these Answer: c Section: 11.3 Transcription in Prokaryotes Difficulty: Medium 28) A base pair sequence preceding the transcription initiation site is known as a/an: a) Upstream sequence b) Downstream sequence c) Promoter d) Operator e) Enhancer Answer: a Section: 11.3 Transcription in Prokaryotes Difficulty: Easy

29) Which statement is true about E. coli promoters: a) Most promoters have a high degree of sequence similarity. b) -10 and -35 regions show the most conservation between promoters. c) The GC-rich -10 region facilitates the localized unwinding of DNA . d) Distance between the -10 and -35 regions is moderately conserved from 10 to 30 nucleotides in length e) None of these statements are correct Answer: b Section: 11.3 Transcription in Prokaryotes Difficulty: Medium 30) The –10 consensus sequence in the nontemplate strand is often recognized by which of the following base pair sequences? a) TATAAT b) TTGACA c) ATTGAC d) CAGTTA e) TAAAAA Answer: a Section: 11.3 Transcription in Prokaryotes Difficulty: Easy 31) The sigma subunit initially recognizes and binds to which of the following sequences? a) -10 b) -35 c) -50 d) +10 e) +35 Answer: b Section: 11.3 Transcription in Prokaryotes Difficulty: Easy 32) The covalent extension of a growing RNA chain takes place in which of the following locations? a) Promoter b) Operator c) Transcription bubble d) Replication bubble e) Enhancer Answer: c Section: 11.3 Transcription in Prokaryotes Difficulty: Easy

33) Which of the following characteristics of transcriptional termination is indicative of rho-dependent activity? a) Hairpin structure b) Protein moving 3'→5' on the growing mRNA c) Six or more GU basepairs d) Protein mediated "pulling" of the nascent RNA chain from the transcription bubble e) All of these Answer: d Section: 11.3 Transcription in Prokaryotes Difficulty: Medium 34) Which of the following is not a major modification that the primary transcript in eukaryotes undergoes before use during translation? a) Methyl guanosine caps are added to the 5′ ends of the primary transcripts b) Poly(A) tails are added to the 3′ ends of the transcripts c) Intron sequences are spliced out of transcripts. d) Exon sequences are spliced out of transcripts. e) All of these are major modifications Answer: d Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Medium 35) Which of the following is the RNA polymerase that is used for transcription in eukaryotes?

1.

RNA Polymerase 1 2.

RNA Polymerase 2 3.

RNA Polymerase 3 a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Medium

36) In eukaryotes, when RNA polymerase II initiates transcription: 1. Basal transcription factors, designated TFIX (X is a letter designating individual factors), are required.

2. TFIID is the first protein to bind the promoter.

3. TFIIA catalyzes the unwinding of the DNA helix a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: d Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Medium 37) Which of the following RNA polymerases catalyze the initiation of transcription in most eukaryotic genes? a) RNA polymerase I b) RNA polymerase II c) RNA polymerase III d) RNA polymerase IV e) RNA polymerase V Answer: b Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Easy 38) The addition of the 5' 7-methyl guanosine (7-MG) cap occurs during which stage of eukaryotic transcription? a) Initiation b) Elongation c) Termination d) Post-transcriptional processing e) It can occur at any time Answer: b Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Easy 39) What is the purpose for the addition of the 5' 7-methyl guanosine cap on eukaryotic pre-mRNA molecules? a) It prevents the degredation of the sequence by nucleases b) It prevents the degredation of the sequence by lysosomes c) It guides the pre-mRNA sequence out of the plasma membrane d) All of these e) None of these Answer: a Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Easy

40) The polyadenylation of the pre-mRNA sequences in eukaryotes occurs during which stage of eukaryotic transcription? a) Initation b) Elongation c) Termination d) Post-transcriptional processing e) It can occur at any time Answer: c Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Easy 41) The 3′ ends of RNA transcripts synthesized by RNA polymerase II are produced by: 1.

Endonucleolytic cleavage of the primary transcripts 2.

Termination of transcription 3.

The addition of the 5' 7-methyl guanosine cap a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: d Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Medium 42) Which of the following is a form of RNA editing?

1. The polyadenylation of the pre-mRNA sequence 2. The changing of the structures of individual bases 3. The insertion or deletion of uridine monophosphate residues a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: e Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Medium

43) Noncoding sequences found in eukaryotic genes that interrupt the coding sequences are known as: a) Introns b) Exons c) Spliceosomes d) Pre-mRNA e) None of these Answer: a Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Medium 44) Introns in tRNA are removed by: a) The concerted action of a splicing endonuclease and Ligase b) Autocatalytical splicing c) Ribonucleoprotein structures called spliceosomes d) All of these e) none of these Answer: a Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Medium 45) Introns in mRNA are removed by: a) The concerted action of a splicing endonuclease and Ligase b) Autocatalytical splicing c) Ribonucleoprotein structures called spliceosomes d) All of these e) none of these Answer: c Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Easy 46) The snRNA responsible for the recognition/binding of the 5' splice site prior to the initial cleavage reaction is: a) U1 b) U2 c) U3 and U4 d) U5 and U1 e) U6 Answer: a Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Easy

Question Type: Essay 47) Briefly describe the process of initiation for RNA transcription in prokaryotes Answer: Initiation of RNA chains involves three steps: (1) binding of the RNA polymerase holoenzyme to a promoter region in DNA; (2) the localized unwinding of the two strands of DNA

by RNA polymerase, providing a template strand free to base-pair with incoming ribonucleotides; and (3) the formation of phosphodiester bonds between the first few ribonucleotides in the nascent RNA chain. The holoenzyme remains bound at the promoter region during the synthesis of the first eight or nine bonds; then the sigma factor is released, and the core enzyme begins the elongation phase of RNA synthesis. During initiation, short chains of two to nine ribonucleotides are synthesized and released. This abortive synthesis stops once chains of 10 or more ribonucleotides have been synthesized and RNA polymerase has begun to move downstream from the promoter.

Section: 11.3 Transcription in Prokaryotes Difficulty: Medium 48) The promoter region in E. coli is comprised of two traditionally recognized sequences.

Describe these sequences, their properties, and their purposes.

Answer: Two short sequences within the E. coli promoters are sufficiently conserved to be recognized, but even these are seldom identical in two different promoters. The midpoints of the two conserved sequences occur at about 10 and 35 nucleotide pairs, respectively, before the transcription-initiation site (Figure 11.9). Thus they are called the –10 sequence and the –35 sequence, respectively. Although these sequences vary slightly from gene to gene, some nucleotides are highly conserved. The nucleotide sequences that are present in such conserved genetic elements most often are called consensus sequences. The –10 consensus sequence in the nontemplate strand is TATAAT; the –35 consensus sequence is TTGACA. The sigma subunit initially recognizes and binds to the –35 sequence; thus, this sequence is sometimes called the recognition sequence. The A:T-rich –10 sequence facilitates the localized unwinding of DNA, which is an essential prerequisite to the synthesis of a new RNA chain. The distance between the –35 and –10 sequences is highly conserved in E. coli promoters, never being less than 15 or more than 20 nucleotide pairs in length. In addition, the first or 5 base in E. coli RNAs is usually (>90%) a purine.

Section: 11.3 Transcription in Prokaryotes Difficulty: Medium

49) Compare and contrast rho-independent termination and rho-dependent termination during transcription in prokaryotes Answer: There are two types of transcription terminators in E. coli. One type results in termination only in the presence of a protein called rho ( ); therefore, such termination sequences are called rho-dependent terminators. The other type results in the termination of transcription without the involvement of rho; such sequences are called rho-independent terminators. Rho-independent terminators contain a G:C-rich region followed by six or more A:T base pairs, with the A's present in the template strand (Figure 11.11, top). The nucleotide sequence of the G:C-rich region is such that regions of the single-stranded RNA can base-pair and form hairpinlike structures (Figure 11.11, bottom). The RNA hairpin structures form immediately after the synthesis of the participating regions of the RNA chain and retard the movement of RNA polymerase molecules along the DNA, causing pauses in chain extension. Since A:U base-pairing is weak, requiring less energy to separate the bases than any of the other standard base pairs, the run of U's after the hairpin region is thought to facilitate the release of the newly synthesized RNA chains from the DNA template when the hairpin structure causes RNA polymerase to pause at this site. The mechanism by which rho-dependent termination of transcription occurs is still uncertain.

Rho-dependent termination sequences are 50 to 90 base pairs long and specify RNA transcripts that are rich in C residues and largely devoid of G's. Beyond that, different rho-dependent termination signals have little in common. The rho protein binds to the growing RNA chain and moves 5′ to 3′ along the RNA, seeming to pursue the RNA polymerase molecule catalyzing the synthesis of the chain. When RNA polymerase slows down or pauses at the rho-dependent termination sequence, rho catches up with the polymerase and pulls the nascent RNA chain from the transcription bubble.

Section: 11.3 Transcription in Prokaryotes Difficulty: Medium 50) How does transcription and RNA processing in eukaryotes differ from the same processes in prokaryotes?

Answer: Although the overall process of RNA synthesis is similar in prokaryotes and eukaryotes, the process is considerably more complex in eukaryotes. In eukaryotes, RNA is synthesized in the nucleus, and most RNAs that encode proteins must be transported to the cytoplasm for translation on ribosomes. Three different enzymes catalyze transcription in eukaryotes, and the resulting RNA transcripts undergo three important modifications, including the excision of noncoding sequences called introns. The nucleotide sequences of some RNA transcripts are modified posttranscriptionally by RNA editing. Moreover, unlike the E. coli enzyme, all three eukaryotic RNA polymerases require the assistance of other proteins called transcription factors in order to initiate the synthesis of RNA chains. Prokaryotic mRNAs often contain the coding regions of two or more genes; such mRNAs are said to be multigenic. In contrast, many of the eukaryotic transcripts that have been characterized contain the coding region of a single gene (are monogenic).

But eukaryotic mRNAs may be either monogenic or multigenic.

Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Hard

51) How does initiation of transcription in eukaryotes differ from initiation of transcription in prokaryotes?

Answer: Unlike their prokaryotic counterparts, eukaryotic RNA polymerases cannot initiate transcription by themselves. All three eukaryotic RNA polymerases require the assistance of protein transcription factors to start the synthesis of an RNA chain. Indeed, these transcription factors must bind to a promoter region in DNA and form an appropriate initiation complex before RNA polymerase will bind and initiate transcription. Different promoters and transcription factors are utilized by the RNA polymerases I, II, and III. The promoters recognized by RNA polymerase II consist of short conserved elements, or modules, located upstream from the transcription startpoint. The conserved element closest to the transcription start site (position +1) is called the TATA box; it has the consensus sequence TATAAAA (reading 5′ to 3′ on the nontemplate strand) and is centered at about position –30. The TATA box plays an important role in positioning the transcription startpoint. The second conserved element is called the CAAT box; it usually occurs near position –80 and has the consensus sequence GGCCAATCT. Two other conserved elements, the GC box, consensus GGGCGG, and the octamer box, consensus ATTTGCAT, often are present in RNA polymerase II promoters; they influence the efficiency of a promoter in initiating transcription. The initiation of transcription by RNA polymerase II requires the assistance of several basal transcription factors.

Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Hard

52) Compare, in detail, the three forms of introns removal in eukaryotes.

Answer: The excision of introns from tRNA precursors occurs in two stages. In stage I, a nuclear membrane-bound splicing endonuclease makes two cuts precisely at the ends of the intron. Then, in stage II, a splicing ligase joins the two halves of the tRNA to produce the mature form of the tRNA molecule. The specificity for these reactions resides in conserved three-dimensional features of the tRNA precursors, not in the nucleotide sequences per se.

The autocatalytic excision of the intron in the rRNA precursor and certain other introns requires no external energy source and no protein catalytic activity. Instead, the splicing mechanism involves a series of phosphoester bond transfers, with no bonds lost or gained in the process. The reaction requires a guanine nucleoside or nucleotide with a free 3′-OH group (GTP, GDP, GMP, or guanosine all work) as a cofactor plus a monovalent cation and a divalent cation. The requirement for the G-3′-OH is absolute; no other base can be substituted in the nucleoside or nucleotide cofactor. The intron is excised by means of two phosphoester bond transfers, and the excised intron can subsequently circularize by means of another phosphoester bond transfer. The autocatalytic circularization of the excised intron suggests that the self-splicing of these rRNA precursors resides primarily, if not entirely, within the intron structure itself. Presumably, the autocatalytic activity is dependent on the secondary structure of the intron or at least the secondary structure of the RNA precursor molecule. The secondary structures of these self-splicing RNAs must bring the reactive groups into close juxtaposition to allow the phosphoester bond transfers to occur. Since the self-splicing phosphoester bond transfers are potentially reversible reactions, rapid degradation of the excised introns or export of the spliced rRNAs to the cytoplasm may drive splicing in the forward direction.

The introns in nuclear pre-mRNAs are excised in two steps like the introns in yeast tRNA precursors and Tetrahymena rRNA precursors that were discussed in the preceding two sections.

However, the introns are not excised by simple splicing nucleases and ligases or autocatalytically, and no guanosine cofactor is required. Instead, nuclear pre-mRNA splicing is carried out by complex RNA/protein structures called spliceosomes (Figure 11.22). These structures are in many ways like small ribosomes. They contain a set of small RNA molecules called snRNAs (small nuclear RNAs) and about 40 different proteins. The two steps in nuclear pre-mRNA splicing are known (Figure 11.23); however, some of the details of the splicing process are still uncertain.

Five snRNAs, called U1, U2, U4, U5, and U6, are involved in nuclear pre-mRNA splicing as components of the spliceosome. (snRNA U3 is localized in the nucleolus and probably is involved in the formation of ribosomes.) In mammals, these snRNAs range in size from 100 nucleotides (U6) to 215 nucleotides (U3). Some of the snRNAs in the yeast S. cerevisiae are much larger. These snRNAs do not exist as free RNA molecules. Instead, they are present in small nuclear RNA–protein complexes called snRNPs (small nuclear ribonucleoproteins). Spliceosomes are assembled from four different snRNPs and protein splicing factors during the splicing process.

Each of the snRNAs U1, U2, and U5 is present by itself in a specific snRNP particle. snRNAs U4 and U6 are present together in a fourth snRNP; U4 and U6 snRNAs contain two regions of intermolecular complementarity that are base-paired in the U4/U6 snRNP. Each of the four types of snRNP particles contains a subset of seven well-characterized snRNP proteins plus one or more proteins unique to the particular type of snRNP particle. All four snRNP complexes are present in the isolated spliceosomes shown in Figure 11.22. The first step in nuclear pre-mRNA splicing involves cleavage at the 5′ intron splice site (GU-intron) and the formation of an intramolecular phosphodiester linkage between the 5′ carbon of the G at the cleavage site and the 2′ carbon of a conserved A residue near the 3′ end of the intron. This step occurs on complete spliceosomes

(Figure 11.23) and requires the hydrolysis of ATP. Evidence indicates that the U1 snRNP must bind at the 5′ splice site prior to the initial cleavage reaction. Recognition of the cleavage site at the 5′ end of the intron probably involves base-pairing between the consensus sequence at this site and a complementary sequence near the 5′ terminus of snRNA U1. However, the specificity of the binding of at least some of the snRNPs to intron consensus sequences involves both the snRNAs and specific snRNP proteins. The second snRNP to be added to the splicing complex appears to be the U2 snRNP; it binds at the consensus sequence that contains the conserved A residue that forms the branch point in the lariat structure of the spliced intron. Thereafter, the U5 snRNP binds at the 3′ splice site, and the U4/U6 snRNP is added to the complex to yield the complete spliceosome (Figures 11.22 and 11.23). When the 5′ intron splice site is cleaved in step 1, the U4 snRNA is released from the spliceosome. In step 2 of the splicing reaction, the 3′ splice site of the intron is cleaved, and the two exons are joined by a normal 5′ to 3′ phosphodiester linkage (Figure 11.23). The spliced mRNA is now ready for export to the cytoplasm and translation on ribosomes.

Section: 11.4 Transcription and RNA Processing in Eukaryotes Difficulty: Medium

Chapter 12 Test Bank Question Type: Multiple Choice 1) The process by which the genetic information stored in mRNA nucleotide sequences is used to specify the amino acid sequences in polypeptide gene products is known as: a) Replication b) Transcription c) Translation d) Transcription and Translation e) None of these Answer: c Section: 12.1 Protein Structure Difficulty: Easy 2) In which part of a eukaryotic cell does translation take place? a) Nucleus b) Ribosomes c) Lysosome d) Nucleolus e) Smooth ER Answer: b Section: 12.1 Protein Structure Difficulty: Easy 3) How many different amino acids can be used in the composition of proteins? a) 5 b) 10 c) 15 d) 20 e) 30 Answer: d Section: 12.1 Protein Structure Difficulty: Easy 4) All of the amino acids, except proline, contain: a) A free amine group b) A free carboxyl group c) A free R group d) A free amine group and a free carboxyl group e) A free carboxyl group and a free R group Answer: d Section: 12.1 Protein Structure Difficulty: Easy

5) Amino acids differ from one another by the _________ that are present. a) Amino group b) Carboxyl group c) R group d) R plasmid e) Core group Answer: c Section: 12.1 Protein Structure Difficulty: Easy 6) Which of the following is a type of R group? a) Hydrophobic b) Hydrophilic c) Acidic d) Basic e) All of these Answer: e Section: 12.1 Protein Structure Difficulty: Easy 7) A compound composed of two or more amino acids is known as a/an: a) Peptide b) Amino c) Nucleotide d) Peptide and Amino e) All of these Answer: A Section: 12.1 Protein Structure Difficulty: Easy 8) Amino acids in polypeptides are joined by what type of bond? a) Hydrogen b) Peptide c) Covalent d) Ionic e) Weak Answer: b Section: 12.1 Protein Structure Difficulty: Easy

9) Which of the following statements is not correct about amino acids and proteins? a) There are 20 different amino acids b) Structures of the amino acids can be modified c) Side chains are designated R-groups, which can belong to one of four classes: hydrophobic, hydrophilic, acidic, and basic d) Peptide bonds are formed between the amino groups of two amino acids e) All of these are correct Answer: e Section: 12.1 Protein Structure Difficulty: Medium 10) Which of the following non-covalent bonds is not involved in the tertiary structure formation of proteins? a) Ionic bonds b) Hydrogen bonds c) Disulfide bridges d) Hydrophobic interactions e) Van der Waal interactions Answer: c Section: 12.1 Protein Structure Difficulty: Medium 11) The amino acid sequence of a polypeptide is the: a) Primary structure b) Secondary structure c) Tertiary structure d) Quaternary structure e) None of these Answer: a Section: 12.1 Protein Structure Difficulty: Easy 12) The association of two or more polypeptides in a multimeric protein is know as the: a) Primary structure b) Secondary structure c) Tertiary structure d) Quaternary structure e) None of these Answer: d Section: 12.1 Protein Structure Difficulty: Easy

13) Proteins that help nascent polypeptides form the correct three dimensional structure are known as: a) Guide proteins b) Chaperone proteins c) Folding proteins d) Tertiary proteins e) Scaffold proteins Answer: b Section: 12.1 Protein Structure Difficulty: Easy 14) Which of the following is a secondary structure often observed in proteins? a) α helix b) β sheet c) α subunit d) α helix and β sheet e) All of these Answer: d Section: 12.1 Protein Structure Difficulty: Easy 15) Which of the following statements about protein structure is correct? a) The primary structure of a protein does not dictate the final shape of the protein. b) Two common types of secondary structure are α helices and β sheets, both of which are held in place via hydrogen bonds. c) Tertiary protein structure is maintained primarily by covalent disulfide bridges. d) All peptides exhibit four levels of structural organization: primary, secondary, tertiary, and quaternary. e) Ionic bonds in tertiary structure maintain a strong attraction between amino acids in the interior of living cells.

Answer: b Section: 12.1 Protein Structure Difficulty: Medium 16) Which type(s) of RNA molecules are involved in the translation process? a) mRNA b) tRNA c) rRNA d) mRNA and tRNA e) All of these Answer: e Section: 12.3 Protein Synthesis: Translation Difficulty: Easy

17) Which set of enzymes is responsible for attaching the correct amino acid to an appropriate tRNA molecule during the charging process? a) RNA gyrases b) RNA polymerases c) Amino-acyl transacetylases d) Aminoacyl tRNA synthetases e) All of these Answer: d Section: 12.3 Protein Synthesis: Translation Difficulty: Easy 18) Of the 64 possible nucleotide codon triplets, how many specify polypeptide chain termination? a) 61 b) 1 c) 2 d) 3 e) 64 Answer: d Section: 12.3 Protein Synthesis: Translation Difficulty: Easy 19) When a mRNA molecule is simultaneously translated by several ribosomes, the result is the formation of a/an: a) Polysome b) Polypeptie c) Ribosome d) Charged tRNA e) None of these Answer: a Section: 12.3 Protein Synthesis: Translation Difficulty: Easy 20) Which of the following statements about RNA involvement in translation is true? a) Aminoacyl-tRNA synthetases are responsible for attaching amino acids to the appropriate tRNAs. b) The nucleolus is responsible for the synthesis of tRNAs catalyzed by RNA Pol II. c) Eukaryotes, because of the complexity of their genome, are able to maintain multiple copies of rRNA genes. In contrast, the highly simplified genome of E. coli maintains a single copy. d) The adapter molecule of protein synthesis, the rRNA, recognizes the anticodons of the mRNA molecule to direct protein synthesis. e) None of these is correct.

Answer: a Section: 12.3 Protein Synthesis: Translation Difficulty: Medium

21) Which of the following is not a true statement about ribosomes? a) They are structured into large and small subunits b) They are comprised of RNA and protein c) The ribosomes in prokaryotes are typically larger than those found in eukaryotes d) The ribosome subunits associate during the initiation of translation e) All of these are false statements Answer: c Section: 12.3 Protein Synthesis: Translation Difficulty: Medium 22) Which of the following is the correct anti-codon sequence that corresponds with the codon sequence AUG? a) AUG b) GUA c) TAC d) UAC e) CAU Answer: d Section: 12.3 Protein Synthesis: Translation Difficulty: Easy 23) Unusual nucleosides that are found in mature tRNA molecules are added during: a) Transcription initiation b) Transcription elongation c) Transcription termination d) Post transcriptional modification e) It is not currently known when the unusual nucleosides are added Answer: d Section: 12.3 Protein Synthesis: Translation Difficulty: Easy 24) Which of the following is an unusual nucleoside that is only found in tRNA? a) Adenine b) Inosine c) Proline d) Phenylalanine e) Guanine Answer: b Section: 12.3 Protein Synthesis: Translation Difficulty: Easy

25) In the process of coupling tRNAs to amino acids, which of the following statements is incorrect? a) Amino acids attach to the 3' end of tRNAs. b) Prior to attachment between tRNA and amino acid, the amino acid becomes linked to AMP by aminoacyl-tRNA synthetase. c) The amino acid-AMP linkage is broken to attach the amino acid to the tRNA and release AMP by aminoacyl-tRNA synthetase. d) Aminoacyl-tRNA synthetase is responsible for matching the correct tRNA with the appropriate amino acid. e) None of these.

Answer: e Section: 12.3 Protein Synthesis: Translation Difficulty: Medium 26) Which of the following is a tRNA binding site on a ribosome? a) P b) A c) E d) P and A e) All of these Answer: e Section: 12.3 Protein Synthesis: Translation Difficulty: Easy 27) The initiation of translation with the 30S ribosomal subunit in E. coli does not involve which of the following: a) tRNAfmet b) Initiation factors IF-1, IF-2, and IF-3 c) ATP d) GTP e) mRNA with the appropriate Shine-Dalgarno sequence complementary to a sequence found within the 16S rRNA Answer: c Section: 12.3 Protein Synthesis: Translation Difficulty: Medium

28) Which of the following is a difference in the process of initiation of translation between prokaryotes and eukaryotes?

1. The amino group of the methionine on the initiator tRNA is not formylated in eukaryotes as it is in prokaryotes.

2. The initiation complex forms at the 5′ terminus of the mRNA, not at the Shine-Dalgarno/AUG translation start site as in prokaryotes.

3. The process takes place at the ribosome in prokaryotes but not in eukaryotes. a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: d Section: 12.3 Protein Synthesis: Translation Difficulty: Medium 29) Which of the following codons is most often recognized as the start codon for translation? a) GUA b) AUG c) CUG d) UUG e) AAG Answer: b Section: 12.3 Protein Synthesis: Translation Difficulty: Easy 30) Eukaryotic initiation of translation is similar to prokaryotes in which of the following ways? a) There is a formyl group on the initiation methionine b) Initiation complex forms at the Shine-Dalgarno/AUG translation start site c) After binding, the ribosome scans the mRNA and begins translation at the first available AUG d) The initiator methionyl-tRNA interacts with a soluble initiation factor and enters the P-site directly e) All of these Answer: d Section: 12.3 Protein Synthesis: Translation Difficulty: Medium

31) The initiation complex moves in which direction along the mRNA molecule? a) 5' → 3' b) 3' → 5' c) 1'→ 4' d) 5' → 3' and 3' → 5' e) None of these Answer: a Section: 12.3 Protein Synthesis: Translation Difficulty: Easy 32) Which of the following is not directly required for chain elongation during translation? a) Ef-Tu b) A site on the ribosome c) GTP d) IF2 e) All of these are required Answer: d Section: 12.3 Protein Synthesis: Translation Difficulty: Easy 33) The reaction in which the growing chain from the tRNA in the P site covalently joins the chain to the tRNA in the A site is catalyzed by which of the following enzymes? a) Gyrase b) Topoisomerase c) Peptidyl transferase d) Aminoacyl-tRNA synthetase e) Ligase Answer: c Section: 12.3 Protein Synthesis: Translation Difficulty: Easy 34) During translocation the peptidyl-tRNA present in the A site of the ribosome is translocated to the P site, and the uncharged tRNA in the P site is translocated to the E site, as the ribosome moves three nucleotides toward the 3′ end of the mRNA molecule. Which of the following soluble factors is required for this process to occur? a) IF-1 b) EF-Tu c) EF-G d) IF-2 e) IF-3 Answer: c Section: 12.3 Protein Synthesis: Translation Difficulty: Medium

35) Polypeptide chain elongation and termination of translation differ in prokaryotes and eukaryotes in which of the following ways? a) Binding of the aminoacyl-tRNA to the A-site b) Number of release factors c) Transfer of the growing peptide from the tRNA in the P-site to the tRNA in the A-site d) Translocation requirement of GTP hydrolysis e) Translocation of the ribosome along the mRNA to position the next codon in the A-site Answer: b Section: 12.3 Protein Synthesis: Translation Difficulty: Medium 36) Which of the following is a chain termination codon? a) UAA b) UAG c) UGA d) UAA and UAG e) All of these Answer: e Section: 12.3 Protein Synthesis: Translation Difficulty: Easy 37) The genetic code is: a) Composed of triplet codons b) Non-overlapping c) Degenerate d) Universal e) All of these Answer: e Section: 12.4 The Genetic Code Difficulty: Easy 38) Most amino acids are specified by two to four different codons. This means that the genetic code is: a) Non-overlapping b) Degenerate c) Ordered d) Non-degenerate e) Universal Answer: b Section: 12.4 The Genetic Code Difficulty: Easy

39) With few exceptions, all 64 codons have the same meaning in all organisms. This means that the genetic code is: a) Non-overlapping b) Degenerate c) Ordered d) Non-degenerate e) Universal Answer: e Section: 12.4 The Genetic Code Difficulty: Easy 40) What type of bond occurs between the nucleotide bases in the codon and the anticodon? a) Ionic b) Covalent c) Hydrogen d) Disulfide bridges e) Peptide Answer: c Section: 12.5 Codon-tRNA Interactions Difficulty: Easy 41) The hydrogen bonding between the bases in the anticodons of tRNAs and the corresponding codons of mRNAs follows strict base-pairing rules only for the first two bases, while the base-pairing for the third base of the codon is less stringent. This is known as the: a) Crick hypothesis b) tRNA hypothesis c) Wobble hypothesis d) Central Dogma e) None of these Answer: c Section: 12.5 Codon-tRNA Interactions Difficulty: Easy 42) Based upon the Wobble hypothesis, when inosine is present at the 5' end of the anitocodon, with which of the following bases will it base pair? a) Adenine b) Uracil c) Cytosine d) Adenine and Uracil e) All of these Answer: e Section: 12.5 Codon-tRNA Interactions Difficulty: Easy

43) Which position on the anticodon is the wobble position? a) 5' end b) 3' end c) Middle of the anticodon d) 5' end or 3' end e) Any position can be the wobble position Answer: a Section: 12.5 Codon-tRNA Interactions Difficulty: Easy 44) Which of the following is a type of mutation that can cause altered codon recognition? a) Suppressor mutations b) Nonsense mutations c) Missense mutations d) Suppressor mutations and Nonsense mutations e) All of these Answer: e Section: 12.5 Codon-tRNA Interactions Difficulty: Easy 45) If a tRNAala alters its anticodon from 3'-AUC-5' to the sequence to 3'-AUU-5', which of the following mutations can be suppressed? a) Missense b) Missense only if alanine does not adversely affect protein structure c) Silent d) Nonsense (only the 5'-UAA-3' codon) e) None of these Answer: d Section: 12.5 Codon-tRNA Interactions Difficulty: Easy Question Type: Essay 46) Briefly summarize the roles that each of the components play in the process of translation.

Answer: The mRNA molecules provide the specifications for the amino acid sequences of the polypeptide gene products. The ribosomes provide many of the macromolecular components required for the translation process. The tRNAs provide the adaptor molecules needed to incorporate amino acids into polypeptides in response to codons in mRNAs. In addition, several soluble proteins participate in the process Section: 12.3 Protein Synthesis: Translation Difficulty: Medium

47) Explain, in detail, how translation is initiated in E. coli.

Answer: In E. coli, the initiation process involves the 30S subunit of the ribosome, a special initiator tRNA, an mRNA molecule, three soluble protein initiation factors: IF-1, IF-2, and IF-3, and one molecule of GTP. Translation occurs on 70S ribosomes, but the ribosomes dissociate into their 30S and 50S subunits each time they complete the synthesis of a polypeptide chain. In the first stage of the initiation of translation, a free 30S subunit interacts with an mRNA molecule and the initiation factors. The 50S subunit joins the complex to form the 70S ribosome in the final step of the initiation process. The synthesis of polypeptides is initiated by a special tRNA, designated tRNAfMet, in response to a translation initiation codon (usually AUG, sometimes GUG). Polypeptide chain initiation begins with the formation of two complexes: (1) one contains initiation factor IF-2 and methionyl-tRNAfMet, and (2) the other contains an mRNA molecule, a 30S ribosomal subunit and initiation factor IF-3. The formation of the 30S subunit/mRNA complex depends in part on base-pairing between a nucleotide sequence near the 3′ end of the 16S rRNA and a sequence near the 5′ end of the mRNA molecule. Prokaryotic mRNAs contain a conserved polypurine tract, consensus AGGAGG, located about seven nucleotides upstream from the AUG initiation codon. This conserved hexamer, called the Shine-Dalgarno sequence after the scientists who discovered it, is complementary to a sequence near the 3′ terminus of the 16S ribosomal RNA. The IF-2/methionyl-tRNAfMet complex and the mRNA/30S subunit/IF-3 complex subsequently combine with each other and with initiation factor IF-1 and one molecule of GTP to form the complete 30S initiation complex. The final step in the initiation of translation is the addition of the 50S subunit to the 30S initiation complex to produce the complete 70S ribosome. Initiation factor IF-3 must be released from the complex before the 50S subunit can join the complex. The addition of the 50S ribosomal subunit to the complex positions the initiator tRNA, methionyl-tRNAfMet, in the peptidyl (P) site with the anticodon of the tRNA aligned with the AUG initiation codon of the mRNA. With the initiator AUG positioned in the P site, the second codon of the mRNA is in register with the A site, dictating the aminoacyl-tRNA binding specificity at that site and setting the stage for the second phase in polypeptide synthesis, chain elongation Section: 12.3 Protein Synthesis: Translation Difficulty: Medium 48) Summarize the steps in chain elongation during translation.

Answer: The addition of each amino acid to the growing polypeptide occurs in three steps: (1) binding of an aminoacyl-tRNA to the A site of the ribosome, (2) transfer of the growing polypeptide chain from the tRNA in the P site to the tRNA in the A site by the formation of a new peptide bond, and (3) translocation of the ribosome along the mRNA to position the next codon in the A site (Figure 12.17). During step 3, the nascent polypeptide-tRNA and the uncharged tRNA are translocated from the A and P sites to the P and E sites, respectively. These three steps are repeated in a cyclic manner throughout the elongation process.

Section: 12.3 Protein Synthesis: Translation Difficulty: Medium

49) Explain how chain termination occurs during translation, including the differences between chin termination in prokaryotes and eukaryotes.

Answer: Polypeptide chain elongation undergoes termination when any of three chain-termination codons (UAA, UAG, or UGA) enters the A site on the ribosome (Figure 12.19). These three stop codons are recognized by soluble proteins called release factors (RFs). In E. coli, there are two release factors, RF-1 and RF-2. RF-1 recognizes termination codons UAA and UAG; RF-2 recognizes UAA and UGA. In eukaryotes, a single release factor (eRF) recognizes all three termination codons. The presence of a release factor in the A site alters the activity of peptidyl transferase such that it adds a water molecule to the carboxyl terminus of the nascent polypeptide.

This reaction releases the polypeptide from the tRNA molecule in the P site and triggers the translocation of the free tRNA to the E site. Termination is completed by the release of the mRNA molecule from the ribosome and the dissociation of the ribosome into its subunits. The ribosomal subunits are then ready to initiate another round of protein synthesis, as previously described.

Section: 12.3 Protein Synthesis: Translation Difficulty: Medium 50) Explain the Wobble hypothesis and discuss the predictions based upon this hypothesis that have proven true.

Answer: The hydrogen bonding between the bases in the anticodons of tRNAs and the codons of mRNAs follows strict base-pairing rules only for the first two bases of the codon. The base-pairing involving the third base of the codon is less stringent, allowing what Crick has called wobble at this site. On the basis of molecular distances and steric (three-dimensional structure) considerations, Crick proposed that wobble would allow several types, but not all types, of base-pairing at the third codon base in the codon–anticodon interaction. His proposal has since been strongly supported by experimental data. Table 12.2 shows the base-pairing predicted by Crick's wobble hypothesis.

The wobble hypothesis predicted the existence of at least two tRNAs for each amino acid with codons that exhibit complete degeneracy, and this has proven to be true. The wobble hypothesis also predicted the occurrence of three tRNAs for the six serine codons. Three serine tRNAs have been characterized: (1) tRNASer1 (anticodon AGG) binds to codons UCU and UCC, (2) tRNASer2 (anticodon AGU) binds to codons UCA and UCG, and (3) tRNASer3 (anticodon UCG) binds to codons AGU and AGC. These specificities were verified by the trinucleotide-stimulated binding of purified aminoacyl-tRNAs to ribosomes in vitro. Finally, several tRNAs contain the base inosine, which is made from the purine hypoxanthine. Inosine is produced by a posttranscriptional modification of adenosine. Crick's wobble hypothesis predicted that when inosine is present at the 5′ end of an anticodon (the wobble position), it would base-pair with uracil, cytosine, or adenine in the codon. In fact, purified alanyl-tRNA containing inosine (I) at the 5′ position of the anticodon (see Figure 12.11) binds to ribosomes activated with GCU, GCC, or GCA trinucleotides (Figure 12.21). The same result has been obtained with other purified tRNAs with inosine at the 5 position of the anticodon. Thus, Crick's wobble hypothesis nicely explains the relationships between tRNAs and codons given the degenerate, but ordered, genetic code Section: 12.4 The Genetic Code Difficulty: Medium

Chapter 13 Test Bank Question Type: Multiple Choice 1) Living cells contain numerous enzymes that constantly scan DNA, searching for damaged or incorrectly paired nucleotides. When detected, these defects are corrected by: a) DNA scanning enzymes b) RNA polymerase c) DNA repair enzymes d) DNA polymerase e) Ligase Answer: c Section: 13.1 Mutation Difficulty: Easy 2) Both the change in the genetic material and the process by which the change occurs is referred to as a/an: a) Mutation b) Error c) Reparation d) Mutation and Error e) All of these Answer: a Section: 13.1 Mutation Difficulty: Easy 3) An organism that exhibits a novel phenotype resulting from a mutation is called a: a) Wild type b) Mutant c) Positive phenotype d) Negative phenotype e) None of these Answer: b Section: 13.1 Mutation Difficulty: Easy 4) Mutations that involve changes at specific sites in a gene are referred to as: a) Frameshift mutations b) Nonsense mutations c) Misssense mutations d) Point mutations e) None of these Answer: d Section: 13.1 Mutation Difficulty: Easy

5) Which of the following is considered a point mutation?

1. Substitution of the base A for the base C 2. Deletion of the base T 3. Insertion of the base G a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 13.1 Mutation Difficulty: Medium 6) Which of the following statements is not true about mutation? a) Mutation refers to both change in the genetic material and the process by which the change occurs b) Point mutations involve changes at specific sites in a gene c) If a mutation occurs in a somatic cell, the mutant phenotype will be passed on to the offspring of that organism d) The intrinsic mistake frequency in nucleotide selection during DNA polymerization is 10-5 e) None of these Answer: c Section: 13.1 Mutation Difficulty: Medium 7) Which of the following organisms has never experienced a genetic mutation? a) E. coli b) Humans c) Drosophila d) T4 virus e) All of these have experienced genetic mutations Answer: e Section: 13.1 Mutation Difficulty: Easy 8) Which cell type experiences germinal mutations? a) Squamous Epithelial b) Chondrocytes c) Gametes d) Cuboidal Epithelial e) Columnar Epithelial Answer: c Section: 13.1 Mutation Difficulty: Easy

9) If a mutation occurs in a somatic cell, the resulting mutant phenotype will occur: a) Only in the individual cell b) Only in the progeny from that individual cell c) Only in the offspring of that organism d) In both the progeny of that individual cell and the individual cell itself e) Neither the progeny from that individual cell or the offspring of the organism Answer: d Section: 13.1 Mutation Difficulty: Easy 10) Mutations that occur without a known cause are known as: a) Spontaneous mutations b) Induced mutations c) Harmful mutations d) Adaptive mutations e) None of these Answer: a Section: 13.1 Mutation Difficulty: Easy 11) Which of the following is a known mutagen? a) UV light b) Thalidomide c) Irradiation d) UV light and Irradiation e) All of these Answer: e Section: 13.1 Mutation Difficulty: Easy 12) A backcross can be used to distinguish between a) A forward versus reverse mutation b) A somatic versus germinal mutation c) A back mutation versus a suppressor mutation d) All of the above e) None of the above Answer: d Section: 13.1 Mutation Difficulty: Medium

13) A mutation that provides a selective advantage to the mutant organism when grown in the environment in which it originated is known as a/an: a) Adaptive mutation b) Forward mutation c) Suppressor mutation d) Missense mutation e) Nonsense mutation Answer: a Section: 13.1 Mutation Difficulty: Easy 14) Gene X undergoes a mutation which converts it from wild-type to mutant. Later, a second mutation in the genome in gene Y occurs which causes the wild-type phenotype of the first gene (gene X) to be restored. Respectively, what are the appropriate designations for the two mutational events? a) Forward mutation, back mutation b) Forward mutation, suppressor mutation c) Reverse mutation, back mutation d) Reverse mutation, suppressor mutation e) Reverse mutation, forward mutation Answer: b Section: 13.1 Mutation Difficulty: Medium 15) Restoration of the wild type phenotype in a mutant organism can result from which of the following types of mutations? a) Back mutation b) Suppressor mutation c) Forward mutation d) Back mutation and Suppressor mutation e) Suppressor mutation and Forward mutation Answer: d Section: 13.1 Mutation Difficulty: Easy

16) Hemoglobin is a model system to illustrate the deleterious effects of mutations. Which of the following statements about deleterious mutations in human hemoglobin genes is false? a) Hemoglobin A and hemoglobin S differ at only one amino acid position b) In hemoglobin S, the substitution of valine for glutamic acid causes an aggregation of hemoglobin molecules not found in hemoglobin A c) The molecular difference between hemoglobin A and hemoglobin S was identified by protein sequence before nucleotide sequence d) The valine to glutamic acid substitution is the only amino acid change so far identified in the  chain of hemoglobin e) Homozygous (Hbsb/Hbsb) leads to severe hemolytic anemia, which is often fatal.

Answer: d Section: 13.1 Mutation Difficulty: Medium 17) Genes containing mutations with no effect on phenotype or small effects that can be recognized only by special techniques are called: a) Isomers b) Isoalleles c) Null alleles d) Lethal alleles e) Forward alleles Answer: b Section: 13.2 The Molecular Basis of Mutation Difficulty: Easy 18) Mutations that result in no gene product or totally nonfunctional gene products are called: a) Isomers b) Isoalleles c) Null alleles d) Lethal alleles e) Forward alleles Answer: c Section: 13.2 The Molecular Basis of Mutation Difficulty: Easy 19) Which of the following mutations is described correctly? a) Phenylketonuria: An autosomal recessive disease resulting in the overproduction of phenyalanine hydroxylase b) Alkaptonuria: An autosomal recessive disease which inactivates homogentistic acid oxidase c) Albinism: An autosomal dominant trait, which blocks the synthesis of melanin from tyrosine d) Tyrosinemia: A lack of a tyrosine catabolic enzyme, which leads to an decrease in tyrosine in the blood and urine e) All of these Answer: b Section: 13.2 The Molecular Basis of Mutation Difficulty: Medium

20) A mutation that has no phenotypic effect on the organism is known as: a) Neutral mutation b) Null mutation c) Isomer mutation d) Lethal allele e) All of these Answer: a Section: 13.2 The Molecular Basis of Mutation Difficulty: Easy 21) Mutations that are lethal in one environment but viable in another are known as: a) Lethal mutations b) Conditional lethal mutations c) Null mutations d) Isoallele mutations e) All of these Answer: b Section: 13.2 The Molecular Basis of Mutation Difficulty: Easy 22) Which of the following is not exhibited Fragile X syndrome? a) An X chromosome which contains up to 1000 copies of the CCG repeat at one site b) Inherited mental retardation c) A correlation between the number of repeats and the severity of the defects d) The phenomenon of "anticipation" e) Decreasing numbers of the CCG repeat in successive generations Answer: e Section: 13.2 The Molecular Basis of Mutation Difficulty: Medium 23) Chemical mutations that change the position of hydrogen atoms in a nitrogenous base are known as: a) Isomer mutations b) Isomer shifts c) Tautomeric shifts d) Isomer mutations and Isomer shifts e) All of these Answer: c Section: 13.2 The Molecular Basis of Mutation Difficulty: Easy

24) Which base pair combinations can form when nitrogenous bases are present in their rare imino or enol states? a) A:T b) C:G c) A:C d) A:T and C:G e) C:G and A:C Answer: c Section: 13.2 The Molecular Basis of Mutation Difficulty: Easy 25) What is the net effect of a tautomeric shift event, and the subsequent replication required to segregate the mismatched base pair?

1. A:T to G:C base pair substitution 2. G:C to A:T base pair substitution 3. A:C to G:T base pair substitution a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: d Section: 13.2 The Molecular Basis of Mutation Difficulty: Medium 26) Base-pair substitutions in which the purine in one strand of DNA is replaced with a different purine, and the pyrimidine in the complementary strand is replaced with a different pyrimidine are known as: a) Isomers b) Transitions c) Transversions d) Inversions e) None of these Answer: b Section: 13.2 The Molecular Basis of Mutation Difficulty: Easy

27) Base-pair substitutions involving the replacement of a purine with a pyrimidine and vice versa are called: a) Isomers b) Transitions c) Transversions d) Inversions e) None of these Answer: c Section: 13.2 The Molecular Basis of Mutation Difficulty: Easy 28) Base-pair additions and deletions that alter the reading frame of all subsequent base-pair triplets are collectively referred to as: a) Inversions b) Transversions c) Frameshift mutations d) Point mutations e) Codon mutations Answer: c Section: 13.2 The Molecular Basis of Mutation Difficulty: Easy 29) Some mutagenic agents only affect replicating DNA; others are mutagenic to replicating and non-replicating DNA. Which of the following mutagens does not belong to the second class? a) Alkylating agents b) Base analogs c) Nitroud acid d) All of these e) None of these Answer: b Section: 13.3 Mutagenesis Difficulty: Medium 30) Which type of mutation can be caused by acridine dyes? a) Transversions b) Transitions c) Frameshift d) Chromosomal e) All of these Answer: c Section: 13.3 Mutagenesis Difficulty: Easy

31) Which of the following statements about radiation induced mutations is true? a) X rays, cosmic rays, and UV rays are all classified as ionizing radiation b) Ionizing radiation results in raising electrons to an atom's outer orbitals, a state referred to as excitation c) In mammals, chronic irradiation is as effective in inducing mutations as acute irradiation d) UV radiation results in the formation of purine dimers and purine hydrates e) X rays can result in gross changes of chromosome structure, such as large deletions, duplications and inversions Answer: e Section: 13.3 Mutagenesis Difficulty: Medium 32) Bruce Ames and coworkers developed an inexpensive and sensitive method for testing the mutagenicity of chemicals with histidine auxotrophic mutants of: a) E. coli b) S. aureus c) B. subtilis d) Salmonella e) C. elegans Answer: d Section: 13.3 Mutagenesis Difficulty: Easy 33) Bruce Ames constructed a mutagenicity test which initially missed carcinogens which were noncarcinogenic to the tester strains. Why? a) Tester strains were more resistant to the mutagenic effects of the carcinogen than eukaryotes b) Not all carcinogens are mutagens c) Potential carcinogens needed to be modified in eukaryotes before they were mutagenic d) Histidine biosynthesis was inhibited by the presence of the carcinogens. e) All of these Answer: c Section: 13.3 Mutagenesis Difficulty: Medium 34) Which of the following mutations or techniques was not used in deducing the pathway of T4 morphogenesis? a) Auxitrophic mutations b) Temperature-sensitive mutations c) Suppressor-sensitive mutations d) Electron micrsoscopy e) Biochemical analysis Answer: a Section: 13.4 Assigning Mutations to Genes by the Complementation Test Difficulty: Easy

35) Which of the following DNA alterations can be corrected by light-dependent repair? a) Methylation b) Thymine dimers c) Mismatched basepairing d) Hydroxylation e) Inversions Answer: b Section: 13.5 DNA Repair Mechanisms Difficulty: Easy 36) Which of the following statements about excision repair is correct? a) Base excision repair is initiated by DNA glycosylases that recognize abnormal deoxyriboses in DNA b) Nucleotide excision repair removes large regions of DNA via an exonuclease which cuts on either side of the damaged bases c) E. coli exonuclease activity is carried out by uvrA, uvrB, and uvrC, and the resulting gap is filled in by DNA Pol III d) DNA glycosylases cleave the altered nucleoside (base and sugar) from the DNA backbone creating an apurinic or apyrimidinic site e) In humans, a mechanism similar to E. coli is carried out where the protein XPA acts as the exonuclease.

Answer: b Section: 13.5 DNA Repair Mechanisms Difficulty: Medium 37) Which of the following statements about the mismatch repair pathway is not correct? a) MutH contains a GATC-specific endonuclease activity b) Cleavage of the unmethylated strand only occurs 3' to the mismatch c) This system of repair utilizes MutH, MutL, MutS, and UvrD d) The repair system can distinguish the template strand from the newly synthesized strand e) All of these are correct Answer: b Section: 13.5 DNA Repair Mechanisms Difficulty: Medium 38) Which DNA repair mechanism can be template-independent and require polymerase activity? a) Post-replication repair b) Mismatch repair c) Light-dependent repair d) Error-prone repair e) Excision repair Answer: d Section: 13.5 DNA Repair Mechanisms Difficulty: Easy

39) Which of the following molecules carries out single-strand assimilation? a) XPA b) MutD c) RecA d) RecBCD e) UvrD Answer: c Section: 13.5 DNA Repair Mechanisms Difficulty: Easy 40) Which of the following enzymes performs light dependent repair? a) DNA gyrase b) DNA polymerase c) DNA photolyase d) RNA polymerase e) DNA helicase Answer: c Section: 13.5 DNA Repair Mechanisms Difficulty: Easy 41) Which of the following disease is not associated with a defect in a DNA repair pathway? a) Bloom syndrome b) Werner syndrome c) Rothmund-Thomson syndrome d) Cystic Fibrosis e) Fanconi anemia Answer: d Section: 13.5 DNA Repair Mechanisms Difficulty: Easy 42) Which of the following proteins is vital to the process of recombination? a) RecA b) CroA c) HemA d) Gyrase e) Topoisomerase Answer: a Section: 13.6 DNA Recombination Mechanisms Difficulty: Easy

43) Many of the currently popular models of crossing over were derived from a model proposed by: a) James Watson b) Robin Holliday c) Francis Crick d) Rosalind Franklin e) None of these Answer: b Section: 13.6 DNA Recombination Mechanisms Difficulty: Easy 44) X-shaped recombination intermediates formed from Holliday model recombination are known as: a) Tai molecules b) Chi forms c) Holliday forms d) X forms e) None of these Answer: b Section: 13.6 DNA Recombination Mechanisms Difficulty: Easy 45) Gene conversion of Neurospora requires recombination and which of the following events? a) Reversion b) Suppression c) Excision repair d) Heteroduplex formation e) Inversion Answer: d Section: 13.6 DNA Recombination Mechanisms Difficulty: Easy

Question Type: Essay 46) Briefly explain how mutant polypeptides can cause blocks in a metabolic pathway. Please use a real-life example.

Answer:

If Gene A is mutated and can not form an active polypeptide form of Enzyme A then Intermediate Y will not be produced thereby blocking the production of the end product Z. The same could be true if Gene B is mutated. An example of this phenomenon in humans is the metabolism of the aromatic amino acids phenylalanine, because some of the early studies of mutations in humans revealed blocks in this pathway. Phenylalanine is an essential amino acid required for protein synthesis; it is not synthesized de novo in humans as in microorganisms.

Thus, it must be obtained from dietary proteins. The best-known inherited defect in phenylalanine-tyrosine metabolism is phenylketonuria, which is caused by the absence of phenylalanine hydroxylase, the enzyme that converts phenylalanine to tyrosine. Newborns with phenylketonuria, an autosomal recessive disease, develop severe mental retardation if not placed on a diet low in phenylalanine. The first inherited disorder in the phenylalanine-tyrosine metabolic pathway to be studied in humans was alkaptonuria, which is caused by autosomal recessive mutations that inactivate the enzyme homogentisic acid oxidase.

Section: 13.1 Mutation Difficulty: Medium

47) Explain, in detail, how the hemoglobin mutants provide excellent examples of deleterious mutations and how they show that mutation is a process in which changes in gene structure, can cause changes in the amino acid sequences of the polypeptide gene products.

Answer: Many different variants of adult hemoglobin have been identified in human populations, and several of them have severe phenotypic effects. Many of the variants were initially detected by their altered electrophoretic behavior. The hemoglobin variants provide an excellent illustration of the effects of mutation on the structures and functions of gene products and, ultimately, on the phenotypes of the affected individuals When the amino acid sequences of the  chains of hemoglobin A and the hemoglobin in patients with sickle-cell anemia (hemoglobin S) were determined and compared, hemoglobin S was found to differ from hemoglobin A at only one position. The sixth amino acid from the amino terminus of the  chain of hemoglobin A is glutamic acid (a negatively charged amino acid). The  chain of hemoglobin S contains valine (no charge at neutral pH) at that position. The  chains of hemoglobin A and hemoglobin S are identical. Thus, the change of a single amino acid in one polypeptide can have severe effects on the phenotype. In the case of hemoglobin S, the substitution of valine for glutamic acid at the sixth position in the  chain allows a new bond to form, which changes the conformation of the protein and leads to aggregation of hemoglobin molecules. This change results in the grossly abnormal (sickle) shape of the red blood cells. The mutational change in the HbA allele that gave rise to HbS was a substitution of a T:A base pair for an A:T base pair, with a T in the transcribed strand in the first case and an A in the transcribed strand in the second case (see Figure 1.9). This A:T → T:A base-pair substitution was first predicted from protein sequence data and the known codon assignments, and was later verified by sequencing the HbA and HbS alleles.

Section: 13.1 Mutation Difficulty: Medium

48) How do tautomeric shifts cause alterations in the pairing potential of nitrogenous bases?

Answer: Watson and Crick pointed out that the structures of the bases in DNA are not static.

Hydrogen atoms can move from one position in a purine or pyrimidine to another position—for example, from an amino group to a ring nitrogen. Such chemical fluctuations are called tautomeric shifts. Although tautomeric shifts are rare, they may be of considerable importance in DNA metabolism because some alter the pairing potential of the bases. The nucleotide structures that were discussed in Chapter 9 are the common, more stable forms, in which adenine always pairs with thymine and guanine always pairs with cytosine. The more stable keto forms of thymine and guanine and the amino forms of adenine and cytosine may infrequently undergo tautomeric shifts to less stable enol and imino forms, respectively. The bases would be expected to exist in their less stable tautomeric forms for only short periods of time. However, if a base existed in the rare form at the moment that it was being replicated or being incorporated into a nascent DNA chain, a mutation would result. When the bases are present in their rare imino or enol states, they can form adenine-cytosine and guanine-thymine base pairs. The net effect of such an event, and the subsequent replication required to segregate the mismatched base pair, is an A:T to G:C or a G:C to A:T base-pair substitution. Mutations resulting from tautomeric shifts in the bases of DNA involve the replacement of a purine in one strand of DNA with the other purine and the replacement of a pyrimidine in the complementary strand with the other pyrimidine. Such base-pair substitutions are called transitions. Base-pair substitutions involving the replacement of a purine with a pyrimidine and vice versa are called transversions. There are three substitutions—one transition and two transversions—possible for every base pair. A total of four different transitions and eight different transversions are possible.

Section: 13.2 The Molecular Basis of Mutation Difficulty: Medium 49) Briefly explain how stationary phase mutagenesis occurs in bacteria such as E. coli.

Answer: It occurs when populations of bacteria quit growing—enter the stationary phase—due to starvation or some other environmental stress. When bacteria such as E. coli enter stationary phase due to starvation or other environmental stress, an error-prone DNA repair pathway called the SOS response is induced. During the SOS response, a large number of genes that encode proteins involved in DNA metabolism—replication, recombination, and repair—are turned on.

Some of the induced proteins are involved in the repair of damaged DNA by recombination; others are error-prone DNA polymerases (IV and V in E. coli) that replicate past damaged segments of DNA, and, in so doing, produce mutations.

Section: 13.3 Mutagenesis Difficulty: Medium

50) How does UV light induce mutations in DNA as it is a non-ionizing form of radiation?

Answer: Ultraviolet (UV) radiation does not possess sufficient energy to induce ionizations.

However, it is readily absorbed by many organic molecules such as the purines and pyrimidines in DNA, which then enter a more reactive or excited state. UV rays penetrate tissue only slightly.

Thus, in multicellular organisms, only the epidermal layer of cells usually is exposed to the effects of UV. However, ultraviolet light is a potent mutagen for unicellular organisms. The maximum absorption of UV by DNA is at a wavelength of 254 nm. Maximum mutagenicity also occurs at 254 nm, suggesting that the UV-induced mutation process is mediated directly by the absorption of UV by purines and pyrimidines. In vitro studies show that the pyrimidines absorb strongly at 254 nm and, as a result, become very reactive. Two major products of UV absorption by pyrimidines (thymine and cytosine) are pyrimidine hydrates and pyrimidine dimers. Thymine dimers cause mutations in two ways. (1) Dimers perturb the structure of DNA double helices and interfere with accurate DNA replication. (2) Errors occur during the cellular processes that repair defects in DNA, such as UV-induced thymine dimers.

Section: 13.3 Mutagenesis Difficulty: Medium

Chapter 14 Test Bank Question Type: Multiple Choice 1) Which of the following techniques allow researchers to isolate and characterize DNA sequences? a) Recombinant DNA techniques b) Gene cloning c) DNA amplification d) Gene cloning and DNA amplification e) All of these Answer: e Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 2) Small self replicating genetic elements used to clone genes are known as: a) Cloner sections b) Cloning vectors c) Cloning sequences d) Restriction enzymes e) Silencers Answer: b Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 3) Which of the following can be determined using a cloned gene?

1. A gene’s nucleotide sequence 2. A gene’s function 3. A gene’s amino acid sequence a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy

4) In a gene cloning procedure the GFP gene is inserted into a plasmid and the plasmid is then taken up through transformation into an E.coli cell where it replicates. In which portion of this gene cloning procedure is the recombinant DNA molecule constructed?

1. The GFP is inserted into the plasmid 2. The plasmid is taken up via transformation 3. The plasmid replicates in the E. coli cell a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: a Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium 5) In a gene cloning procedure the GFP gene is inserted into a plasmid and the plasmid is then taken up through transformation into an E.coli cell where it replicates. In which portion of this procedure is gene cloning actually taking place?

1. The GFP is inserted into the plasmid 2. The plasmid is taken up via transformation 3. The plasmid replicates in the E. coli cell a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: c Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium 6) In a certain procedure, short DNA strands that are complementary to DNA sequences on either side of the gene or DNA sequence of interest are synthesized and used to initiate its amplification in vitro by a special (heat-stable) DNA polymerase. The procedure is known as: a) Telomerase Chain Reaction b) Polymerase Chain Reaction c) Gene cloning d) DNA fingerprinting e) Restriction Fragment Length Polymorphism Answer: b Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy

7) Which of the following enzymes is required for sequencing a gene? a) RNA polymerases b) DNA gyrases c) Restriction endonucleases d) Lactases e) None of these Answer: c Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 8) Which of the following types of restriction enzymes cleave DNA at only specific nucleotide sequences known as restriction sites, regardless of the source of the DNA? a) Type I b) Type II c) Type III d) Type IV e) Type V Answer: b Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 9) What is the natural biological function of restriction endonucleases? a) To clone genes in eukaryotic sequences b) To protect viruses from bacterial invasion c) To protect bacteria from viral invasion d) To protect eukaryotes from bacterial invasion e) To clone genes in prokaryotic sequences Answer: c Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 10) What method allows bacteria to protect endogenous restriction sites from being cleaved by restriction enzymes? a) Methylation b) Amylation c) Glycosylation d) Methylation and Amylation e) None of these Answer: a Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy

11) Which of the following is not a feature of restriction enzymes? a) They recognize palindromic sequences b) They make staggered cuts c) They cut at random sites d) They are named after the species in which the enzyme is produced e) All of these are true Answer: c Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium 12) Which enzyme is used to join two complementary DNA fragments after a cut has been made? a) restriction endocnucleases b) RNA polymerase c) DNA gyrase d) DNA ligase e) Helicase Answer: d Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 13) Which of the following is true regarding restriction enzymes?

1. Methylation protects an organism’s DNA from cutting by restriction enzymes 2. Restriction enzymes are exonucleases 3. Restriction enzymes can make staggered or blunt cuts a) 1 b) 2 c) 3 d) 1 and 2 e) 1 and 3 Answer: e Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium

14) Why is it possible to join a piece of DNA from a human that has been digested using EcoRI and a piece of DNA from a bacterium that has been digested using EcoRI? a) They have been cut with the same restriction enzyme therefore the staggered ends are the same and are complementary b) Bacteria and humans exist in a symbiotic relationship and therefore share DNA and can be joined together c) They can join together only in the presence of high heat and pressure, such as is found in an autoclave. d) They cannot join together because the DNA has come from two different species and even though they are digested with the same restriction enzyme the cuts are at different locations. e) All of these Answer: a Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium 15) A DNA molecule containing DNA fragments from two or more different sources is known as a: a) Recombinant DNA molecule b) Mutated DNA molecule c) Biologically normal DNA molecule d) Wild type DNA molecule e) Homeric DNA molecule Answer: a Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 16) Which of the following is an essential component in a cloning vector? a) Origin of replication b) Selectable marker gene c) Unique restriction enzyme site d) Origin of replication and Unique restriction enzyme site e) All of these Answer: e Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 17) A cluster of unique restriction sites in a cloning vector is often referred to as a/an: a) Polylinker b) Polycloning site c) Unilinker d) Polylinker and Polycloning site e) All of these Answer: d Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy

18) Which of the following is not commonly used as a cloning vector? a) Plasmid b) Cosmid c) Phagemid d) Bacteriphage e) All of these are commonly used vectors Answer: e Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 19) Which of the following was used to create many of today's plasmid cloning vectors? a) pBR322 b) pGLO c) pBAC d) pGFP e) pBLU Answer: a Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 20) Most of today's bacteriophage cloning vectors are derived from which phage? a) SV40 b) T4 c) Lambda d) T2 e) pGLO Answer: c Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 21) Which portion of the lambda phage can be excised and used for the insertion of foreign DNA? a) The external portions required for lytic growth b) The central portion required for lysogenic growth c) The external portions required for lysogenic growth d) The central portion required for lytic growth e) No part can be excised, but extra DNA can be added in Answer: b Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy

22) Which type of vector would be most effective for inserting a foreign piece of DNA that is 13kb in size into E. coli? a) Plasmid b) Bacteriophage c) BAC d) Cosmid e) YAC Answer: b Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium 23) A cosmid can best be described as a hybrid of: a) A plasmid and a bacteriophage b) A plasmid and a BAC c) A bacteriophage and a phagemid d) A phagemid and a BAC e) None of these Answer: a Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 24) Which of the following is an advantage that the cosmid vector possesses?

1. Can replicate autonomously in E. coli 2. Has the packaging capacity of lambda chromosome 3. Efficient transformation in E. coli a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: e Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium 25) Which of the following statements about cosmid vectors is false? a) They can accept DNA inserts of 35-45kb in size b) They can replicate autonomously like a plasmid c) Their DNA is packaged in the heads of lambda phage particles d) They do not contain an origin of replication or a selectable marker gene e) All of these Answer: d Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium

26) Which of the following statements about phagemid vectors is incorrect? a) They contain components from both phage chromosomes and plasmids b) They require a helper phage to replicate and package single stranded DNA particles in phage heads c) An example of a phagemid vector is pUC118 d) Can allow researchers to perform directional cloning e) All of these Answer: e Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium 27) Which type of vector can replicate in both E. coli and another species? a) Plasmids b) Lambda phage c) Cosmids d) Shuttle vectors e) None of these Answer: d Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 29) Which of the following is found in a YAC? a) Yeast origin of replication b) Two yeast telomeres at the end of the minichromosome c) Yeast centromere d) Polylinker site e) All of these Answer: e Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 30) A disadvantage of cloning vectors is: a) An inability to accommodate eukaryotic DNA b) The absence of unique restriction sites c) An inability to be packaged in vitro d) A strict limit on the insert size (no larger than 10kb) e) The instability of large foreign DNA inserts Answer: e Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium

31) Filamentous single-stranded DNA phages infect cells by absorbing to which of the following structures? a) Cell membrane b) F pili c) Cell wall d) Nucleus e) Mitochondria Answer: b Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 32) Which of the following is the correct order of steps in PCR? a) Extension of DNA, denaturation of DNA, annealing of primers b) Denaturation of DNA, extension of DNA, annealing of primers c) Denaturation of DNA, annealing of primers, extension of DNA d) Annealing of primers, extension of DNA, denaturation of DNA e) Annealing of primers, denaturation of DNA, extension of DNA Answer: c Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium 33) The purpose of the primer sequence in PCR is to: a) Provide a promoter sequence for DNA polymerase b) Provide a free 3'-OH end required for covalent extension c) Allow for efficient denaturation of the double stranded DNA d) Keep the DNA denatured after the temperature is reduced e) All of these Answer: b Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Easy 34) If a PCR reaction begins using only one double stranded fragment of DNA, how many fragments will be produced after 12 cycles? a) 2 b) 12 c) 256 d) 1024 e) 4096 Answer: e Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium

35) Why is Taq polymerase used more commonly in PCR than DNA polymerase I from E. coli?

1. Taq polymerase remains active during the high heat denaturation step unlike DNA polymerase I 2. Taq polymerase is less expensive than DNA polymerase I 3. Taq polymerase must be added after each deanturation step unlike DNA polymerase I a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: a Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium 36) Which of the following is true about Taq polymerase?

1. It does not contain a built-in 3′ → 5′ proofreading activity 2. It produces a higher than normal frequency of replication errors 3. It is derived from a bacteria known as Thermus aquaticus a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: e Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium 37) Which of the following is a disadvantage of the PCR process? a) Fragments longer than 35kb cannot be efficiently amplified b) It replicates double stranded DNA exponentially c) It requires DNA to be denatured d) All of these e) None of these Answer: a Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium 38) What does the enzyme reverse transcriptase do? a) Degrades foreign DNA b) Synthesizes RNA complementary to a DNA template c) Synthesizes DNA complementary to an RNA template d) Inactivates restriction endonucleases e) Adds nucleotides to the end of DNA molecules Answer: c Section: 14.2 Construction and Screening of DNA Libraries Difficulty: Easy

39) DNA molecules that are synthesized from an RNA template are known as: a) cDNA molecules b) sDNA molecules c) gDNA molecules d) rDNA molecules e) All of these Answer: a Section: 14.2 Construction and Screening of DNA Libraries Difficulty: Easy 40) cDNA libraries contain only: a) The non coding regions of expressed genes b) The coding regions of expressed genes c) The non coding and coding regions of non-expressed genes d) All of these e) None of these Answer: c Section: 14.2 Construction and Screening of DNA Libraries Difficulty: Easy 41) Which of the following is a method by which a gene can be isolated from a DNA library? a) In situ colony hybridization b) Genetic complementation c) Genetic selection d) In situ plaque hybridization e) All of these Answer: e Section: 14.2 Construction and Screening of DNA Libraries Difficulty: Medium 42) When mutations are available in a gene of interest, the wild-type allele of the gene can be identified by a process called: a) In vitro hybridization b) Plaque hybridization c) Southern hybridization d) Complementation screening e) In vitro packaging Answer: d Section: 14.2 Construction and Screening of DNA Libraries Difficulty: Easy

43) If you were constructing the first strand of a cDNA library, you would need all of the following except: a) DNA ligase b) DNA template c) RNA template d) Poly-T oligomers e) Reverse transcriptase Answer: b Section: 14.2 Construction and Screening of DNA Libraries Difficulty: Medium 44) All of the following enzymes are used in the construction of cDNA libraries, except: a) Terminal transferase b) Ribonuclease H c) DNA polymerase I d) DNA Ligase e) Exonuclease Answer: e Section: 14.2 Construction and Screening of DNA Libraries Difficulty: Easy 45) Site specific mutagenesis is commonly and efficiently performed by which procedure? a) PCR b) Southern Blotting c) Nothern Blotting d) Western Blotting e) Complementation Testing Answer: a Section: 14.2 Construction and Screening of DNA Libraries Difficulty: Easy 46) A researcher is using gel electrophoresis to separate very small DNA fragments. What type of gel should the researcher use to perform this separation? a) Agarose b) Blood agar c) Acrymalide d) Agarose or Blood agar e) All of these are appropriate Answer: c Section: 14.3 The Molecular Analysis of DNA, RNA, and Protein Difficulty: Easy

47) Which of the following techniques is most appropriate for analyzing RNA fragments? a) Souther Blotting b) Northern Blotting c) RT-PCR d) Southern Blotting and Northern Blotting e) Northern Blotting and RT-PCR Answer: e Section: 14.3 The Molecular Analysis of DNA, RNA, and Protein Difficulty: Medium 48) If you wanted to identify and characterize a protein in a cellular extract, which of the following techniques would you use? a) Southern Blotting b) Northern Blotting c) Western Blotting d) Southern Blotting and Western Blotting e) All of these Answer: c Section: 14.3 The Molecular Analysis of DNA, RNA, and Protein Difficulty: Medium 49) A physical map of a chromosome can be created by: a) Mapping the restriction cleavage sites in the chromosome b) Using electron microscopy c) Using western blotting d) Using recombination experiments e) All of these Answer: a Section: 14.4 The Molecular Analysis of Genes and Chromosomes Difficulty: Medium

50) What is the genetic sequence shown on the autoradiogram below?

a) AATGTCAGCCTT b) TTCCGACTGTAA c) AAATTTTCCCGG d) GGCCCTTTTAAA e) TTTAAAAGGGCC Answer: a Section: 14.4 The Molecular Analysis of Genes and Chromosomes Difficulty: Medium

Question Type: Essay 51) Draw all of the DNA fragments that would result after 3 rounds of PCR if you start with: linear template DNA that is 2kb in length, two primers each of which have complementary sequences located 0.5 kb in from each end of the template and which are properly oriented for PCR, and all other components necessary for PCR. Label all DNA ends with regards to polarity and give fragment sizes where applicable.

Answer:

Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Hard

52) What is a restriction endonuclease? Name and give a drawing of the types of ends that can be produced after restriction enzyme digestion. Why are these enzymes named restriction endonucleases?

Answer: A restriction endonuclease is an enzyme that digests DNA from an internal site specific location. All type II restriction endonucleases have unique digestion sites and when the DNA is cut, staggered or blunt ends may be produced. These enzymes are named restriction endonucleases because they cut the DNA internally, instead of from the ends, and it restricts the host range of the virus that would naturally infect the bacteria. In nature the restriction endonucleases act as the immune system of the bacteria in that it will cut the foreign DNA but not the self methylated bacterial DNA. Section: 14.1 Basic Techniques Used to Identify, Amplify, and Clone Genes Difficulty: Medium 53) Explain how genomic libraries are constructed and the two methods that are used to insert the DNA into the cloning vector.

Answer: Genomic DNA libraries are usually prepared by isolating total DNA from an organism, digesting the DNA with a restriction endonuclease, and inserting the restriction fragments into an appropriate cloning vector. Two different procedures are used to insert the DNA fragments into the cloning vector. If the restriction enzyme that is used makes staggered cuts in DNA, producing complementary single-stranded ends, the restriction fragments can be ligated directly into vector DNA molecules cut with the same enzyme (Figure 15.15). An advantage of this procedure is that the foreign DNA inserts can be precisely excised from the vector DNA by cleavage with the restriction endonuclease used to prepare the genomic DNA fragments for cloning. If the restriction enzyme cuts both strands of DNA at the same position, producing blunt ends, complementary single-stranded tails must be added to the DNA fragments in vitro.

This is accomplished by using the enzyme terminal transferase to add nucleotides to the 3′ termini of the DNA strands after the 5′ ends are cut back with phage exonuclease. Usually, poly(A) tails are added to the cleaved vector DNA, and poly(T) tails are added to the genomic DNA fragments, or vice versa. Then, the T-tailed genomic DNA fragments are inserted into the A-tailed vector DNA molecules with DNA ligase. Since the T and A tails will not always be the same length, the E. coli enzymes exonuclease III and DNA polymerase I are used to cut back overhangs and fill in gaps, respectively. DNA ligase will only seal nicks between adjacent nucleotides; it will not add nucleotides if gaps are present. Once the genomic DNA fragments are ligated into vector DNA, the recombinant DNA molecules must be introduced into host cells for amplification by replication in vivo. This step usually involves transforming antibioticsensitive recipient cells under conditions where a single recombinant DNA molecule is introduced per cell (for most cells) (Chapter 8). When E. coli is used, the bacteria must first be made permeable to DNA by treatment with chemicals or a short pulse of electricity. Transformed cells are then selected by growing the cells under conditions where the selectable marker gene of the vector is essential for growth. A good genomic DNA library contains essentially all of the DNA sequences in the genome of interest. For large genomes, complete libraries contain hundreds of thousands of different recombinant clones.

Section: 14.2 Construction and Screening of DNA Libraries Difficulty: Medium

54) How can site-specific mutagenesis be conducted using PCR? Answer: Site-specific mutagenesis is accomplished by using a mutagenic PCR primer. The mutagenic primer is an oligonucleotide 12 to 15 nucleotides in length, which is largely complementary to one strand of the DNA sequence of interest, but which contains one or more noncomplementary or “mismatched” bases. The mismatched bases will provide the desired mutant sequence. The synthetic oligonucleotide primer is used in conjunction with another primer complementary to the other strand of the DNA sequence of interest. The two primers are used to amplify the intervening DNA sequence as in standard PCR (see Figure 15.14). After many cycles of amplification by PCR, the PCR products will consist almost entirely of mutant DNA fragments.

Section: 14.3 The Molecular Analysis of DNA, RNA, and Protein Difficulty: Medium 55) How is a Southern Blot performed and why does this technique provide more information than gel electrophoresis?

Answer: While gel electrophoresis allows one to separate DNA fragments based upon molecular size, Southern blots allow investigators to identify the locations of genes and other DNA sequences on restriction fragments separated by gel electrophoresis. The essential feature of this technique is the transfer of the DNA molecules that have been separated by gel electrophoresis onto nitrocellulose or nylon membranes (Figure 15.20). Such transfers of DNA to membranes are called Southern blots after the scientist who developed the technique. The DNA is denatured either prior to or during transfer by placing the gel in an alkaline solution. After transfer, the DNA is immobilized on the membrane by drying or UV irradiation. A radioactive DNA probe containing the sequence of interest is then hybridized (Chapter 9) with the immobilized DNA on the membrane. The probe will hybridize only with DNA molecules that contain a nucleotide sequence complementary to the sequence of the probe. Nonhybridized probe is then washed off the membrane, and the washed membrane is exposed to X-ray film to detect the presence of the radioactivity. After the film is developed, the dark bands show the positions of DNA sequences that have hybridized with the probe.

Section: 14.3 The Molecular Analysis of DNA, RNA, and Protein Difficulty: Medium 56) Describe the Sanger method of DNA sequencing.

Answer: 2′,3′-Dideoxyribonucleoside triphosphates are the chain-terminators most frequently used in the Sanger sequencing procedure. If a 2′,3′-dideoxynucleotide is added to the end of a chain, it will block subsequent extension of that chain since the 2′,3′-dideoxynucleotides have no 3 -OH. By using (1) 2′,3′-dideoxythymidine triphosphate (ddTTP), (2) 2′,3′-dideoxycytidine triphosphate (ddCTP), (3) 2′,3′-dideoxyadenosine triphosphate (ddATP), and (4) 2 ,3 dideoxyguanosine triphosphate (ddGTP) as chain-terminators in four separate DNA synthesis reactions, four populations of fragments can be generated, and each population will contain chains that all terminate with the same base (T, C, A, or G) (Figure 15.28). In a given reaction, the ratio of dXTP:ddXTP (where X can be any one of the four bases) is kept at approximately 100:1, so that the probability of termination at a given X in the nascent chain is about 1/100. This yields a population of fragments terminating at all potential (X) termination sites within a distance of a few hundred nucleotides from the original primer terminus. After the DNA fragments generated in the four parallel reactions are released from the template strands by

denaturation, they are separated by polyacrylamide gel electrophoresis and their positions in the gel are detected by autoradiography. The bands on the autoradiograms correspond to radioactive chains of different lengths; they produce a “ladder” defining the nucleotide sequence of the longest chain that has been synthesized (Figure 15.29). The shortest fragment will migrate the greatest distance and give rise to the band nearest the anode (the positive electrode). Each successive band will contain chains that are one nucleotide longer than the chains in the preceding band of the ladder. The 3′-terminal nucleotide of the chain in each band will be the dideoxynucleotide chain-terminator present in the reaction mixture (1, 2, 3, or 4) in which that specific chain was produced (Figure 15.28). By reading the ladder produced by autoradiography of the polyacrylamide gels used to separate the fragments generated in each of the four parallel reactions, the complete nucleotide sequence of a DNA chain can be determined. This is illustrated in Figure 15.28 for a hypothetical nucleotide sequence. An autoradiogram of an actual dideoxynucleotide chain-terminator sequencing gel is shown in Figure 15.29. Under optimal conditions, long sequences of several hundred nucleotides can be determined from a single sequencing gel.

Section: 14.4 The Molecular Analysis of Genes and Chromosomes Difficulty: Medium 57) When pASH1 is digested by various restriction enzymes and combinations thereof, the following bands are observed. Construct a restriction enzyme map with this information.

EcoR1

BamH1

Hind3

6.0 kbp

6.0 kbp

6.0 kbp

Hae2

EcoR1 & Hae2

EcoR1 & Hind3

EcoR1 & BamH1 & BamH1 Hind3 5.0 kbp 4.5 kbp

3.5 kbp 3.0 kbp 2.5 kbp 2.0 kbp

2.0 kbp

1.0 kbp

1.0 kbp

1.5 kbp

Answer:

1.0 kbp

Section: 14.4 The Molecular Analysis of Genes and Chromosomes Difficulty: Hard

Chapter 15 Test Bank Question Type: Multiple Choice 1) The subdiscipline of genetics that focuses on the structure and function of entire genomes is known as: a) Recombinant DNA techniques b) Gene cloning c) DNA amplification d) Genomics e) Proteomics Answer: d Section: 15.1 Genomics: An Overview Difficulty: Easy 2) Which subdiscipline based technology is known to fall under the term genomics? a) Mapping b) Sequencing c) Analyzing the functions of entire genomes d) All of these e) None of these Answer: d Section: 15.1 Genomics: An Overview Difficulty: Easy 3) The study of genome evolution is also known as: a) Comparative genomics b) Functional genomics c) Structural genomics d) Phylogenetic genomics e) Comparative genomics and Phylogenic genomics Answer: a Section: 15.1 Genomics: An Overview Difficulty: Easy

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4) Which of the following is a method used for mapping the chromosomal location of genes and other molecular markers?

1. Recombination frequency mapping 2. Mapping based on positions relative to cytological features 3. Mapping based on physical distance a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Medium 5) The ability of scientists to identify and isolate genes based on information about their location in the genome is known as: a) Functional mapping b) Positional cloning c) Positional mapping d) Microarray analysis e) None of these Answer: b Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Easy 6) Which of the following is used to construct a genetic map of a chromosome?

1. Recombination frequencies 2. Banding patterns of chromosomes 3. Molecular distances separating sites on a DNA fragment a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: a Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Medium

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7) Which of the following is used to construct a physical map of a chromosome?

1. Recombination frequencies 2. Banding patterns of chromosomes 3. Molecular distances separating sites on a DNA fragment a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: c Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Medium 8) Overlapping genomic clones found in physical chromosome maps are also known as: a) STSs b) Contigs c) ORI sites d) Polylinkers e) All of these Answer: b Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Easy 9) Markers that have been mapped both physically and genetically on a chromosome are known as: a) Anchor markers b) Attachment markers c) Hybridized markers d) Anchor markers and Attachment markers e) All of these Answer: a Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Easy

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10) Which of the following is a method for correlating a physical map of a chromosome with a genetic map and/or a cytological map?

1. Genes that have been cloned can be positioned on the cytological map by in situ hybridization 2. Locating clones of genetically mapped genes or RFLPs on the physical map 3. Using PCR to amplify short unique genomic DNA sequences and Southern blots to relate these sequences to overlapping clones on physical maps a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Medium 11) Short unique anchor sequences that correlate physical, genetic, and cytological maps of chromosomes are known as: a) STSs b) ORIs c) PCRs d) STRs e) ESTs Answer: a Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Easy 12) Short cDNA sequences used as hybridization probes to anchor physical maps to genetic and cytological maps of chromosomes are known as: a) STSs b) ORIs c) PCRs d) STRs e) ESTs Answer: e Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Easy

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13) Within the euchromatic regions of human chromosomes one CentiMorgan is equivalent, on average, to what physical distance of DNA? a) 1 mb b) 1 kb c) 1 mm d) 1 cm e) 10 mb Answer: a Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Easy 14) Variations in the lengths of DNA fragments produced by restriction enzyme digestion are known as: a) RFLPs b) ESTs c) PCRs d) VNTRs e) STRs Answer: a Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Easy 15) Which of the following is a type of RFLP, particularly useful for mapping in humans, that is not based upon the cleavage pattern of a particular restriction enzyme but on the difference of the number of copies of a repeated sequence between restriction sites? a) RFLP b) EST c) PCR d) FISH e) VNTR Answer: e Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Easy 16) Polymorphic tandem repeats that are only two to five nucleotide pairs long and are extremely valuable in the creation of high density maps of eukaryotic chromosomes are known as: a) RFLP b) EST c) Microsatellites d) Macrosatellites e) FISH Answer: c Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Easy

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17) Which of the following techniques can be used to position clones on a cytological map? a) Southern blotting b) Northern blotting c) In situ hybridization d) Western blotting e) PCR Answer: c Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Easy 18) A map that integrates the cytological, genetic and physical maps of a chromosome is referred to as: a) High density maps b) High correlation maps c) Correlated map structure d) All of these e) None of these Answer: a Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Easy 19) Which of the following criteria would allow a researcher to perform chromosome walking in an organism?

1. Small genome size 2. A small amount of dispersed repetitive DNA 3. A large amount of dispersed repetitive DNA a) 1 b) 2 c) 3 d) 1 and 2 e) 1 and 3 Answer: d Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Medium 20) Which of the following organisms would be best suited for undergoing chromosome walking to clone a gene based upon map position? a) C. elegans b) Human c) Mouse d) Rhesus Monkey e) C. elegans and Human Answer: a Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Medium

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21) Which of the following is true for the procedures of both chromosome walking and chromosome jumping?

1. It is initiated by using a molecular probe such as an RFLP as a starting point 2. DNA fragments are prepared by partial digestion with a restriction enzyme 3. Genomic fragments are circularized using DNA Ligase a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: a Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Medium 22) If a molecular marker such as a restriction fragment-length polymorphism (RFLP) maps close to a gene, the gene can usually be isolated by which of the following? a) Chromosome walking b) Chromosome jumping c) Chromosome dancing d) Chromosome walking and Chromosome jumping e) All of these Answer: d Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Medium 23) Which of the following genes have been identified using chromosome jumping? a) Cystic fibrosis gene b) Vestigial wing gene c) Blue eye gene d) Cystic fibrosis and Blue eye gene e) All of these Answer: a Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Easy 24) Which of the following was not an initial goal of the Human Genome Project when it was organized in 1990? a) Map all the human genes b) Construct a detailed physical map of the entire human genome c) Determine nucleotide sequence of all 24 human chromosomes d) Map the gene sequence of Francis Crick e) All of these were initial goals Answer: d Section: 15.3 The Human Genome Project Difficulty: Easy

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25) Which of the following technologies allowed researchers to create a high density/ higher resolution map of the human genome? a) FISH b) PCR c) Radiation hybrid mapping d) Recombination mapping e) None of these Answer: c Section: 15.3 The Human Genome Project Difficulty: Easy 26) Which of the following men are responsible for the rapid progress of the sequencing activities of the Human Genome Project? a) Francis Collins b) Craig Venter c) James Watson d) Francis Collins and Craig Venter e) Francis Collins and James Watson Answer: d Section: 15.3 The Human Genome Project Difficulty: Easy 27) Based on the sequencing data acquired from the Human Genome Project, how many genes are in the human genome? a) 20,000-35,000 b) 50,000-100,000 c) 100,000-120,000 d) 150,000-200,000 e) 220,000-250,000 Answer: a Section: 15.3 The Human Genome Project Difficulty: Easy 28) Based on the sequencing data acquired from the Human Genome Project, what percentage of the human genome is comprised of exons, which will be translated into proteins? a) 1.1% b) 24% c) 54% d) 75% e) 90% Answer: a Section: 15.3 The Human Genome Project Difficulty: Easy

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29) Which of the following could be a future application of the technology, skills, and data acquired from the Human Genome Project?

1. Individuals can have their DNA sequenced 2. Researchers can determine genetic variations among different populations of humans 3. Drugs can be developed based upon an individual’s genetic makeup a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 15.3 The Human Genome Project Difficulty: Easy 30) The most common changes in human genomes are single nucleotide-pair substitutions such as A:T to G:C or G:C to A:T substitutions. These are known as: a) NSPs b) VNTRs c) STRs d) SNPs e) ESTs Answer: d Section: 15.3 The Human Genome Project Difficulty: Easy 31) Single nucleotide polymorphisms can be identified in the human genome via the use of what technology? a) PCR b) Microarray Hybridization c) Southern blots d) Northern blots e) Western blots Answer: b Section: 15.3 The Human Genome Project Difficulty: Easy

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32) Which of the following could be a result of studying the SNPs in the human genome?

1. It may be possible to trace important genetic events in the evolutionary history of our species 2. It may be possible to predict a person's susceptibility to diseases like cancer and heart disease 3. It may be possible to track human ancestry a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 15.3 The Human Genome Project Difficulty: Medium 33) The SNPs on a chromosome segment that tend to be inherited together define a genetic unit called a: a) Phenotype b) Genotype c) Haplotype d) Maplotype e) Genome sequence Answer: c Section: 15.3 The Human Genome Project Difficulty: Easy 34) What is the purpose of the International HapMap Project? a) To sequence the genomes of each individual on earth b) To characterize the similarities and differences in human genomes worldwide c) To create physical chromosome maps of each individual on earth d) To correlate protein function with gene structure e) All of these Answer: b Section: 15.3 The Human Genome Project Difficulty: Easy 35) In what type of hybridization are the gene-specific nucleotide sequences bound to membranes in specific patterns, and hybridized to radioactive, or fluorescent, RNA or cDNA preparations? a) Southern blots b) FISH c) Dot-blot arrays d) Northern blots e) Western blots Answer: c Section: 15.4 RNA and Protein Assays of Genome Functions Difficulty: Easy

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36) Which of the following is a disadvantage of using gene chips to study gene expression?

1) Gene chips cannot provide information on whether a sequence has been transcribed 2) Gene chips cannot provide information on whether a sequence has been translated 3) Gene chips hold sequences for entire genomes so it makes it harder to study individual sequences a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: b Section: 15.4 RNA and Protein Assays of Genome Functions Difficulty: Easy 37) Which of the following proteins is being used to monitor the synthesis and localization of proteins in vivo? a) Lactase b) GFP c) Kinase d) Lactase and GFP e) Lactase and Kinase Answer: b Section: 15.4 RNA and Protein Assays of Genome Functions Difficulty: Easy 38) Which of the following is a subdiscipline of genetics that compares nucleotide sequences of the genomes of various organisms in order to study evolutionary relationships? a) Functional genetics b) Structural genetics c) Comparative genetics d) Evolutionary genetics e) Bioinformatics Answer: c Section: 15.5 Genome Diversity and Evolution Difficulty: Easy 39) The science of gathering, manipulating, storing, retrieving, and classifying recorded biological information is known as: a) Functional genetics b) Structural genetics c) Comparative genetics d) Evolutionary genetics e) Bioinformatics Answer: e Section: 15.5 Genome Diversity and Evolution Difficulty: Easy

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40) Homologous genes within a species are known as: a) Paralogues b) Orthologues c) Heterologues d) Similogues e) None of these Answer: a Section: 15.5 Genome Diversity and Evolution Difficulty: Easy 41) Homologous genes within a species are known as: a) Paralogues b) Orthologues c) Heterologues d) Similogues e) None of these Answer: a Section: 15.5 Genome Diversity and Evolution Difficulty: Easy 41) Homologous genes present within different species are known as: a) Paralogues b) Orthologues c) Heterologues d) Similogues e) None of these Answer: b Section: 15.5 Genome Diversity and Evolution Difficulty: Easy 42) Which of the following is not true regarding mtDNA? a) It is found in the mitochondria b) Most are circular c) It is the same as nuclear DNA d) It is differently sized in different species e) All of these are true Answer: c Section: 15.5 Genome Diversity and Evolution Difficulty: Easy

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43) As eukaryotic organisms have increased in complexity, the proportion of their genomes that encodes proteins has: a) Increased b) Decreased c) Stayed the same d) Increased in complexity e) Decreased in complexity Answer: b Section: 15.5 Genome Diversity and Evolution Difficulty: Easy 44) Which of the following is a bioinformatics tool? a) BLAST search b) GENBANK c) Microarray hybridization d) BLAST search and GENBANK e) All of these Answer: d Section: 15.5 Genome Diversity and Evolution Difficulty: Easy 45) Which of the following procedures is a variation on FISH and has been used for comparative genome analysis on some species? a) Recombination mapping b) Chromosome jumping c) Chromosome painting d) Chromosome fluorescence e) Chromosome walking Answer: c Section: 15.5 Genome Diversity and Evolution Difficulty: Easy

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Question Type: Essay 46) How are physical maps of chromosomes/genomes created using clone banks?

Answer: The RFLP mapping procedure has been used to construct detailed genetic maps of chromosomes, which, in turn, have made positional cloning feasible. These genetic maps have been supplemented with physical maps of chromosomes. By isolating and preparing restriction maps of large numbers of genomic clones, overlapping clones can be identified and used to construct physical maps of chromosomes and even entire genomes. In principle, this procedure is simple (Figure 16.5). However, in practice, it is a formidable task, especially for large genomes.

The restriction maps of large genomic clones in YAC, PAC, or BAC vectors (Chapter 15) are analyzed by computer and organized in overlapping sets of clones called contigs. As more data are added, adjacent contigs are joined; when the physical map of a genome is complete, each chromosome is represented by a single contig map.

Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Medium 47) How is positional cloning conducted in order to clone a gene that is moderately expressed?

Answer: The gene is first mapped to a specific region of a given chromosome by genetic crosses or, in the case of humans, by pedigree analysis, which usually requires large families. The gene is next localized on the physical map of this region of the chromosome. Candidate genes in the segment of the chromosome identified by physical mapping are then isolated from mutant and wild-type individuals and sequenced to identify mutations that would result in a loss of gene function. In species where transformation is possible, copies of the wild-type alleles of candidate genes are introduced into mutant organisms to determine whether the wild-type genes will restore the wild phenotype. Restoration of the wild phenotype to a mutant organism provides strong evidence that the introduced wild-type gene is the gene of interest. Positional cloning is accomplished by mapping the gene of interest, identifying an RFLP or other molecular marker near the gene, and then “walking” or “jumping” along the chromosome until the gene is reached.

Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Medium

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48) How does the technique of chromosome walking compare with the technique of chromosome jumping?

Answer: When the distance from the closest molecular marker to the gene of interest is large, a technique called chromosome jumping can be used to speed up an otherwise long walk. Each jump can cover a distance of 100 kb or more. The chromosome jumping procedure is illustrated in Figure 16.8. Like a walk, a jump is initiated by using a molecular probe such as an RFLP as a starting point. However, with chromosome jumps, large DNA fragments are prepared by partial digestion of genomic DNA with a restriction endonuclease. The large genomic fragments are then circularized with DNA ligase. A second restriction endonuclease is used to excise the junction fragment from the circular molecule. This junction fragment will contain both ends of the long fragment; it can be identified by hybridizing the DNA fragments on Southern blots to the initial molecular probe. A restriction map of the junction fragment is prepared, and a restriction fragment that corresponds to the distal end of the long genomic fragment is cloned and used to initiate a chromosome walk or a second chromosome jump. Chromosome jumping has proven especially useful in work with large genomes such as the human genome.

Section: 15.2 Correlated Genetic, Cytological, and Physical Maps of Chromosomes Difficulty: Hard 49) Why was it necessary to use radiation hybrid mapping to help construct a map of the human genome, and how is radiation hybrid mapping conducted?

Answer: Unfortunately, the resolution of genetic mapping in humans is quite low—in the range of 1–10 mb. The resolution of fluorescent in situ hybridization (FISH) is also approximately 1 mb. Higher resolution mapping (down to 50 kb) can be achieved by radiation hybrid mapping, a modification of the somatic-cell hybridization mapping procedure. Standard somatic-cell hybridization involves the fusion of human cells and rodent cells growing in culture and the correlation of human gene products with human chromosomes retained in the hybrid cells.

Radiation hybrid mapping is performed by fragmenting the chromosomes of the human cells with heavy irradiation prior to cell fusion. The irradiated human cells are then fused with Chinese hamster (or other rodent) cells growing in culture, usually in the presence of a chemical such as polyethylene glycol to increase the efficiency of cell fusion. The human–Chinese hamster somatic-cell hybrids are then identified by growth in an appropriate selection medium.

Many of the human chromosome fragments become integrated into the Chinese hamster chromosomes during this process and are transmitted to progeny cells just like the normal genes in the Chinese hamster chromosomes. The polymerase chain reaction (PCR; Chapter 15) is then used to screen a large panel of the selected hybrid cells for the presence of human genetic markers. Chromosome maps are constructed based on the assumption that the probability of an X ray-induced break between two markers is directly proportional to the distances separating them in chromosomal DNA. Section: 15.3 The Human Genome Project Difficulty: Medium

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50) What are the advantages and disadvantages to using gene chip or microarray technology to study gene expression? How are microarrays created and used? What can be done to overcome the disadvantages of this technology?

Answer: New technologies now allow scientists to produce microarrays that contain thousands of hybridization probes on a single membrane or other solid support.

Microarrays of oligonucleotide or cDNA probes are produced in several ways: (1) microsynthesis of oligonucleotides in situ (on the chip), (2) spotting prefabricated oligonucleotide probes on solid supports, and (3) spotting DNA fragments or cDNAs on the supports. The probes on the microarrays are then hybridized to labeled (usually with a fluorescent tag) RNA or cDNA samples. The amount of RNA or cDNA hybridized to each probe in a microarray is quantified by using sophisticated scanners with micrometer resolution and appropriate computer software. As knowledge in the field has advanced, geneticists have been able to study the expression of more and more genes. Now, for the first time, they can study the expression of all the genes of an organism simultaneously. The ability to analyze the expression of entire genomes will enhance the current explosion of new information in biology and will eventually lead to an understanding of the normal process of human development and the causes of at least some human diseases. Array hybridizations and gene chips can be used to determine whether genes are transcribed, but they provide no information about the translation of the gene transcripts. Thus, biologists often use antibodies to detect the protein products of genes of interest. Western blots are used to detect proteins separated by electrophoresis, and antibodies coupled to fluorescent compounds are used to detect the location of proteins in vivo. However, both of these approaches provide only a single time-point assay of a protein in a cell, tissue, or organism. The discovery of a naturally occurring fluorescent protein, the green fluorescent protein (GFP) of the jellyfish Aequorea victoria, has provided a powerful tool that can be used to study gene expression at the protein level. GFP is now being used to monitor the synthesis and localization of specific proteins in a wide variety of living cells. These studies entail constructing fusion genes that contain the nucleotide sequence encoding GFP, coupled in frame to the nucleotide sequence encoding the protein of interest; introducing the chimeric gene into cells by transformation; and studying the fluorescence of the fusion protein in transgenic cells exposed to blue or UV light. Because GFP is a small protein, it can often be coupled to proteins without interfering with their activity or interaction with other cellular components.

Section: 15.4 RNA and Protein Assays of Genome Functions Difficulty: Medium

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Chapter 16 Test Bank Question Type: Multiple Choice 1) The gene for which of the following diseases was identified by positional cloning? a) Huntington’s disease b) Cystic fibrosis c) Breast Cancer d) Huntington's disease and Cystic fibrosis e) All of these Answer: d Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Easy 2) Which of the following can be attributed to the presence of unstable trinucleotide repeats? a) Huntington’s disease b) Fragile X syndrome c) Mytonic Dystrophy d) Huntington's Disease and Fragile X Syndrome e) All of these Answer: d Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Easy 3) How many CAG nucleotide repeats are commonly found in patients with Huntington's disease? a) 1-10 b) 11-34 c) 42-100 d) 200-500 e) 1000-2000 Answer: c Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Easy

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4) The genetic cause of Huntington's disease can best be described as a/an: a) Increased trinucleotide repeat in the huntingtin gene b) Increased trinuclotide repeat in the fibrosin gene c) Decreased trinucleotide repeat in the huntingtin gene d) Decreased trinucleotide repeat in the fibrosin gene e) None of these Answer: a Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Easy 5) In Huntington's disease the age of onset is: a) Proportionally correlated with the number of CAG repeats b) Inversely correlated with the number of CAG repeats c) Always at 30 years of age d) Always between 30 and 40 years of age e) None of these Answer: b Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Easy 6) On which chromosome is the huntingtin gene located? a) Chromosome 1 b) Chromosome 4 c) Chromosome 7 d) Chromosome 21 e) Chromosome 22 Answer: b Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Easy 7) On which chromosome is the CF gene located? a) Chromosome 1 b) Chromosome 4 c) Chromosome 7 d) Chromosome 21 e) Chromosome 22 Answer: c Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Easy

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8) Why was the use of the sweat gland cDNA library critical in identifying the CF gene?

1. The CF gene is only expressed in epithelial cells of the lungs, pancreas, salivary glands, sweat glands, intestine, and reproductive tract 2. The CF gene is not expressed in any other tissue than sweat glands 3. The CF gene is not expressed in sweat glands a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: a Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Medium 9) Seventy percent of all Cystic Fibrosis cases are caused by which of the following mutations? a) ΔF508 trinucleotide deletion b) ΔF508 trinucleotide repeat c) CAG trinucleotide repeat d) CAG trinuclotide deletion e) None of these Answer: a Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Easy 10) Which of the following was critical in identifying the CF gene for cystic fibrosis? a) Use of a sweat gland cDNA library b) Unique structure of the CF gene product c) Important clues from biochemical analysis d) Characteristic symptoms of the disease e) None of these Answer: a Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Easy

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11) The characterization of the huntingtin and CF genes has led to which of the following? a) DNA tests for the mutations which cause the respective diseases b) DNA tests for the protein that is formed in the respective diseases c) Treatments for the respective diseases d) Treatments for all neurological degenerative diseases e) None of these Answer: a Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Easy 12) Which of the following techniques is used to test for unstable trinucleotide repeats associated with Huntington's disease? a) Positional cloning b) DNA sequencing c) PCR d) In situ hybridization e) Northern blotting Answer: c Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Easy 13) The mutation that causes sickle cell anemia can be tested for by: 1. Testing for the presence or absence of a specific restriction enzyme cleavage site in DNA 2. Testing for the presence of a trinucleotide repeat in DNA 3. Testing for the absence of a trinucleotide repeat in DNA a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: a Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Medium

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14) When testing for the presence of the sickle cell anemia HbS allele, a Southern blot exhibits two small bands for the HbA allele and one band for the HbS allele. What do these results mean? a) The subject is homozygous for the sickle cell allele b) The subject is heterozygous c) The subject is homozygous for the normal allele d) The results are inconclusive and the test will need to be run again e) Southern blots cannot test for the presence of the sickle cell allele Answer: b Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Medium 15) The practice of introducing functional gene copies into an individual with two nonfunctional copies is known as: a) Gene cloning b) Gene therapy c) Gene diagnostics d) Southern blotting e) None of these Answer: b Section: 16.2 Human Gene Therapy Difficulty: Easy 16) A gene that has been introduced into a cell or an organism is known as a: a) Polygene b) Cisgene c) Transgene d) Xenogene e) Autogene Answer: c Section: 16.2 Human Gene Therapy Difficulty: Easy 17) Which of the following is true regarding somatic cell gene therapy?

1. The diseased gene will continue to be present in germ line cells 2. It will treat disease symptoms in an individual 3. It is less complex than organ transplantation a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 16.2 Human Gene Therapy Difficulty: Medium

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18) Which of the following statements regarding germ line cell gene therapy is false?

1. It is currently being practiced on humans.

2. There are no moral or ethical considerations with this type of gene therapy 3. It stops the diseased gene from being passed down to offspring a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: d Section: 16.2 Human Gene Therapy Difficulty: Medium 19) Which of the following is a common method of delivery used for somatic cell gene therapy? a) Viral vectors b) Enzyme linked immunoglobin vectors c) Cosmid vectors d) Viral vectors and Enzyme linked immunoglobulin vectors e) All of these Answer: a Section: 16.2 Human Gene Therapy Difficulty: Easy 20) Which of the following requirements must be fulfilled before a gene therapy procedure will be approved?

1. The gene must be cloned and well characterized 2. The risks of gene therapy to the patient must have been carefully evaluated and shown to be minimum 3. The disease must not be treatable by other strategies a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 16.2 Human Gene Therapy Difficulty: Easy

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21) The first use of gene therapy occurred in 1990 on a patient with which of the following diseases? a) Huntington’s disease b) Cystic fibrosis c) ADA-SCID d) AIDS e) Systemic lupus Answer: c Section: 16.2 Human Gene Therapy Difficulty: Easy 22) Why might a patient with ADA-SCID be a good candidate for somatic cell gene therapy?

1. The ADA gene was one of the first human disease genes to be cloned and characterized.

2. White blood cells can easily be obtained from ADA - SCID patients and reintroduced after functional copies of the ADA gene are added 3. Even a small amount of functional ADA will restore partial immune function a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 16.2 Human Gene Therapy Difficulty: Medium 23) Presently, somatic cell gene therapy for ADA-SCID presents challenges. What are those challenges? a) The therapy is too expensive b) The therapy is transient and the gene promoters are silenced by the host quickly c) The functional gene copy is lost during introduction d) The lifespan of red blood cells is short e) None of these Answer: b Section: 16.2 Human Gene Therapy Difficulty: Medium

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24) Currently white blood cells have been used in the somatic cell gene therapy treatment for patients with ADA-SCID, but their short lifespan is prohibitive in the treatment plan. What other type of cell could be used, and is being tested, with better results? a) Bone marrow stem cells b) Red blood cells c) Epithelial cells d) T lymphocytes e) B lymphocytes Answer: a Section: 16.2 Human Gene Therapy Difficulty: Medium 25) Somatic cell gene therapy requires a retroviral vector used. As a result, use of the therapy to treat _____ has led to leukemia. a) Huntington’s disease b) Cystic Fibrosis c) X-linked SCID d) ADA-SCID e) ALS Answer: c Section: 16.2 Human Gene Therapy Difficulty: Easy 26) Recorded patterns of DNA polymorphisms that can provide strong evidence of an individual’s identity are known as: a) DNA STRs b) DNA fingerprints c) RFLP fingerprints d) Genetic identification e) None of these Answer: b Section: 16.3 DNA Profiling Difficulty: Easy

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27) Why would it be extremely rare for two individuals to share the same genetic profile?

1. Many base-pair substitutions are silent 2. Many base-pair substitutions occur in non-coding regions of DNA 3. Duplications and deletions of DNA sequences and other genome rearrangements contribute to the evolutionary divergence of genomes a) 1 b) 2 c) 3 d)1 and 3 e) All of these Answer: e Section: 16.3 DNA Profiling Difficulty: Medium 28) An accurate estimate of the probability that two individuals will have matching DNA fingerprints would require reliable information about: 1. The frequency of polymorphisms in the population in question 2. The frequency of polymorphisms in all populations 3. The frequency of polymorphisms in a different population than the one in question a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: a Section: 16.3 DNA Profiling Difficulty: Medium 29) Which of the following is a possible use for DNA fingerprinting?

1. Identifying the father of a child 2. Identifying the presence of a person at a scene of a crime 3. Identifying an unknown soldier a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 16.3 DNA Profiling Difficulty: Easy

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30) Why is it necessary to examine DNA from a child’s mother in order to determine paternity?

1. Approximately half of the bands in the child’s DNA print result from maternal DNA sequences, therefore it is necessary to account for the presence of these bands when comparing the DNA to the potential fathers 2. Approximately half of the bands in the child’s DNA print result from maternal DNA sequences, and it is in the child’s best interest to confirm maternity as well as paternity.

3. Approximately half of the bands in the child’s DNA print result from maternal DNA sequences, and it is cheaper to determine maternity than it is paternity. a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: a Section: 16.3 DNA Profiling Difficulty: Medium 31) When confirming paternity it would be best to use: 1. Multiple numbers of probes so that the genome can be categorized more accurately 2. Multiple potential father’s DNA samples so as to rule out infidelity 3. Unknown samples of DNA to prevent bias a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: a Section: 16.3 DNA Profiling Difficulty: Medium 32) When was DNA fingerprinting first used as a forensic application in a court of law? a) 1987 b) 1988 c) 1989 d) 1990 e) 1991 Answer: b Section: 16.3 DNA Profiling Difficulty: Easy

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33) Which of the following statements regarding DNA fingerprinting is true? a) DNA fingerprinting has the drawback that it requires extremely large quantities of DNA for analysis. b) Inbreeding within a population decreases the probability of finding identical fingerprints between individuals. c) An advantage of the technique is that data obtained from one population can easily be extrapolated to another population. d) DNA fingerprinting is applicable in all cases of questionable identity except in paternity cases. e) A prerequisite of the technique is the availability of information about the frequency of the polymorphisms in the population.

Answer: e Section: 16.3 DNA Profiling Difficulty: Easy 34) Which of the following statements regarding paternity testing is correct? a) Blood type data can be used to positively identify the father of a child b) Blood type data has several advantages over DNA fingerprinting in paternity testing c) For paternity testing, only DNA samples from the possible fathers and the child are required. d) Approximately half of the bands in the child's DNA print match those of the father. e) Increasing the number of hybridizing probes increases the ambiguity of the paternity test.

Answer: d Section: 16.3 DNA Profiling Difficulty: Medium 35) Which of the following types of fragments are commonly used to create DNA fingerprints? a) VNTRs b) STRs c) ESTs d) VNTRs and STRs e) All of these Answer: d Section: 16.3 DNA Profiling Difficulty: Easy 36) Which of the following human proteins has been produced in microorganisms? a) Insulin b) Dystrophin c) Apo-dystrophin d) COVERT protein e) Huntingtin gene product Answer: a Section: 16.4 DNA Production of Eukaryotic Proteins in Bacteria Difficulty: Easy

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37) Which of the following was the first pharmaceutical protein produced by bacteria approved for human use? a) Insulin b) hGH c) Erythropoietin d) Rennin e) Interferon Answer: a Section: 16.4 DNA Production of Eukaryotic Proteins in Bacteria Difficulty: Easy 38) Which of the following was the second pharmaceutical protein produced by bacteria approved for human use? a) Insulin b) hGH c) Erythropoietin d) Rennin e) Interferon Answer: b Section: 16.4 DNA Production of Eukaryotic Proteins in Bacteria Difficulty: Easy 39) How can bacterial produced eukaryotic proteins can be used? a) Pharmaceuticals b) Digestive aids for animals c) Cleaning aids in detergents d) All of these e) None of these Answer: d Section: 16.4 DNA Production of Eukaryotic Proteins in Bacteria Difficulty: Easy 40) Which of the following is used to make cheese? a) Insulin b) hGH c) Erythropoietin d) Renning e) Interferon Answer: d Section: 16.4 DNA Production of Eukaryotic Proteins in Bacteria Difficulty: Easy

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41) Which statement is false concerning Agrobacterium tumefaciens: a) It causes the crown gall disease in plants. b) It carries the Ti plasmid which induces tumors in plants. c) The vir region of the Ti plasmid integrates into the plant genome d) The T-DNA segment of the Ti plasmid encodes enzymes for phytohormone synthesis. e) It is an important vector for making transgenic plants.

Answer: d Section: 16.5 Transgenic Animals and Plants Difficulty: Medium 42) The most commonly used method for making transgenic plants is: a) Electroporation b) Microprojectile bombardment c) A. tumefaciens-mediated transformation d) Retrovirus-mediated genetic transfer. e) Transposon-mediated genetic transfer Answer: c Section: 16.5 Transgenic Animals and Plants Difficulty: Easy 43) Totipotency refers to: a) The ability of a mammalian cell to develop into a differentiated cell. b) The ability of a plant cell to accept foreign DNA. c) The ability of a bacterial cell to express foreign DNA. d) The ability of a plant or animal cell to produce all of the differentiated cells of the mature plant. e) The ability of a mammalian cell to express all of the genes encoded by the genome.

Answer: d Section: 16.5 Transgenic Animals and Plants Difficulty: Easy 44) How do antisense mRNA molecules work? a) Annealing to DNA to prevent replication b) Annealing to DNA to prevent transcription c) Being translated into non-functional protein molecules d) Being translated into functional protein molecules that interact with normal protein molecules e) Annealing to sense mRNA to prevent translation.

Answer: e Section: 16.6 Reversing Genetics: Dissecting Biological Processes by Inhibiting Gene Expression Difficulty: Easy

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45) Which of the following is incorrect concerning RNAi? a) Double-strand RNA is injected into a cell b) A vector can be introduced to produce a double-stranded RNA hairpin c) Fragments of the RNA anneal with a promoter and silence gene transcription d) Fragments of the RNA anneal with the target mRNA. e) The RiSC system degrades the target mRNA and the annealed RNA fragment.

Answer: c Section: 16.6 Reversing Genetics: Dissecting Biological Processes by Inhibiting Gene Expression Difficulty: Easy Question Type: Essay 46) How does the trinucleotide repeat in the huntingtin gene cause the clinical manifestations of Huntington's Disease?

Answer: The expanded CAG repeat region in the mutant huntingtin gene encodes an abnormally long polyglutamine region near the amino terminus of the protein. The elongated polyglutamine region fosters protein-protein interactions that lead to the accumulation of aggregates of the huntingtin protein in brain cells. These protein aggregates are thought to cause the clinical symptoms of HD. Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Medium 47) How can the DNA test for Huntington's disease potentially diminish human suffering because of this disease?

Answer: While the test for the HD mutation has not led to a treatment or cure for this disease, it can lead to a diminishing “spread” of the disease. Given the availability of the DNA test for the HD mutation, individuals who are at risk of transmitting the defective gene to their children can determine whether they carry it before starting a family. Each person with a heterozygous parent has a 50 percent chance of not carrying the defective gene. If the test is negative, she or he can begin a family with no concern about transmitting the mutation. If the test is positive, the fetus can be tested prenatally, or the couple can consider in vitro fertilization, as did the parents we discussed at the beginning of this chapter. If the eight-cell pre-embryo tests negative for the HD mutation, it can be implanted in the mother's uterus with the knowledge that it carries two normal copies of the huntingtin gene. If used conscientiously, the DNA test for the HD mutation should diminish human suffering from this dreaded disease.

Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Medium

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48) How does the CF gene product cause clinical manifestations of the disease Cystic Fibrosis?

Answer: The CF gene product, called the cystic fibrosis transmembrane conductance regulator, or CFTR protein, forms ion channels (Figure 17.4) through the membranes of cells that line the respiratory tract, pancreas, sweat glands, intestine, and other organs and regulates the flow of salts and water in and out of these cells. Because the mutant CFTR protein does not function properly in CF patients, salt accumulates in epithelial cells and mucus builds up on the surfaces of these cells. The presence of mucus on the lining of the respiratory tract leads to chronic, progressive infections by Pseudomonas aeruginosa, Staphylococcus aureus, and related bacteria.

These infections, in turn, frequently result in respiratory failure and death. However, the mutations in the CF gene are pleiotropic; they cause a number of distinct phenotypic effects.

Malfunctions of the pancreas, liver, bones, and intestinal tract are common in individuals with CF. Although CFTR forms chloride channels (Figure 17.4), it also regulates the activity of several other transport systems such as potassium and sodium channels. Some work suggests that CFTR may play a role in regulating lipid metabolism and transport. CFTR interacts with a number of other proteins and undergoes phosphorylation/dephosphorylation by kinases and phosphatases. Thus, CFTR should be considered multifunctional. Indeed, some of the symptoms of CF may result from the loss of CFTR functions other than the chloride channels.

Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Medium 49) What is the molecular basis for Huntington's Disease? Describe a simple molecular technique that is often used to test for this defect.

Answer: Huntington's disease is associated with the HD mutation, which contains 42 to 100 copies of a (CAG)n trinucleotide repeat on chromosome 4. The trinucleotide repeat regions of HD chromosomes are unstable and the age of onset of HD is inversely correlated with the number of copies of the trinucleotide repeat. A simple and accurate DNA test for the mutation can be performed by designing oligonucleotide primers based on sequences on either side of the repeat region and amplifying the region by PCR. The number of repeats can then be determined by polyacrylamide gel electrophoresis.

Section: 16.1 Use of Recombinant DNA Technology to Identify Human Genes and Diagnose Genetic Diseases Difficulty: Medium 50) What types of viral vectors are commonly used to conduct somatic cell gene therapy? What are the advantages and disadvantages of each?

Answer: To perform somatic-cell gene therapy, wild-type genes must be introduced into and expressed in cells homozygous or hemizygous for a mutant allele of the gene. In principle, the wild-type gene could be delivered to the mutant cells by any of several different procedures.

Most commonly, viruses are used to carry the wild-type gene into cells. In the case of retroviral vectors, the wild-type transgene is integrated—along with the retroviral DNA—into the DNA of the host cell. Thus, when retroviral vectors are used, the transgene is transmitted to all progeny cells in the affected cell lineage.

One advantage of retroviral vectors is that they insert themselves into the chromosomes of host cells and, therefore, are transmitted to progeny cells during cell division. However, like transposable elements, they can cause mutation by inserting themselves into genes of host cells.

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In addition, some retroviral DNAs up-regulate the expression of genes close to their sites of integration. Scientists have known for many years that there was some risk that the retroviral vectors used in gene therapy might cause mutations by integrating within genes. With other viral vectors, such as those derived from adenoviruses, the transgenes are present only transiently in host cells, because the genomes of these viruses replicate autonomously and persist only until the immune system eliminates the viruses along with the infected cells. The advantage of these vectors over retroviral vectors is that no potentially harmful mutations are induced during the integration step. However, they have two major disadvantages: (1) transgene expression is transient, lasting only as long as the viral infection persists, and (2) most humans exhibit strong immune responses to these viruses, presumably because of prior exposure to the same or closely related viruses.

Section: 16.2 Human Gene Therapy Difficulty: Medium 51) How is hGH produced for pharmaceutical use by using E. coli as a producer?

Answer: To obtain expression in E. coli, the hGH coding sequence must be placed under the control of E. coli regulatory elements. Therefore, the hGH coding sequence was joined to the promoter and ribosome-binding sequences of the E. coli lac operon. To accomplish this, a HaeIII cleavage site in the nucleotide-pair triplet specifying codon 24 of hGH was used to fuse a synthetic DNA sequence encoding amino acids 1–23 to a partial cDNA sequence encoding amino acids 24–191. This unit was then inserted into a plasmid carrying the lac regulatory signals and introduced into E. coli by transformation. The hGH produced in E. coli in these first experiments contained methionine at the amino terminus. Native hGH has an amino-terminal phenylalanine: a methionine is initially present but is then enzymatically removed. E. coli also removes many amino-terminal methionine residues posttranslationally. However, the excision of the terminal methionine is sequence-dependent, and E. coli cells do not excise the aminoterminal methionine residue from hGH. Nevertheless, the hGH synthesized in E. coli was found to be fully active in humans despite the presence of the extra amino acid. More recently, a DNA sequence encoding a signal peptide (the amino acid sequence required for transport of proteins across membranes) has been added to an HGH gene construct similar to the one shown in Figure 17.13. With the signal sequence added, hGH is both secreted and correctly processed; that is, the methionine residue is removed with the rest of the signal peptide during the transport of the primary translation product across the membrane. This product is identical to native hGH. Section: 16.4 DNA Production of Eukaryotic Proteins in Bacteria Difficulty: Medium

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Chapter 17 Test Bank Question Type: Multiple Choice 1) Which of the following is a mechanism by which transcription is regulated?

1. Rapid turn-on and turn-off of gene expression in response to environmental changes 2. Preprogrammed circuits or cascades of gene expression 3. Light-switch method a) 1 b) 2 c) 3 d) 1 and 2 e) 1 and 3 Answer: d Section: 17.1 Strategies for Regulating Genes in Prokaryotes Difficulty: Easy 2) Genes that are continually being expressed in most cells are referred to as a) inducible genes. b) repressible genes. c) constitutive genes. d) tRNA genes. e) mRNA genes.

Answer: c Section: 17.1 Strategies for Regulating Genes in Prokaryotes Difficulty: Easy 3) Which of the following molecules would be made by a constitutive gene? a) tRNA b) RNA polymerase c) Lactose d) tRNA and RNA polymerase e) All of these are correct.

Answer: d Section: 17.1 Strategies for Regulating Genes in Prokaryotes Difficulty: Easy 4) Genes whose expression is turned on in response to a substance in the environment are known as a) inducible genes. b) repressible genes. c) constitutive genes. d) inducible genes and repressible genes. e) All of these are correct.

Answer: a Section: 17.1 Strategies for Regulating Genes in Prokaryotes Difficulty: Easy

5) Enzymes that are involved in catabolic pathways are characteristically controlled by what type of genes? a) Inducible genes b) Repressible genes c) Constitutive genes d) Inducible genes and constitutive genes e) All of these are correct.

Answer: a Section: 17.1 Strategies for Regulating Genes in Prokaryotes Difficulty: Easy 6) Genes whose expression is turned off in response to a substance in the environment are known as a) inducible genes. b) repressible genes. c) constitutive genes. d) inducible genes and repressible genes. e) All of these are correct.

Answer: b Section: 17.1 Strategies for Regulating Genes in Prokaryotes Difficulty: Easy 7) Which type of genes typically control the enzymes involved in anabolic pathways? a) Inducible genes b) Repressible genes c) Constitutive genes d) Inducible genes and repressible genes e) All of these are correct.

Answer: b Section: 17.1 Strategies for Regulating Genes in Prokaryotes Difficulty: Easy 8) Which of the following would most likely be controlled by a repressible gene? a) Tryptophan synthesis b) Lactose catabolism c) Arabinose catabolism d) Tryptophan synthesis and lactose catabolism e) Lactose catabolism and arabinose catabolism Answer: a Section: 17.1 Strategies for Regulating Genes in Prokaryotes Difficulty: Easy

9) Which level of gene regulation occurs via the action of inducible and repressible genes? a) Replication b) Transcription c) Translation d) Replication and transcription e) Replication and translation Answer: b Section: 17.1 Strategies for Regulating Genes in Prokaryotes Difficulty: Easy 10) The process of turning on the expression of genes in response to a substance in the environment is called a) repression. b) catabolite repression. c) glucose effect. d) induction. e) positive feedback.

Answer: d Section: 17.1 Strategies for Regulating Genes in Prokaryotes Difficulty: Easy 11) Which of the following statements best describes a constitutive gene? a) A gene that is expressed in the presence of an inducer b) A gene that is expressed in the absence of a corepressor c) A gene that is expressed only in the presence of lactose d) A gene that is expressed continuously. e) A gene that is under temporal regulation Answer: d Section: 17.1 Strategies for Regulating Genes in Prokaryotes Difficulty: Easy 12) Which of the following statements is false? a) In an inducible operon, free repressor binds to the operator. b) In an inducible operon, free repressor cannot bind to the operator. c) In a repressible operon, free repressor binds to the operator. d) In an inducible operon, only a repressor-corepressor complex can bind to the operator. e) In a repressible operon, only the free corepressor can bind to the operator.

Answer: a Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Medium

13) Genes encoding products that regulate the expression of other genes are known as a) positive control genes. b) negative control genes. c) regulator genes. d) transcriptional genes. e) translatable genes.

Answer: c Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Easy 14) What type of regulation occurs when the product of the regulator gene is required to turn on the expression of one or more structural genes? a) Constitutive control b) Positive control c) Negative control d) Feedback control e) None of these are correct.

Answer: b Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Easy 15) What type of regulation occurs when the product of the regulator gene is necessary to shut off the expression of structural genes? a) Constitutive control b) Positive control c) Negative control d) Feedback control e) None of these are correct.

Answer: c Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Easy 16) Where does the regulator bind in order to control transcription? a) Promoter b) Operator c) Regulator protein binding site d) Enhancer e) Silencer Answer: c Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Easy

17) Which of the following types of molecules controls whether or not a regulator can bind at the appropriate site? a) Effector molecules b) Activator molecules c) Derepressor molecules d) Effector molecules and derepressor molecules e) All of these are correct.

Answer: a Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Easy 18) In a negative, repressible regulatory mechanism, transcription of the structural gene(s) occurs in a) the absence of the co-repressor, but not in its presence. b) the absence of the inducer but not in its presence. c) the absence or presence of the co-repressor. d) the presence of the co-repressor, but not in its absence. e) None of these are correct.

Answer: a Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Easy 19) In prokaryotes, genes with related functions often are present in coordinately regulated genetic units called what? a) Operators b) Operons c) Codons d) Anticodons e) Regulatory units Answer: b Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Easy 20) Which of the following is not a constituent of an operon? a) Operator b) Structural genes c) Promoter d) Intron e) All of these are correct.

Answer: d Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Easy

21) What happens when a repressor is bound to the operator? a) RNA polymerase is prevented from transcribing the structural genes in the operon. b) RNA polymerase transcribes the structural genes in the operon. c) RNA polymerase increases the rate of transcription. d) RNA polymerase decreases the rate of transcription. e) None of these are correct.

Answer: a Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Easy 22) Which of the following is false with regard to the operator segment in an operon?

1. Operator regions are contiguous with promoter regions.

2. Operator regions are often located between the promoters and the structural genes that they regulate.

3. Repressors bind to the operator and turn on transcription. a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: c Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Easy 23) In which type of operon does the free repressor binds to the operator, turning off transcription? a) Inducible operon b) Repressible operon c) Constitutive operon d) Inducible operon and repressible operon e) All of these are correct.

Answer: a Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Easy 24) Which of the following best describes the lac operon? a) Negatively controlled repressible operon b) Negatively controlled inducible operon c) Negatively controlled constitutive operon d) Positively controlled repressible operon e) Positively controlled inducible operon Answer: b Section: 17.3 The Lactose Operon in E.coli: Induction and Catabolite Repression Difficulty: Easy

25) Which of the following statements is true regarding the lac operon? a) The repressor binds to the promoter element. b) Transcription is initiated by the binding of the inducer to the promoter. c) The lac operon is unique because it is not subject to catabolite repression. d) The promoter element is not essential for the transcription of the operon. e) lacZ, lacY and lacA gene products are synthesized at low levels in the uninduced state.

Answer: e Section: 17.3 The Lactose Operon in E.coli: Induction and Catabolite Repression Difficulty: Medium 26) The enzyme -galactosidase is encoded by a) lacA. b) lacy. c) lacZ. d) lacI. e) both lacA and lacY. Answer: c Section: 17.3 The Lactose Operon in E.coli: Induction and Catabolite Repression Difficulty: Easy 27) The lac repressor (I+ gene product) controls the expression of structural genes located where? a) Cis to the lacI+ allele b) Trans to the lacI+ allele c) Cis to the lacO+ allele d) Cis to the lacI+ allele and Trans to the lacI+ allele e) All of these are correct.

Answer: d Section: 17.3 The Lactose Operon in E.coli: Induction and Catabolite Repression Difficulty: Easy 28) Which is true about an operator constitutive (Oc) mutation? a) An Oc mutation on the chromosome can affect the expression of structural genes on a plasmid. b) The Oc mutation prevents the expression of a diffusible product that regulates the lac operon. c) An F' I+P+ Oc Z-Y+A-/I+P+O+Z+Y+A+ merozygote expresses the enzymes of the lac operon constitutively. d) The Oc mutations act only in cis. e) An F' I+P+ Oc Z+Y+A+/I+P+O+Z-Y-A- merozygote is inducible for the three enzymes of the lac operon.

Answer: d Section: 17.3 The Lactose Operon in E.coli: Induction and Catabolite Repression Difficulty: Medium

29) How do promoter mutations change the expression of the lac operon genes?

1. They change the inducibility of the lac operon.

2. They modify the levels of gene expression in the induced and uninduced state.

3. They change the frequency of initiation of lac operon transcription. a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: e Section: 17.3 The Lactose Operon in E.coli: Induction and Catabolite Repression Difficulty: Medium 30) Which of the following is a constituent of the lac operon promoter?

1. RNA polymerase binding site 2. CAP binding site 3. lac I gene a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: d Section: 17.3 The Lactose Operon in E.coli: Induction and Catabolite Repression Difficulty: Easy 31) What happens in the lac operon when both lactose and glucose are present in the environment? a) Transcription of the lac operon structural genes will occur. b) Transcription of the lac operon structural genes will occur but at a very slow rate. c) Transcription of the lac operon structural genes will occur at an increased rate. d) Transcription of the lac operon will not occur because the inducer is not present in the environment. e) Transcription of the lac operon will not occur because of catabolite repression.

Answer: e Section: 17.3 The Lactose Operon in E.coli: Induction and Catabolite Repression Difficulty: Medium 32) In the lac operon, what type of control is exhibited by the CAP/cAMP complex? a) Negative control b) Positive control c) Constitutive control d) Negative control and positive control e) All of these are correct.

Answer: b Section: 17.3 The Lactose Operon in E.coli: Induction and Catabolite Repression Difficulty: Easy

33) Which statement about catabolite repression is false? a) Glucose is the preferred energy source for E. coli and represses the utilization of other sugars. b) The CAP protein is involved in catabolite repression of the lac operon. c) RNA polymerase is the effector molecule that enables CAP to bind to the promoter. d) The CAP protein and its effector molecule exert positive control over the transcription of the lac operon. e) High glucose concentrations decrease the intracellular concentration of the effector molecule of the CAP protein.

Answer: c Section: 17.3 The Lactose Operon in E.coli: Induction and Catabolite Repression Difficulty: Medium 34) Which of the following substances is the inducer for the lac operon? a) Lactose b) Tryptophan c) cAMP d) Glucose e) CAP Answer: a Section: 17.3 The Lactose Operon in E.coli: Induction and Catabolite Repression Difficulty: Easy 35) When are the structural genes in the tryptophan operon transcribed?

1. Tryptophan is in abundance in the environment.

2. Tryptophan is absent from the environment.

3. Tryptophan is present in low amounts in the environment. a) 1 b) 2 c) 3 d) 1 and 3 e) 2 and 3 Answer: e Section: 17.4 The Tryptophan Operon in E.coli: Repression and Attenuation Difficulty: Easy

36) Which of the following is true regarding the regulation of the trp operon? a) trpR mutants that lack functional repressor have a tenfold elevated rate of synthesis of the tryptophan biosynthetic enzymes in the presence of tryptophan. b) Tryptophan is an inducer of the trp operon. c) trpL deletion mutants in which there is no 162-nucleotide-long leader sequence in the mRNA result in relief of repression and derepression of the trp operon. d) Premature termination of the trp operon occurs in the presence of tryptophan charged tRNAtrp. e) In the presence of tryptophan, the ribosome cannot translate past the trp codons to the leader-peptide termination codon.

Answer: d Section: 17.4 The Tryptophan Operon in E.coli: Repression and Attenuation Difficulty: Easy 37) Which of the following is the best example of de-repression? a) Switching on the trp operon in the presence of tryptophan b) Switching off the trp operon in the presence of tryptophan c) Switching off the trp operon in the absence of tryptophan d) Switching off the lac operon by the presence of glucose in the medium e) Switching on the lac operon in the presence of lactose Answer: a Section: 17.4 The Tryptophan Operon in E.coli: Repression and Attenuation Difficulty: Medium 38) Which characteristic is exhibited by transcription-termination signals found at the ends of most bacterial operons, such as the trp operon? a) They encode a nascent RNA with the potential to form hydrogen-bonded hairpin structures. b) They contain the operator region to which the repressor binds to shut off transcription. c) They contain the site for the binding of the CAP-cAMP complex to affect catabolite repression. d) They contain the binding site for the repressor-corepressor complex. e) They begin with the sequence ATG. Answer: a Section: 17.4 The Tryptophan Operon in E.coli: Repression and Attenuation Difficulty: Medium 39) What type of operon is the trp operon? a) Positive repressible operon b) Negative repressible operon c) Positive inducible operon d) Negative inducible operon e) Constitutive inducible operon Answer: b Section: 17.4 The Tryptophan Operon in E.coli: Repression and Attenuation Difficulty: Easy

40) When temperate bacteriophages, such as phage , enter the lysogenic pathway, covalently inserting their chromosome into the chromosome of the host, their lytic genes must be a) turned on. b) turned off. c) induced. d) turned on and induced. e) None of these are correct.

Answer: b Section: 17.5 Posttranscriptional Control of Gene Expression Difficulty: Easy 41) Which of the following genes of phage  encodes the repressor? a) CI b) OL c) OR d) CI and OL e) OL and OR Answer: a Section: 17.5 Posttranscriptional Control of Gene Expression Difficulty: Easy 42) Where does the phage  repressor bind to prevent transcription of the lytic genes? a) CI b) OL c) OR d) CI and OL e) OL and OR Answer: e Section: 17.5 Posttranscriptional Control of Gene Expression Difficulty: Easy 43) Which of the following is a constituent of the phage  repressor? a) DNA-binding domain b) Dimerization domain c) Connector region d) DNA-binding domain and Dimerization domain e) All of these are correct.

Answer: e Section: 17.5 Posttranscriptional Control of Gene Expression Difficulty: Easy

44) UV light induces nondefective  prophages to enter the lytic pathway by which of the following mechanisms? a) SOS response b) Conversion of RecA to a protease c) Cleavage of the connector region of the  repressor d) Prevention of dimerization of the DNA-binding domain of the repressor e) All of these are correct.

Answer: c Section: 17.5 Posttranscriptional Control of Gene Expression Difficulty: Easy 45) Which of the following is true about the life cycle of phage ? a) When phage  injects the Cro protein into E. coli it enters the lytic cycle. b) When phage  injects the repressor into E. coli it enters the lysogenic pathway. c) Phage  enters the lytic cycle when the repressor occupies the OL and OR sites. d) Phage  enters the lysogenic pathway when the Cro protein occupies the OL and OR sites. e) Translation of the cro mRNA is blocked by cro antisense RNA. Answer: e Section: 17.5 Posttranscriptional Control of Gene Expression Difficulty: Easy 46) How is sequential gene expression controlled in bacterial viruses like E. coli phages T7 and T4 and Bacillus subtilis phage SP01? a) Modifying the DNA topology b) Sequential binding of a series of repressor molecules c) Modifying the specificity of RNA polymerase for different promoter sequences d) Sequential binding of inducer molecules to the repressor e) None of these are correct.

Answer: c Section: 17.5 Posttranscriptional Control of Gene Expression Difficulty: Medium 47) Negative autogenous regulation refers to a) inhibition of translation of a specific mRNA by a protein it encodes. b) inhibition of transcription of a gene by a specific mRNA it encodes. c) inhibition of the first enzyme of a pathway by the end product of the biosynthetic pathway. d) inhibition of the synthesis of ribosomes by the accumulation of protein molecules. e) inhibition of the growth of an organism by the accumulation of toxic waste products.

Answer: a Section: 17.5 Posttranscriptional Control of Gene Expression Difficulty: Easy

48) Which of the following is true about regulation of prokaryotic gene expression? a) Genes such as rRNAs and tRNAs, which specify housekeeping functions, can be repressed or induced as required. b) Genes that encode enzymes involved in catabolic pathways are always expressed constitutively. c) Genes that encode anabolic enzymes are not repressible. d) Transcriptional regulation is the most common mechanism of control in prokaryotes. e) Translational regulation is the most common mechanism of control in prokaryotes.

Answer: d Section: 17.5 Posttranscriptional Control of Gene Expression Difficulty: Medium 49) Regulatory fine-tuning frequently occurs at the level of translation by: 1. Modulation of the rate of polypeptide chain initiation 2. Modulation of the rate of chain elongation 3. Modulation of the amino acid sequence a) 1 b) 2 c) 3 d) 1 and 2 e) All of these are correct.

Answer: d Section: 17.5 Posttranscriptional Control of Gene Expression Difficulty: Medium 50) Which of the following statements regarding regulation of prokaryotic gene expression is incorrect? a) Accumulation of an end product of a biosynthetic pathway may result in the inhibition of the first enzyme in the pathway. b) Different regions of an mRNA molecule may degrade at different rates. c) E. coli cells devote a larger share of energy to the production of ribosomes under starvation conditions than in conditions favorable to growth. d) Hairpins in the mRNA molecule can decrease translation rates by impeding the migration of ribosomes. e) Unequal efficiencies of translational initiation are known to occur at the ATG start codons of different genes.

Answer: c Section: 17.5 Posttranscriptional Control of Gene Expression Difficulty: Medium

51) What form of inhibition occurs when the product of a biosynthetic pathway inhibits the activity of the first enzyme in the pathway, rapidly arresting the biosynthesis of the product? a) Competitive inhibition b) Feedback inhibition c) Non-allosteric inhibition d) Competitive inhibition and feedback inhibition e) None of these are correct.

Answer: b Section: 17.5 Posttranscriptional Control of Gene Expression Difficulty: Medium Question Type: Essay 52) Based upon what is known about the regulation of transcription, regulatory mechanisms fall into two general categories. What are these categories and how do they work to aid organisms in cellular growth?

Answer: (1) Mechanisms that involve the rapid turn-on and turn-off of gene expression in response to environmental changes. Regulatory mechanisms of this type are important in microorganisms because of the frequent exposure of these organisms to sudden changes in environment. They provide microorganisms with considerable “plasticity,” an ability to adjust their metabolic processes rapidly in order to achieve maximal growth and reproduction under a wide range of environmental conditions. (2) Mechanisms referred to as preprogrammed circuits or cascades of gene expression. In these cases, some event triggers the expression of one set of genes. The product(s) of one or more of these genes functions by turning off the transcription of the first set of genes or turning on the transcription of a second set of genes. Then, one or more of the products of the second set acts by turning on a third set, and so on. In these cases, the sequential expression of genes is genetically preprogrammed, and the genes cannot usually be turned on out of sequence.

Section: 17.1 Strategies for Regulating Genes in Prokaryotes Difficulty: Medium 53) Compare how a negative inducible control mechanism and a positive inducible control mechanism work in the presence and absence of an inducer.

Answer: In a negative, inducible control mechanism (Figure 19.3a, left), the free repressor binds to the RPBS and prevents the transcription of the structural gene(s) in the absence of inducer. When inducer is present, it is bound by the repressor, and the repressor/inducer complex cannot bind to the RPBS. With no repressor bound to the RPBS, RNA polymerase binds to the promoter and transcribes the structural gene(s). In a positive, inducible control mechanism (Figure 19.3a, right), the activator cannot bind to the RPBS unless inducer is present, and RNA polymerase cannot transcribe the structural gene(s) unless the activator/inducer complex is bound to the RPBS. Thus, transcription of the structural genes is turned on only in the presence of inducer.

Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Medium

54) Compare how a negative repressible control mechanism and a positive repressible control mechanism work in the presence and absence of a co-repressor.

Answer: In a negative, repressible regulatory mechanism, transcription of the structural gene(s) occurs in the absence of the co-repressor, but not in its presence. When the repressor/co-repressor complex is bound to the RPBS, it prevents RNA polymerase from transcribing the structural genes.

In the absence of co-repressor, free repressor cannot bind to the RPBS; thus, RNA polymerase can bind to the promoter and transcribe the structural genes. In a positive, repressible control mechanism, the product of the regulator gene, the activator, must be bound to the RPBS in order for RNA polymerase to bind to the promoter and transcribe the structural gene(s). When co-repressor is present, it forms a complex with the activator protein, and this activator/co-repressor complex is unable to bind to the RPBS; consequently, RNA polymerase cannot bind to the promoter and transcribe the structural gene(s).

Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Medium 55) Draw a sketch of a simplistic operon with 4 structural genes.

Answer: --Promoter--Operator---Gene 1—Gene 2—Gene 3—Gene 4 Section: 17.2 Operons: Coordinately Regulated Units of Gene Expression Difficulty: Easy 56) How does catabolite repression stop transcription of the lac operon in the presence of glucose?

Answer: The catabolite repression of the lac operon and several other operons is mediated by a regulatory protein called CAP and a small effector molecule called cyclic AMP. Because CAP binds cAMP when this mononucleotide is present at sufficient concentrations, it is sometimes called the cyclic AMP receptor protein. The lac promoter contains two separate binding sites, one for RNA polymerase and one for the CAP/cAMP complex. The CAP/cAMP complex must be present at its binding site in the lac promoter in order for the operon to be induced normally. The CAP/cAMP complex thus exerts positive control over the transcription of the lac operon. It has an effect exactly opposite to that of repressor binding to an operator. Although the precise mechanism by which CAP/cAMP stimulates RNA polymerase binding to the promoter is still uncertain, its positive control of lac operon transcription is firmly established by the results of both in vivo and in vitro experiments. CAP functions as a dimer; thus, like the lac repressor, it is multimeric in its functional state. Only the CAP/cAMP complex binds to the lac promoter; in the absence of cAMP, CAP does not bind. Thus, cAMP acts as the effector molecule, determining the effect of CAP on lac operon transcription. The intracellular cAMP concentration is sensitive to the presence or absence of glucose. High concentrations of glucose cause sharp decreases in the intracellular concentration of cAMP. Glucose prevents the activation of adenylcyclase, the enzyme that catalyzes the formation of cAMP from ATP. Thus, the presence of glucose results in a decrease in the intracellular concentration of cAMP. In the presence of a low concentration of cAMP, CAP cannot bind to the lac operon promoter. In turn, RNA polymerase cannot bind efficiently to the lac promoter in the absence of bound CAP/cAMP. Thus, in the presence of glucose, lac operon transcription never exceeds 2 percent of the induced rate observed in the absence of glucose.

Section: 17.3 The Lactose Operon in E.coli: Induction and Catabolite Repression Difficulty: Hard

57) How does attenuation help regulate gene expression in the trp operon?

Answer: Attenuation occurs by control of the termination of transcription at a site near the end of the mRNA leader sequence. This “premature” termination of trp operon transcription occurs only in the presence of tryptophan-charged tRNATrp. When this premature termination or attenuation occurs, a truncated (140 nucleotides) trp transcript is produced. The attenuator region has a nucleotide-pair sequence essentially identical to the transcription-termination signals found at the ends of most bacterial operons. Transcription of these termination signals yields a nascent RNA with the potential to form a hydrogen-bonded hairpin structure followed by several uracils.

When a nascent transcript forms this hairpin structure, it causes a conformational change in the associated RNA polymerase, resulting in termination of transcription within the following, more weakly hydrogen-bonded (A:U)n region of DNA-RNA base-pairing. The 162-nucleotide-long leader sequence of the trp operon mRNA contains sequences that can base-pair to form alternate stem-and-loop or hairpin structures. The presence or absence of tryptophan determines which of these alternative structures will form. The leader peptide contains two contiguous tryptophan residues. The two trp codons are positioned such that in low concentrations of tryptophan, the ribosome will stall before it encounters the base-paired structure formed by leader regions 2 and 3.

Because the pairing of regions 2 and 3 precludes the formation of the transcription-termination hairpin by the base-pairing of regions 3 and 4, transcription will continue past the attenuator into the trpE gene in the absence of tryptophan. In the presence of sufficient tryptophan, the ribosome can translate past the trp codons to the leader-peptide termination codon. In the process, it will disrupt the base-pairing between leader regions 2 and 3. This disruption leaves region 3 free to pair with region 4, forming the transcription-termination hairpin. Thus, in the presence of sufficient tryptophan, transcription frequently (about 90 percent of the time) terminates at the attenuator, reducing the amount of mRNA for the trp structural genes.

Section: 17.4 The Tryptophan Operon in E.coli: Repression and Attenuation Difficulty: Hard

Chapter 18 Test Bank Question Type: Multiple Choice 1) At which level is the gene expression of eukaryotes regulated? a) Transcription b) Post-transcriptional processing c) Translation d) Transcription and post-transcriptional processing e) All of these are correct.

Answer: e Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Easy 2) If gene expression is regulated at the transcriptional level, where in a eukaryotic cell does this regulation take place? a) Nucleus b) Mitochondria c) Rough ER d) Ribosome e) Cytoplasm Answer: a Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Easy 3) Positive and negative regulator proteins that bind to specific regions of the DNA and stimulate or inhibit transcription in eukaryotes are known as a) transcription inhibitors. b) transcription factors. c) unit factors. d) translation regulators. e) None of these are correct.

Answer: b Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Easy 4) Which of the following statements is correct? a) Prokaryotes undergo processing after transcription and eukaryotes do not. b) Eukaryotes undergo processing after transcription and prokaryotes do not. c) Prokaryotes undergo transcription but eukaryotes do not. d) Eukaryotes undergo transcription but prokaryotes do not. e) There are no differences between the expression of genes in prokaryotes and eukaryotes.

Answer: b Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Medium

5) Where does RNA transcript modification occur in eukaryotic cells? a) Nucleus b) Mitochondria c) Rough ER d) Lysosome e) Ribosome Answer: a Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Easy 6) Which component of a eukaryotic cell can serve as the location for regulation of gene expression? a) Nucleus b) Cytoplasm c) Mitochondria d) Nucleus and cytoplasm e) All of these are correct.

Answer: d Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Easy 7) The protein-DNA interactions that control whether or not a gene is accessible to RNA polymerase occur during which phase of gene expression? a) Replication b) Transcription c) Translation d) Replication and transcription e) Transcription and translation Answer: b Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Easy

8) Why is transcriptional control of gene expression more difficult in eukaryotes than in prokaryotes?

1. Genes are sequestered in the nucleus.

2. Eukaryotic cells need fairly elaborate internal signaling systems to control the transcription of DNA. 3. Environmental cues may have to pass through layers of cells in order to have an impact on the transcription of genes in a particular tissue. a) 1 b) 2 c) 3 d) 1 and 2 e) All of these are correct.

Answer: e Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Medium 9) How is it possible for a single gene to encode different polypeptides?

1. Post translational processing 2. Alternative splicing during post transcriptional processing 3. Abnormal spliceosome activity during post transcriptional processing a) 1 b) 2 c) 3 d) 1 and 2 and 3 e) It is not possible.

Answer: b Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Easy 10) In a gene with multiple introns present, what will happen if two successive introns, separated by one exon are removed together at the same time by a spliceosome? a) The exon will be removed with the introns. b) The exon will be removed but later reinserted in the same place. c) The exon will be removed but later reinserted at the end of the gene. d) The exon will remain in the sequence. e) The exon will remain in the sequence but will be altered.

Answer: a Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Medium

11) How does alternative splicing modify an RNA transcript to allow for the production of different polypeptides? a) It modifies the coding sequence of an RNA by deleting some of its exons. b) It modifies the coding sequence of an RNA by deleting all of its introns. c) It modifies the coding sequence of an RNA by deleting only some of the introns. d) It modifies the non-coding sequence of an RNA by deleting all of its exons. e) None of these are correct.

Answer: a Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Easy 12) Which of the following is a good example of alternative splicing in eukaryotes?

1. Lac operon 2. Troponin T gene in rats 3. Sex-lethal (Sxl) gene in Drosophila a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: e Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Medium 13) Which of the following can affect RNA transcript stability in the cytoplasm? a) 3 Poly A tails b) The sequence of the 3 untranslated region (3 UTR) preceding a poly(A) tail c) Chemical factors, such as hormones d) Small interfering RNAs (siRNAs) e) All of these are correct.

Answer: e Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Easy 14) Which of the following would not induce eukaryotic gene expression? a) Heat b) Light c) Hormones d) siRNAs e) Growth factors Answer: d Section: 18.2 Induction of Transcriptional Activity by Environmental and Biological Factors Difficulty: Easy

15) A group of proteins that help to stabilize the internal cellular environment when organisms are subjected to the stress of high temperature are known as a) cold shock proteins. b) heat shock proteins. c) chaperone proteins. d) transcription proteins. e) hormones.

Answer: b Section: 18.2 Induction of Transcriptional Activity by Environmental and Biological Factors Difficulty: Easy 16) Which level of gene regulation controls the expression of heat shock proteins? a) Transcription b) Post-transcription processing c) Translation d) Post-transcription processing and translation e) All of these are correct.

Answer: a Section: 18.2 Induction of Transcriptional Activity by Environmental and Biological Factors Difficulty: Easy 17) Which of the following genes is expressed in reaction to the exposure of light? a) Troponin T b) HSPs c) rbcS d) Sxl e) None of these are correct.

Answer: c Section: 18.2 Induction of Transcriptional Activity by Environmental and Biological Factors Difficulty: Easy 18) Which of the following is not true regarding steroid hormones? a) They are small, lipid-soluble molecules derived from cholesterol. b) Insulin is an example. c) Testosterone and estrogen are examples. d) The have little or no trouble passing through cell membranes. e) They interact with cytoplasmic or nuclear proteins called hormone receptors.

Answer: b Section: 18.2 Induction of Transcriptional Activity by Environmental and Biological Factors Difficulty: Medium

19) Hormone-induced gene expression is mediated by specific sequences in the DNA known as A) HREs. B) HSPs. C) ESTs. D) STRs. E) PCRs. Answer: a Section: 18.2 Induction of Transcriptional Activity by Environmental and Biological Factors Difficulty: Easy 20) Which of the following is true regarding hormone response elements? a) The more HREs present, the more vigorous transcription. b) The fewer HREs present, the more vigorous transcription. c) The HREs are specific sequences present in the tRNA molecule. d) The HREs are situated far away from the genes they regulate. e) The HREs are part of post-transcriptional control mechanisms.

Answer: a Section: 18.2 Induction of Transcriptional Activity by Environmental and Biological Factors Difficulty: Medium 21) Which of the following is true about signal transduction by hormones? a) All hormones are linear chains of amino acids (peptides). b) Hormones can diffuse freely in and out of cells. c) Steroid hormone-receptor complexes can enter the cell nucleus. d) Peptide hormones directly act on DNA to stimulate transcription. e) Binding of a peptide hormone to its receptor causes a conformational change in the hormone molecule.

Answer: c Section: 18.2 Induction of Transcriptional Activity by Environmental and Biological Factors Difficulty: Medium 22) Which of the following is true regarding eukaryotic transcription factors? a) Most eukaryotes have only one or a few transcription factors. b) Many transcription factors have dimerization motifs. c) Eukaryotic transcriptional activation does not require protein-protein interactions. d) Transcription factors can be active only when they form homodimers. e) Homeodomains are a class of transcription factors.

Answer: b

23) Which of the following is a component of eukaryotic and prokaryotic gene regulation? a) Alternate splicing b) Heat shock proteins c) Hormone responsive elements d) Euchromatin e) Gene-dosage compensation.

Answer: b Section: 18.2 Induction of Transcriptional Activity by Environmental and Biological Factors Difficulty: Easy 24) Enhancers are a) domains in proteins that enhance gene transcription. b) special cis regions in the vicinity of a gene to which transcription factors bind. c) also capable of inhibiting gene transcription. d) regions along a promoter to which basal transcription factors bind. e) not able to act over long distances away from the gene.

Answer: b Section: 18.3 Molecular Control of Transcription in Eukaryotes Difficulty: Easy 25) Which of the following is true of enhancer sequences? a) They can act over very large distances. b) They can only act in one specific orientation. c) They are usually between 10 and 35 bp long. d) They can only function if located upstream of a gene. e) They can only function if located downstream of a gene.

Answer: a Section: 18.3 Molecular Control of Transcription in Eukaryotes Difficulty: Medium 26) Which of the following motifs is not characteristic of a eukaryotic transcription factor? a) Transmembrane domain b) Helix-turn-helix c) Leucine zippers d) Helix-loop-helix e) All of these are correct.

Answer: a Section: 18.3 Molecular Control of Transcription in Eukaryotes Difficulty: Easy

27) The use of small RNA molecules to interfere with gene expression is known as a) DNA interference. b) RNA interference. c) hormone regulation. d) HRE regulation. e) None of these are correct.

Answer: b Section: 18.4 Posttranscriptional Regulation of Gene Expression by RNA Interference Difficulty: Easy 28) Which of the following molecules participates in RNA interference? a) siRNA b) tRNA c) rRNA d) dsDNA e) All of these are correct.

Answer: a Section: 18.4 Posttranscriptional Regulation of Gene Expression by RNA Interference Difficulty: Easy 29) Which type of enzyme produces siRNA molecules in eukaryotes? a) Restriction exonucleases b) DNA restriction endonucleases c) Dicer enzymes d) Regulator enzymes e) RNA polymerase Answer: c Section: 18.4 Posttranscriptional Regulation of Gene Expression by RNA Interference Difficulty: Easy 30) An interaction mediated by base pairing between the single strand of RNA in the RNA-protein complex and a complementary sequence in the messenger RNA molecule prevents the expression of the gene that produced the mRNA. This is known as a) RNA Induced Silencing Complex. b) RNA Introduction Salination Complex. c) Ribosomal Induced Silencing Complex. d) Ribsomal Interfering Single Complex. e) RNA Interference Silencing Complex.

Answer: a Section: 18.4 Posttranscriptional Regulation of Gene Expression by RNA Interference Difficulty: Easy

31) RISC-associated RNAs that result in mRNA cleavage are usually termed a) siRNAs. b) miRNAs. c) SNRPs. d) mtDNAs. e) STRs. Answer: a Section: 18.4 Posttranscriptional Regulation of Gene Expression by RNA Interference Difficulty: Easy 32) Whenever the RNA within the RISC pairs imperfectly with its target sequence, the mRNA is usually not cleaved; instead, translation of the mRNA is inhibited. RISC-associated RNAs that have this effect are usually termed a) siRNAs. b) miRNAs. c) SNRPs. d) mtDNAs. e) STRs. Answer: b Section: 18.4 Posttranscriptional Regulation of Gene Expression by RNA Interference Difficulty: Medium 33) Which of the following could be a source for a siRNA or miRNA molecule? a) Mir gene b) Transposon c) RNA virus d) Transgene e) All of these are correct.

Answer: e Section: 18.4 Posttranscriptional Regulation of Gene Expression by RNA Interference Difficulty: Easy 34) Which of the following gives support to the idea that transcription takes place in “open” regions of the chromosome?

1. The fact that 3H-uridine incorporated into newly synthesized RNA is localized around the lateral loops of the lampbrush chromosomes rather than around the condensed axes in amphibian oocytes.

2. The study of polytene chromosomes in Drosophila and other Dipteran insects 3. The study of lambrush chromosomes in human oocytes a) 1 b) 2 c) 3 d) 1 and 2 e) All of these are correct.

Answer: d Section: 18.5 Gene Expression and Chromatin Organization Difficulty: Medium

35) The temporal sequence of puffing in polytene chromosomes is controlled by a) ecdysone. b) testosterone. c) estrogen. d) progesterone. e) insulin.

Answer: a Section: 18.5 Gene Expression and Chromatin Organization Difficulty: Easy 36) The nuclease sensitivity of transcriptionally active genes depends on at least two small nonhistone proteins known as a) HMG 14. b) HMG 17. c) HSP 13. d) HMG 14 and HMG 17. e) All of these are correct.

Answer: d Section: 18.5 Gene Expression and Chromatin Organization Difficulty: Easy 37) DNA regions which are locally unwound because transcription has begun, and which are sensitive to low doses of DNAse 1 are known as a) RNase hypersensitive sites. b) RNase hyposensitive sites. c) DNase 1 hypersensitive sites. d) DNase 1 hyposensitive sites. e) None of these are correct.

Answer: c Section: 18.5 Gene Expression and Chromatin Organization Difficulty: Easy 38) In the human genes for -globin, a locus control region is characterized by a) regulating the expression of the -globin gene. b) containing several DNAse I hypersensitive sites. c) being about 15 kb in length. d) acting upstream of the genes it affects. e) All of these are correct.

Answer: e Section: 18.5 Gene Expression and Chromatin Organization Difficulty: Easy

39) In transcribed DNA, nucleosomes are altered by multiprotein complexes that ultimately facilitate the action of RNA polymerase in a process known as a) chromosome reworking. b) chromatin remodeling. c) chromatin editing. d) chromosome altering. e) None of these are correct.

Answer: b Section: 18.5 Gene Expression and Chromatin Organization Difficulty: Easy 40) Which of the following statements is true regarding heterochromatin? a) It stains lightly when stained with Feulgen reagent. b) It represents transcriptionally active regions of the chromosome. c) Specialized chromatin structures (scs) can insulate genes from the effects of heterochromatin. d) Euchromatic genes that are artificially placed next to heterochromatin are transcribed at abnormally high levels. e) Heterochromatin is hypersensitive to DNAse digestion.

Answer: c Section: 18.5 Gene Expression and Chromatin Organization Difficulty: Medium 41) In one individual, euchromatic genes are artificially transposed to a heterochromatic environment, resulting in a mixture of normal and mutant characteristics. This is called a) position-effect variegation. b) gene amplification. c) gene-dosage compensation. d) inactivation of whole chromosomes. e) signal transduction.

Answer: a Section: 18.5 Gene Expression and Chromatin Organization Difficulty: Medium 42) Which of the following can be responsible for silencing a gene?

1. Protein complexes like the Polycomb complex 2. Methylation of DNA 3. Telomeric variant surface glycoprotein like vsg a) 1 b) 2 c) 3 d) 1 and 2 e) All of these are correct.

Answer: e Section: 18.5 Gene Expression and Chromatin Organization Difficulty: Medium

43) Which of the following can increase gene expression? a) Silencing a gene b) Amplifying a gene c) Imprinting a gene d) Silencing a gene and amplifying a gene e) All of these are correct.

Answer: b Section: 18.5 Gene Expression and Chromatin Organization Difficulty: Medium 44) All of the following statements are true regarding inactivation of X-chromosome in mammals, except: a) X-chromosome inactivation begins at the X-inactivation center. b) All of the genes on an inactivated X-chromosome are transcriptionally silent. c) Inactive X-chromosomes can be easily identified in mammalian cells. d) The Barr body represents an inactive X-chromosome. e) Inactive X-chromosomes have a different pattern of distribution of acetylated histone.

Answer: b Section: 18.6 Activation and Inactivation of Whole Chromosomes Difficulty: Medium 45) Which of the following statements is true regarding dosage compensation?

A) In Drosophila, one of the two X-chromosomes is randomly inactivated in females.

B) In mammals, the genes on the single X-chromosome in a male are transcribed at a higher level.

C) The X-inactive specific transcript is a functional RNA molecule that results in X-inactivation in Drosophila.

D) In C. elegans, dosage compensation results from partial repression of X-linked genes in hermaphrodites.

E) The gene product of the gene maleless is responsible for X-inactivation in mammals.

Answer: d Section: 18.6 Activation and Inactivation of Whole Chromosomes Difficulty: Medium

Question Type: Essay 46) How does alternative splicing help to modify RNA transcripts in eukaryotes. Explain your answer by using a real life example such as that of the troponin T gene in rats.

Answer: One example of alternate splicing occurs during the expression of the gene for troponin T, a protein found in the skeletal muscles of vertebrates; the size of this protein ranges from about 150 to 250 amino acids. In the rat, the troponin T gene is more than 16 kb long and contains 18 different exons (Figure 20.2). Transcripts of this gene are spliced in different ways to create a large array of mRNAs. When these are translated, many different troponin T polypeptides are produced. All these polypeptides share amino acids from exons 1–3, 9–15, and 18. However, the regions encoded by exons 4–8 may be present or absent, depending on the splicing pattern, and apparently in any combination. Additional variation is provided by the presence or absence of regions encoded by exons 16 and 17; if 16 is present, 17 is not, and vice versa. These different forms of troponin T presumably function in slightly different ways within the muscles, contributing to the variability of muscle cell action.

Section: 18.1 Ways of Regulating Eukaryotic Gene Expression: An Overview Difficulty: Medium 47) How can an external environmental factor cause the regulation of gene expression in eukaryotes? Explain your answer by providing a real life example such as the heat shock proteins in Drosophila.

Answer: The expression of the heat-shock proteins is regulated at the transcriptional level; that is, heat stress specifically induces the transcription of the genes encoding these proteins (Figure 20.4).

In Drosophila, for example, one of the heat-shock proteins called HSP70 (for heat-shock protein, molecular weight 70 kilodaltons) is encoded by a family of genes located in two nearby clusters on one of the autosomes. Altogether, there are five to six copies of these hsp70 genes in the two clusters. When the temperature exceeds 33°C, as it does on hot summer days, each of the genes is transcribed into RNA, which is then processed and translated to produce HSP70 polypeptides.

This heat-induced transcription of the hsp70 genes is mediated by a polypeptide called the heat-shock transcription factor, or HSTF, which is present in the nuclei of Drosophila cells. When Drosophila are heat stressed, the HSTF is chemically altered by phosphorylation. In this altered state, it binds specifically to nucleotide sequences upstream of the hsp70 genes and makes the genes more accessible to RNA polymerase II, the enzyme that transcribes most protein-encoding genes. The transcription of the hsp70 genes is then vigorously stimulated. The sequences to which the phosphorylated HSTF binds are called heat-shock response elements (HSEs).

Section: 18.2 Induction of Transcriptional Activity by Environmental and Biological Factors Difficulty: Medium

48) Compare and contrast the actions of steroid hormones with peptide hormones.

Answer: The first class, the steroid hormones, are small, lipid-soluble molecules derived from cholesterol. Because of their lipid nature, they have little or no trouble passing through cell membranes. Once these hormones have entered a cell, they interact with cytoplasmic or nuclear proteins called hormone receptors. The receptor/hormone complex that is formed then interacts with the DNA where it acts as a transcription factor to regulate the expression of certain genes.

The second class of hormones, the peptide hormones, are linear chains of amino acids. Like all other polypeptides, these molecules are encoded by genes. Because peptide hormones are typically too large to pass freely through cell membranes, the signals they convey must be transmitted to the interior of cells by membrane-bound receptor proteins. When a peptide hormone interacts with its receptor, it causes a conformational change in the receptor that eventually leads to changes in other proteins inside the cell. Through a cascade of such changes, the hormonal signal is transmitted through the cytoplasm of the cell and into the nucleus, where it ultimately has the effect of regulating the expression of specific genes. This process of transmitting the hormonal signal through the cell and into the nucleus is called signal transduction.

Section: 18.2 Induction of Transcriptional Activity by Environmental and Biological Factors Difficulty: Medium 49) How does the phenomenon of RNA interference regulate gene expression on a post-transcriptional level?

Answer: The phenomenon of RNA interference, which is summarized in Figure 20.11, involves small RNA molecules called short interfering RNAs (siRNAs) or microRNAs (miRNAs). These molecules, 21 to 28 base pairs long, are produced from larger, double-stranded RNA molecules by the enzymatic action of proteins that are double-stranded RNA-specific endonucleases. Because these endonucleases “dice” large RNA into small pieces, they are called Dicer enzymes. The siRNAs and miRNAs produced by Dicer activity are base-paired throughout their lengths except at their 3 ends, where two nucleotides are unpaired. In the cytoplasm, siRNAs and miRNAs become incorporated into ribonucleoprotein particles. The double-stranded siRNA or miRNA in these particles is unwound, and one of its strands is preferentially eliminated. The surviving single strand of RNA is then able to interact with specific messenger RNA molecules. This interaction is mediated by base pairing between the single strand of RNA in the RNA-protein complex and a complementary sequence in the messenger RNA molecule. Because this interaction prevents the expression of the gene that produced the mRNA, the RNA-protein particle is called an RNA-Induced Silencing Complex (RISC).

Section: 18.4 Posttranscriptional Regulation of Gene Expression by RNA Interference Difficulty: Medium

50) Null mutations in male-specific lethal loci in Drosophila lead to male-specific lethality. Why does this happen and what is known about the function of these genes?

Answer: Male-specific lethal loci are dosage compensation genes in Drosophila. In this organism, dosage compensation is achieved by transcribing the genes on the single X-chromosome more vigorously to equal the transcriptional activity of the two X-chromosomes in the female. Current evidence indicates that msl genes are responsible for dosage compensation and their gene products form some sort of protein complex that hyper activates X-linked genes by altering chromatin structure.

Section: 18.6 Activation and Inactivation of Whole Chromosomes Difficulty: Hard

Chapter 19 Test Bank Question Type: Multiple Choice 1) The genetic basis of complex traits is demonstrated by: 1. Resemblance between relatives 2. Similar appearance of monozygotic twins 3. Responses to selective breeding such as the increase in resistance to disease a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 19.1 Complex Traits Difficulty: Easy 2) Which of the following is an example of a complex trait? a) Body size b) Height c) Weight d) All of these e) None of these Answer: d Section: 19.1 Complex Traits Difficulty: Easy 3) Traits in which phenotypic variation can be measured in a sample of individuals from the population are known as a) qualitative traits. b) quantitative traits. c) unique traits. d) statistical traits. e) standard traits.

Answer: b Section: 19.1 Complex Traits Difficulty: Easy

4) Who was one of the first people to show that variation in a quantitative trait is due to a combination of genetic and environmental factors? a) Wilhelm Johannsen b) James Watson c) Francis Crick d) Edward East e) Joshua Lederberg Answer: a Section: 19.1 Complex Traits Difficulty: Easy 5) Which organism was studied by Herman Nilsson-Ehle, in his analysis of multiple genetic contributions to trait variability? a) Broad bean b) Wheat grains c) E. coli d) Garden peas e) Drosophila f) Tobacco Answer: b Section: 19.1 Complex Traits Difficulty: Easy 6) Underlying risk factors that contribute to a variable are called a) the liability. b) the risk. c) the threshold. d) the quantity. e) None of these Answer: a Section: 19.1 Complex Traits Difficulty: Easy 7) A trait that appears when the liability exceeds a certain level is known as a(n) a) liability trait. b) quantitative trait. c) threshold trait. d) qualitative trait. e) phenotypic trait.

Answer: c Section: 19.1 Complex Traits Difficulty: Easy

8) Similarity with respect to a threshold trait is assessed by determining the a) liability rate. b) threshold rate. c) concordance rate. d) zygomatic rate. e) standard deviation.

Answer: c Section: 19.1 Complex Traits Difficulty: Easy 9) Which of the following could be regarded as a threshold trait? a) Cleft lip b) Schizophrenia c) Bipolar disorder d) Cleft lip and Schizophrenia e) All of these Answer: e Section: 19.1 Complex Traits Difficulty: Medium 10) How can the number of genes influencing a trait be estimated? a) Phenotypic segregation b) DNA hybridization c) Northern blotting d) Allelism e) Ames Test Answer: a Section: 19.2 Statistics of Quantitative Traits Difficulty: Easy 11) The evidence that threshold traits in humans have a genetic basis has come from the study of a) non-related individuals. b) marriage partners. c) parents and children. d) twins. e) None of these Answer: d Section: 19.2 Statistics of Quantitative Traits Difficulty: Easy

12) Which of the following is not true of threshold traits? a) Expression is significantly affected by environment b) Genetic liability must cross a critical value for expression to occur c) Monozygotic twins much more likely to both express such traits d) Nature of genetic factors involved are poorly understood e) They often show continuous variation in phenotype.

Answer: e Section: 19.2 Statistics of Quantitative Traits Difficulty: Medium 13) What type of evidence suggests the genetic basis of traits like cleft lip? a) Concordance for monozygotic twins is much greater than for dizygotic twins. b) Environment significantly affects such traits. c) Genes predispose individuals to express the trait. d) Risk increases as genetic relationships become more distant. e) Some kind of threshold must be reached before trait can be expressed.

Answer: a Section: 19.2 Statistics of Quantitative Traits Difficulty: Medium 14) Which of the following is not a feature of quantitative traits? a) Environment influences phenotype b) Effect of environment can vary with genotype c) Mendel's laws do not apply to these traits d) Two or more genes are often involved e) All of these Answer: c Section: 19.2 Statistics of Quantitative Traits Difficulty: Medium 15) The hallmark of quantitative traits is that a) Mendel's laws do not apply to these traits. b) the environment does not influence phenotype. c) there is never more than one gene involved. d) they vary continuously in a population of individuals. e) All of these Answer: d Section: 19.2 Statistics of Quantitative Traits Difficulty: Easy

16) The small fraction of all the individuals in the population that can be measured is known as the a) group. b) sample. c) population. d) set. e) None of these Answer: b Section: 19.2 Statistics of Quantitative Traits Difficulty: Easy 17) A statistic that gives the “center” or average of a distribution is known as the a) mean. b) mode. c) deviation. d) frequency. e) χ2 value.

Answer: a Section: 19.2 Statistics of Quantitative Traits Difficulty: Easy 18) In a sample, the class that contains the most observations is known as the a) mean class. b) modal class. c) normal class. d) hyper class. e) None of these Answer: b Section: 19.2 Statistics of Quantitative Traits Difficulty: Easy 19) Which of the following statistics is useful for describing quantitative traits? a) Correlation coefficient b) Frequency distribution c) Mode of the frequency distribution d) Mean e) All of these Answer: e Section: 19.2 Statistics of Quantitative Traits Difficulty: Easy

20) Which of the following is the formula for variance? a) [(Xi - X)(Yi - Y)]/(n-1)SxSy b) [(Xi - X)(Yi - Y)]/(n-1) c) (Xi - X)2/(n-1) d) Xi/n e) None of these Answer: c Section: 19.3 Statistical Analysis of Quantitative Traits Difficulty: Easy 21) Standard deviation can be more useful than variance in describing a quantitative trait because it a) is always positive. b) is a more accurate measure. c) is the square of the variance. d) has the same units of measure as the mean. e) has useful mathematical properties.

Answer: d Section: 19.3 Statistical Analysis of Quantitative Traits Difficulty: Medium 22) Populations with a small standard deviation for a particular quantitative trait are a) usually prone to extinction. b) usually more homozygous for that trait. c) usually more heterozygous for that trait. d) not very common. e) always normally distributed.

Answer: b Section: 19.3 Statistical Analysis of Quantitative Traits Difficulty: Medium 23) Which of the following statistics indicates the extent to which data are scattered around the mean in a frequency distribution? a) Mode b) Variance c) Standard deviation d) Mode and Variance e) Variance and Standard deviation Answer: e Section: 19.3 Statistical Analysis of Quantitative Traits Difficulty: Easy

24) A trait that is controlled by many genes is referred to as a) pleiotrophic. b) polygenic. c) multigenic. d) dominant. e) codominant.

Answer: b Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Easy 25) Quantitative traits are controlled by many different factors in the environment and in the genotype. This is the ___________ hypothesis. a) Multiple Factor Hypothesis b) Polygenic Factor Hypothesis c) Pleiotropic Factor Hypothesis d) Dominant Factor Hypothesis e) Majority Factor Hypothesis Answer: a Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Easy 26) What is the total phenotypic variance for a wheat population if the environmental variance is 12.3 days2 and the genetic variance is 3.7 days2? a) 12.3 days2 b) 3.7 days2 c) 16.0 days2 d) 45.5 days2 e) 8.6 days2 Answer: c Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Medium 27) The proportion of the total phenotypic variance that is due to genetic differences among individuals in a population is known as a) narrow sense heritability. b) broad sense heritability. c) heritability. d) broad sense inheritance. e) narrow sense inheritance.

Answer: b Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Easy

28) If the broad sense heritability is close to 0 for a population then a) much of the observed variability in the population is due to genetic differences. b) much of the observed variability in the population is due to heritable traits. c) little of the observed variability is due to genetic differences. d) little of the observed variability is due to environmental factors. e) None of these Answer: c Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Medium 29) The additive genetic variance, Va,, as a fraction of the total phenotypic variance, is called the a) total phenotypic variance. b) broad sense heritability. c) narrow sense heritability. d) broad sense inheritance. e) narrow sense inheritance.

Answer: c Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Easy 30) The difference between the mean of the selected parents and the mean of the population from which they were selected is known as a) selection quotient. b) selection differential. c) selection additive. d) selection quotient and selection differential. e) All of these Answer: b Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Easy 31) Knowledge of heritability is important for a) developing conservation strategies for natural populations. b) developing goal oriented selective breeding programs. c) developing health care programs (for high blood pressure, etc.). d) understanding how natural populations evolve. e) All of these Answer: e Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Easy

32) Which of the following is a component of narrow sense heritability? a) Additive effects of alleles b) Dominance interactions among alleles c) Environmental effects d) Epistatic relationships among alleles e) All of these Answer: a Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Easy 33) Which of the following is not a component of broad sense heritability? a) Additive effects of alleles b) Dominance interactions among alleles c) Environmental effects d) Epistatic relationships among alleles e) All of these Answer: c Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Medium 34) Which of the following is a component of the total variability of a trait? a) Additive effects of alleles b) Dominance interactions among alleles c) Environmental effects d) Epistatic relationships among alleles e) All of these Answer: e Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Easy 35) Which of the following has been instrumental in discovering genetic links to human multifactorial diseases? a) Crossing studies b) Estimating narrow sense heritability c) QTL mapping d) Selection studies e) Studies in different environments Answer: c Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Easy

36) Which statistic quantitatively summarizes a pattern of resemblance? a) Regression analysis b) Correlation coefficient c) Standard deviation d) Mean e) Mode Answer: b Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Easy 37) When the correlation coefficient is zero, a) the measurements are uncorrelated. b) the measurements are weakly correlated. c) the measurements are moderately correlated. d) the measurements are highly correlated. e) None of these Answer: a Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Easy 38) The proportion of heritability that is due to shared environmental factors is known as a) rearing coefficient b) environmentality c) environment coefficient d) rearing coefficient and environment coefficient e) All of these Answer: b Section: 19.5 Correlations Between Relatives Difficulty: Easy 39) The correlation between monozygotic twins reared apart provides an estimate of a) narrow sense heritability. b) narrow sense correlation. c) broad sense heritability. d) broad sense correlation. e) None of these Answer: c Section: 19.5 Correlations Between Relatives Difficulty: Easy

40) The correlation between dizygotic twins reared apart provides a maximum estimate of a) narrow sense heritability. b) narrow sense correlation. c) broad sense heritability. d) broad sense correlation. e) None of these Answer: a Section: 19.5 Correlations Between Relatives Difficulty: Easy 41) Which of the following are indicators that genetic factors can influence human behavior? a) Huntington's disease b) Phenylketonuria c) Down Syndrome d) Huntington's disease and Phenylketonuria e) All of these Answer: e Section: 19.5 Correlations Between Relatives Difficulty: Easy 42) IQ had a high heritability in twin studies a) however, environmental effects were negligible. b) however, we cannot assume differences among human population are genetic. c) therefore, differences among human populations must be genetic. d) therefore, intelligence is highly dependent on genotype. e) therefore, environment has little affect on intelligence.

Answer: b Section: 19.5 Correlations Between Relatives Difficulty: Easy 43) In a quantitative genetic analysis of personality traits among monozygotic twins, it was found that: 1. Genetic differences play a small role in determining personality traits.

2. Genetic differences play a significant role in determining personality traits.

3. The broad-sense heritability is reasonably high in the 0.34-0.5 range. a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: e Section: 19.5 Correlations Between Relatives Difficulty: Medium

44) What fraction of the variation among IQ scores is attributable to genetic differences among people? a) 0.1 b) 0.3 c) 0.5 d) 0.7 e) 0.9 Answer: d Section: 19.5 Correlations Between Relatives Difficulty: Easy 45) The most thorough genetic analysis of personality in the general population has come from a) Minnesota Study of Dizygotic Twins Raised Together. b) Minnesota Study of Monozygotic Twins Raised Together. c) Minnesota Study of Monozygotic Twins Raised Apart. d) Minnesota Study of Dizogotic Twins Raised Apart. e) Minnesota Study of Non-Biological Children.

Answer: c Section: 19.5 Correlations Between Relatives Difficulty: Easy Question Type: Essay 46) How did Nilsson-Ehle and East provide evidence that the genetic component of this variation could involve the contributions of several different genes?

Answer: Nilsson-Ehle studied color variation in wheat grains. When he crossed a white-grained variety with a dark red-grained variety, he obtained an F1 with an intermediate red phenotype (Figure 23.1). Self-fertilization of the F1 produced an F2 with seven distinct classes, ranging from white to dark red. The number of F2 classes and the phenotypic ratio that Nilsson-Ehle observed suggested that three independently assorting genes were involved in the determination of grain color. Nilsson-Ehle hypothesized that each gene had two alleles, one causing red grain color and the other white grain color, and that the alleles for red grain color contributed to pigment intensity in an additive fashion. Based on this hypothesis, the genotype of the white-grained parent could be represented as aa bb cc, and the genotype of the red-grained parent could be represented as AA BB CC. The F1 genotype would be Aa Bb Cc, and the F2 would contain an array of genotypes that would differ in the number of pigment-contributing, alleles present. Each phenotypic class in the F2 would carry a different number of these pigment-contributing alleles. The white class, for example, would carry none, the intermediate red class would carry three, and the dark red class would carry six. Nilsson-Ehle's work, published in 1909, showed that a complex inheritance pattern could be explained by the segregation and assortment of multiple genes.

The American geneticist Edward M. East extended Nilsson-Ehle's studies to a trait that did not show simple Mendelian ratios in the F2. East studied the length of the corolla in tobacco flowers (Figure 23.2a). In one pure line, the corolla length averaged 41 mm; in another, it averaged 93 mm.

Within each pure line, East observed some phenotypic variation—presumably the result of environmental influences (Figure 23.2b). By crossing the two lines, East obtained an F1 that had intermediate corolla length and approximately the same amount of variation that he had seen

within each of the parental strains. When East intercrossed the F1 plants, he obtained an F2 with about the same corolla length, on average, that he saw in the F1; however, the F2 plants were much more variable than the F1. This variability was due to two sources: (1) the segregation and independent assortment of different pairs of alleles controlling corolla length, and (2) environmental factors. East inbred some of the F2 plants to produce an F3and observed less variation within the different F3 lines than in the F2. The reduced amount of variation within the F3 lines was presumably due to the segregation of fewer allelic differences. Thus, the complex inheritance pattern that East observed with corolla length could be explained by a combination of genetic segregation and environmental influences.

Section: 19.1 Complex Traits Difficulty: Medium 47) On a class exam three students earned an A, twelve students earned a B, 48 students earned a C, twelve students earned a D and four students earned an F. Draw the frequency distribution of this sample of students.

Answer:

Section: 19.2 Statistics of Quantitative Traits Difficulty: Medium 48) Determine the mode, mean, variance and standard deviation for the following data set.

3,5,6,6,8,1,12 Answer: Mode = 6; Mean = 5.857; Variance = 12.476; SD = 3.532 Section: 19.2 Statistics of Quantitative Traits Difficulty: Medium

49) How does the correlation between monozygotic twins reared apart provides an estimate of the broad-sense heritability?

Answer: Correlation coefficients calculated by the formula given in the previous section can be interpreted in terms of broad- and narrow-sense heritabilties. Geneticists have analyzed the relationships among these quantities, beginning with the pioneering work of R. A. Fisher. This analysis assumes that T, the value of a trait in an individual, is equal to the mean of the population () plus genetic (g) and environmental (e) deviations from the mean: T =m+g+e =m+a+d+i+e The terms a, d, and i in this expression are, respectively, the additive, dominance, and epistatic components of the genetic deviation from the mean. It is also necessary to assume that the genetic factors influencing the phenotype are independent of the environmental factors. Under these assumptions, the correlation coefficient for a pair of relatives equals the proportion of the total variance in the trait that is due to the genetic and environmental factors shared by the relatives.

Table 23.3 presents theoretical interpretations of correlation coefficients for different types of human twins.

Monozygotic twins reared apart (MZA) have identical genotypes. Thus, these twins share all the genetic factors that contribute to the term g in the expression for the value of a quantitative trait, including the additive effects of alleles, the effects of dominance, and the effects of epistasis.

However, because MZA have had separate upbringings, they do not share the environmental effects represented by the term e in the expression. Consequently, a correlation between MZA depends only on their identical genotypes. In the theory of quantitative genetics, this correlation equals the proportion of the total phenotypic variance that is due to genetic differences among the twin pairs—that is, it equals the broad-sense heritability, H2.

Section: 19.4 Molecular Analysis of Complex Traits Difficulty: Medium 50) How do the studies between monozygotic twins show that there is a genetic component to intelligence?

Answer: For IQ test scores, the correlation coefficients of MZ twins, reared together or apart, are very high—in the range of 0.7–0.8 (Table 23.4). By comparison, the correlation coefficients of DZ twins tend to be lower—presumably because they share only half their genes, and the correlation coefficients for unrelated individuals reared together are essentially zero. Such analyses strongly suggest that whatever an IQ test measures, it has a large genetic component. This conclusion is supported by other correlation analyses. For example, the IQs of adopted children are more strongly correlated with the IQs of their biological parents than with those of their adoptive parents. Thus, in the determination of IQ, the biological (that is, genetic) link between parents and children seems to be more influential than the environmental one.

Section: 19.5 Correlations Between Relatives Difficulty: Medium

Chapter 20 Test Bank Question Type: Multiple Choice 1) What is explained by a major population genetics theory?

1. The number of individuals in a population 2. The allelic frequencies within a population 3. The number of individuals outside a population a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: b Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy 2) Which of the following can be calculated using a major population genetics theory?

1. The frequency of homozygous dominant and recessive individuals 2. The frequency of heterozygous individual 3. The frequency of a dominant allele a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy 3) There are 300 individuals in sample group #1 taken from population X. How many total alleles for the trait in question are present in the sample group, assuming the trait is controlled by a single codominant gene? a) 150 b) 300 c) 600 d) 900 e) 1200 Answer: c Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium

4) There are 300 individuals in sample group #1 taken from population X. Seventy-five individuals of sample group are homozygous for the trait in question (IAIA). Seventy five individuals are homozygous for the trait (IBIB). Everyone else is heterozygous (IAIB) for the trait.

What is the frequency of allele A in the sample group, assuming the trait is controlled by a single gene? a) 0 b) 0.375 c) 0.5 d) 0.75 e) 1 Answer: c Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 5) There are 300 individuals in sample group #1 taken from population X. Seventy-five individuals of sample group are homozygous for the trait in question (IAIA). Seventy five individuals are homozygous for the trait (IBIB). Everyone else is heterozygous (IAIB) for the trait.

What is the frequency of allele B in the sample group, assuming the trait is controlled by a single gene? a) 0 b) 0.375 c) 0.5 d) 0.75 e) 1 Answer: c Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 6) Which of the following is true regarding determining the number of alleles in a population sample?

1. It may be impossible to count the number of alleles if one allele is dominant.

2. When counting X-linked alleles it is only important to count the different alleles in males.

3. Codominant alleles can always be counted because they produce unique phenotypes. a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: c Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy

7) In a sample of 400 men, 48 have X-linked color blindness and all the others have normal color vision. If men are hemizygous for the allele, what is the frequency for the X-linked color blindness allele? a) 0.1 b) 0.12 c) 0.24 d) 0.48 e) 0.88 Answer: b Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 8) In a sample of 500 men, 240 have X-linked hemophilia and all the others have normal blood clotting. If men are hemizygous for the allele, what is the frequency for the X-linked hemophilia allele? a) 0.1 b) 0.12 c) 0.24 d) 0.48 e) 0.88 Answer: d Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 9) In a population, whenever the second most frequent allele of a gene has a frequency greater than 0.01, we refer to the situation as a a) genetic polymorphism. b) genetic anomaly. c) genetic mutation. d) genetic abnormality. e) None of these Answer: a Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy 10) Which of the following individuals described a mathematical relationship between allele frequencies and genotype frequencies? a) G.H. Hardy b) Wilhelm Weinberg c) Edward East d) G.H. Hardy and Wilhelm Weinberg e) Wilhelm Weinberg and Edward East Answer: d Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy

11) Which of the following allows one to predict a population's genotype frequencies from its allele frequencies? a) Hardy-Weinberg principle b) East-Hayflick principle c) Hooke-Van Leuwenhoek principle d) Morgan-Hunt principle e) None of these Answer: a Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy 12) Which of the following accurately describes the Hardy-Weinberg genotype frequency expression? a) P2 + 2p +q2 b) P2 + 2pq + q3 c) P2+2pq+q2 d) P3 + 3pq+q3 e) (p+q)3 Answer: c Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy 13) Which of the following is a key assumption underlying the Hardy-Weinberg principle?

1. The members of the population mate at random with respect to the gene under study.

2. The members of the population are artificially selected with respect to the gene under study.

3. The members of the population must live on an island. a) 1 b) 2 c) 3 d) 1 and 3 e) 2 and 3 Answer: a Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 14) Which of the following is a force that could upset the Hardy-Weinberg equilibrium? a) Mutation b) Migration c) Natural selection d) Genetic drift e) All of these Answer: e Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy

15) The frequency of sickle cell anemia, caused by a homozygous condition (HbSHbS) is approximately 0.0016. Assuming the Hardy-Weinberg equilibrium applies, what is the frequency of the HbS allele? a) 0.01 b) 0.00008 c) 0.08 d) 0.04 e) 0.4 Answer: d Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 16) The frequency of sickle cell anemia, caused by a homozygous condition (HbSHbS) is approximately 0.0016. The normal condition is also a homozygous condition (HbAHbA).

Assuming the Hardy-Weinberg equilibrium applies, what is the frequency of the HbA allele? a) 0.0004 b) 0.04 c) 0.4 d) 0.9216 e) 0.96 Answer: e Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 17) The frequency of sickle cell anemia, caused by a homozygous condition (HbSHbS) is approximately 0.0016. The normal condition is also a homozygous condition (HbAHbA).

Assuming the Hardy-Weinberg equilibrium applies, what is the frequency of the carrier state (HbAHbS)? a) 0.0228 b) 0.0432 c) 0.0768 d) 0.0987 e) 0.0384 Answer: c Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium

18) In a sample of 600 individuals (400 men and 200 women) 48 men have X-linked color blindness (Xc) and all the others have normal color vision (XC). What is the frequency of women who are colorblind?

A) 0.12 B) 0.88 C) 0.346 D) 0.02 E) 0.21 Answer: d Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 19) In a sample of 600 individuals (400 men and 200 women) 48 men have X-linked color blindness (Xc) and all the others have normal color vision (Xc). What is the frequency of women who are carriers? a) 0.12 b) 0.88 c) 0.346 d) 0.02 e) 0.21 Answer: e Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 20) The A–B–O blood types are determined by three alleles IA, IB, and i. In one population the frequency of individuals who are homozygous for type A blood is 0.063, the frequency of individuals who are homozygous for type B blood is 0.015, and the frequency of individuals who are blood type O is 0.44. What is the frequency of individuals are blood type AB? a) 0.251 b) 0.122 c) 0.031 d) 0.061 e) 0.5 Answer: d Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium

21) The A–B–O blood types are determined by three alleles IA, IB, and i. In one population the frequency of individuals who are homozygous for type A blood is 0.063, the frequency of individuals who are homozygous for type B blood is 0.015, and the frequency of individuals are blood type O is 0.44. What is the frequency of individuals who are heterozygous for type A blood? a) 0.251 b) 0.663 c) 0.166 d) 0.333 e) 0.061 Answer: d Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 22) The A–B–O blood types are determined by three alleles IA, IB, and i. In one population the frequency of individuals who are homozygous for type A blood is 0.063, the frequency of individuals who are homozygous for type B blood is 0.015, and the frequency of individuals who are blood type O is 0.44. What is the frequency of the i allele? a) 0.251 b) 0.663 c) 0.166 d) 0.333 e) 0.061 Answer: b Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 23) Which of the following is a type of non-random mating? a) Consanguinous mating b) Assortative mating c) Discoursive mating d) Consanguinous mating and assortative mating e) Assortative mating and discoursive mating Answer: d Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy

24) What effect does consanguineous mating and assortative mating have on genotypic frequencies in populations?

1. Reduce the frequency of homozyotes 2. Increase the frequency of homozygotes 3. Reduce the frequency of heterozygotes a) 1 b) 2 c) 3 d) 1 and 3 e) 2 and 3 Answer: e Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy 25) Which of the following prevent the Hardy-Weinberg principle from applying to a population? a) The population experiences random mating for the gene in question. b) The zygotes produced by random mating have different survival rates. c) The population is panmitic. d) The population is not subject to migration. e) All of these Answer: b Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy 26) When a population is a single interbreeding unit, it is said to be a) panmitic. b) non-panmitic. c) heterozygous. d) homozygous. e) None of these Answer: a Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy 27) Which assumption of the Hardy-Weinberg principle is violated by the concept of population subdivision? a) Random-mating b) Allele frequencies are uniform throughout the population c) Individuals are diploid d) Individuals are haploid e) None of these Answer: b Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy

28) In a population that experiences migration (i.e. the merging of two separate populations), why are observed allelic and genotypic frequencies different than what is predicted by the Hardy-Weinberg equilibrium? a) The observed genotype frequencies were not created by random mating within the entire merged population. b) The observed genotype frequencies were created by random mating within the entire merged population. c) The observed genotype frequencies were subject to inbreeding. d) The populations experienced unequal survival rates. e) None of these Answer: a Section: 20.1 The Theory of Allele Frequencies Difficulty: Easy 29) Which of the following occurs when genotypes differ in the ability to survive and reproduce? a) Unequal population division b) Population subdivision c) Natural selection d) Inbreeding e) None of these Answer: c Section: 20.2 Natural Selection Difficulty: Easy 30) The intensity of natural selection is quantified by the a) correlation coefficient. b) selection coefficient. c) cosanguinous coefficient. d) fitness number. e) All of these Answer: b Section: 20.2 Natural Selection Difficulty: Easy 31) At the level of the gene, natural selection changes the frequencies of a) genotypes in biomes. b) genotypes in populations. c) alleles in biomes. d) alleles in populations. e) All of these Answer: d Section: 20.2 Natural Selection Difficulty: Easy

32) At the level of the phenotype, natural selection influences the a) distribution of alleles. b) distribution of quantitative traits. c) distribution of genotypes. d) distribution of alleles and distribution of genotypes. e) None of these Answer: b Section: 20.2 Natural Selection Difficulty: Easy 33) Which of the following is most effective at fixing the recessive allele in a population? a) Selection for a recessive allele b) Selection against a recessive allele c) Selection against homozygous recessive individuals d) Selection against heterozygotes e) All of these Answer: a Section: 20.2 Natural Selection Difficulty: Medium 34) Which type of selection can affect the distribution of a quantitative trait? a) Directional selection b) Disruptive selection c) Stabilizing selection d) Directional selection and Disruptive selection e) All of these Answer: e Section: 20.2 Natural Selection Difficulty: Easy 35) Selection that favors extreme values of a trait at the expense of intermediate values is known as a) directional selection. b) disruptive selection. c) stabilizing selection. d) directional selection and disruptive selection. e) All of these Answer: b Section: 20.2 Natural Selection Difficulty: Easy

36) Selection that favors intermediate values of a trait at the expense of extreme values is known as a) directional selection. b) disruptive selection. c) stabilizing selection. d) directional selection and disruptive selection. e) All of these Answer: c Section: 20.2 Natural Selection Difficulty: Easy 37) What causes allele frequencies to change unpredictably in populations because of reproductive uncertainties? a) Random genetic abnormalities b) Random genetic drift c) Random mating d) Natural selection e) Artificial selection Answer: b Section: 20.3 Random Genetic Drift Difficulty: Easy 38) What is the ultimate source of all genetic variability? a) Natural selection b) Artificial selection c) Mutation d) Natural selection and artificial selection e) None of these Answer: c Section: 20.3 Random Genetic Drift Difficulty: Easy 39) How does random genetic drift affect large populations? a) It may be the primary evolutionary force. b) It has a moderate effect. c) It has very little effect. d) It is not currently known what kind of effect it has on large populations. e) It does not affect large populations at all.

Answer: c Section: 20.3 Random Genetic Drift Difficulty: Easy

40) What general effect(s) does random genetic drift have on populations?

1. It reduces the dominant homozygosity.

2. It reduces the heterozgosity.

3. It reduces the recessive homozygosity. a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: b Section: 20.3 Random Genetic Drift Difficulty: Medium 41) In diploid organisms, the rate at which genetic variability is lost by random genetic drift is a) 1/2N, where N is the population size. b) 1/3N, where N is the population size. c) 1/4N, where N is the population size. d) 1/5N, where N is the population size. e) None of these Answer: a Section: 20.3 Random Genetic Drift Difficulty: Easy 42) Which type of selection creates a dynamic equilibrium in which different alleles are retained in a population despite their being harmful in homozygotes? a) Balancing selection b) Disruptive selection c) Stabilizing selection d) Directional selection e) None of these Answer: a Section: 20.4 Populations in Genetic Equilibrium Difficulty: Easy 43) Which of the following human diseases is associated with balancing selection? a) Tay Sachs b) Down Syndrome c) Sickle Cell Anemia d) Lesch-Nyhan Syndrome e) Alzheimers Disease Answer: c Section: 20.4 Populations in Genetic Equilibrium Difficulty: Easy

44) Selection against a deleterious recessive allele is replenished in the population by mutation, and leads to a dynamic equilibrium in which the frequency of the recessive allele is a simple function of the mutation rate and the selection coefficient. Which formula represents this situation? a) p2+2pq+q2 b) q = us

ˆ c) H = 4 Nu /(4 Nu + 1) d) 1/2N e) None of these Answer: c Section: 20.4 Populations in Genetic Equilibrium Difficulty: Medium 45) A population's acquisition of selectively neutral alleles through mutation is balanced by the loss of these alleles through genetic drift. At equilibrium, the frequency of heterozygotes involving these alleles is a function of the population's size and the mutation rate is a) p2+2pq+q2 b) q = us

ˆ c) H = 4 Nu /(4 Nu + 1) d) 1/2N e) None of these Answer: c Section: 20.4 Populations in Genetic Equilibrium Difficulty: Medium Question Type: Essay 46) There are 800 individuals in sample group #1 taken from population X that were examined to the codominant trait R. When sampled 30 individuals were homozygous for allele Z (RZRZ), 100 individuals were homozygous for allele K (RKRK) and the rest were heterozygous (RZRK)).

Determine how many total alleles for the trait in question are present in the sample group, the frequency of allele Z, and the frequency of allele Q, assuming the trait is controlled by a single codominant gene?

Answer: Total alleles = 1600; Frequency of Z = 0.45625; Frequency of K = 0.54375 Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 47) In a sample of 3000 individuals, 1000 men and 2000 women, 200 men have hemophilia and all the others have normal clotting function. All the women have normal clotting function. What are the frequencies for the hemophilia allele (Xh) and normal allele (XH)?

Answer: Hemophila allele (Xh) = 0.2; Normal allele (XH) = 0.8 Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium

48) The frequency of cystic fibrosis found in those of European descent, caused by a homozygous recessive gene is approximately 0.04. Assuming the Hardy-Weinberg equilibrium applies, what are the frequencies of the recessive/affected allele, the dominant/normal allele, the homozygous normal condition, and the carrier/heterozygous genotype?

Answer: Recessive Allele = 0.2 ; Dominant Allele = 0.8 ; Heterozygous genotype = 0.32 Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 49) The A–B–O blood types are determined by three alleles IA, IB, and i. In a population the frequency of individuals who are homozygous for type A blood is 0.063, the frequency of individuals who are homozygous for type B blood is 0.015, and the frequency of individuals are blood type O is 0.44. What is the frequency of all genotypes?

Answer: IAIA = 0.063; IAi = 0.333; IBIB = 0.015; IBi =0.162 ; IAIB = 0.061 ; ii = 0.44 Section: 20.1 The Theory of Allele Frequencies Difficulty: Medium 50) Given the following data set, calculate the proportional contributions of each genotype to the next generation. What type of selection is demonstrated by this genetic trait?

Genotype: AA Aa aa Relative fitness: 1 1 0.6 Frequency at fertilization: 0.25 0.50 0.25 Answer: Genotype: Relative contribution:

AA 1 × 0.25 = 0.25 Proportional contribution: 0.25/0.90 = 0.278

Aa 1 × 0.50 = 0.50 0.50/0.90 = 0.556

aa 0.6 × 0.25 = 0.15 0.15/0.90 = 0.166

This is “Directional Selection”, selecting against the aa genotype.

Section: 20.2 Natural Selection Difficulty: Medium

Chapter 21 Test Bank Question Type: Multiple Choice 1) Eukaryotic DNA sequences that can move from one position to another are known as a) DNA fingerprints. b) plasmids. c) transposons. d) jumping chromosomes. e) walking chromosomes.

Answer: c Section: 21.1 Transposable Elements: An Overview Difficulty: Easy 2) Which of the following describes the category of transposons in which transposition is accomplished by excising an element from its position in a chromosome and inserting it into another position? a) Cut and paste transposons b) Replicative transposons c) Retrotransposons d) Cut and paste transposons and replicative transposons e) All of these Answer: a Section: 21.1 Transposable Elements: An Overview Difficulty: Easy 3) Which enzyme catalyzes the excision and insertion events conducted by transposons? a) Transformase b) Transposase c) Ligase d) Helicase e) Polymerase Answer: b Section: 21.1 Transposable Elements: An Overview Difficulty: Easy 4) Which of the following describes the category of transposons in which transposition is accomplished through a process that involves replication of the transposable element's DNA? a) Cut and paste transposons b) Replicative transposons c) Retrotransposons d) Cut and paste transposons and replicative transposons e) All of these Answer: b Section: 21.1 Transposable Elements: An Overview Difficulty: Easy

5) Which of the following is not true regarding replicative transposition? a) A transposase encoded by the element mediates an interaction between the element and a potential insertion site. b) The element is replicated and one copy of it is inserted at the new site while one copy remains at the original site. c) There is a net gain of one copy of the element. d) All of these are true. e) All of these are false.

Answer: d Section: 21.1 Transposable Elements: An Overview Difficulty: Medium 6) Which category of transposons accomplishes transposition through a process that involves the insertion of copies of an element that were synthesized from the element's RNA? a) Cut and paste transposons b) Replicative transposons c) Retrotransposons d) Cut and paste transposons and Replicative transposons e) All of these Answer: c Section: 21.1 Transposable Elements: An Overview Difficulty: Easy 7) Which of the following enzymes is responsible for the synthesis of DNA from RNA which is then inserted into the new position in retrotransposition? a) Transformase b) Transposase c) Reverse transcriptase d) Ligase e) Helicase Answer: c Section: 21.1 Transposable Elements: An Overview Difficulty: Easy 8) Which organism exhibits retrotransposition? a) C. elegans b) E. coli c) S. aureus d) E. coli and S. aureus e) All of these Answer: a Section: 21.1 Transposable Elements: An Overview Difficulty: Easy

9) Cut and paste transposition can occur in which of the following organisms? a) C. elegans b) E. coli c) S. aureus d) E. coli and S. aureus e) All of these Answer: e Section: 21.1 Transposable Elements: An Overview Difficulty: Easy 10) Replicative transposition can occur in which of the following organisms? a) C. elegans b) E. coli c) S. aureus d) E. coli and S. aureus e) All of these Answer: d Section: 21.1 Transposable Elements: An Overview Difficulty: Easy 11) Which type of transposon is not found in prokaryotes? a) IS elements b) Composite transposons c) Tn3 d) Tac e) All of these are found in prokaryotes.

Answer: d Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy 12) Which transposable element found in bacteria is the simplest, containing only genes involved in transposition? a) IS elements b) Composite transposons c) Tn3 d) Tac e) None of these Answer: a Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy

13) Which of the following statements about IS elements is not correct? a) IS elements are responsible for the formation of Hfr strains of E. coli. b) IS elements are less than 2,500 nucleotide pairs in size and contain only genes involved with promoting or regulating transcription. c) IS elements are demarcated by short, inverted terminal repeats which have no role in the transpositional process. d) IS elements are capable of creating target site duplications. e) All of these are correct.

Answer: c Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy 14) Short identical sequences at the ends of an IS sequence are known as a) transposon telemorases. b) inverted terminal repeats. c) correlated terminal repeats. d) transposable terminal repeats. e) All of these f) None of these Answer: b Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy 15) What type of transposon is an IS element? a) Cut and paste transposon b) Replicative transposon c) Retrovirus-like transposon d) Retrotransposon e) All of these Answer: a Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy 16) When IS elements insert into chromosomes or plasmids, they create a duplication of part of the DNA sequence at the site of the insertion, with one copy of the duplication on each side of the element known as a) inverted terminal repeats. b) target site duplications. c) target site deletions. d) inverted terminal duplications. e) inverted terminal deletions.

Answer: b Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy

17) Which type of transposon is a composite transposon? a) Cut and paste transposon b) Replicative transposon c) Retrotransposon d) Retrovirus-like transposon e) All of these Answer: a Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy 18) What is formed when two IS elements insert near each other? a) Replicative transposon b) Composite transposon c) Inverted terminal repeats d) Inverted terminal duplications e) Inverted terminal deletions Answer: b Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy 19) Which of the following composite transposons exhibits differences (non-identical nature) in its flanking IS elements? a) Tn 10 b) Tn 5 c) Tn 3 d) Tn 2 e) Tn 1 Answer: b Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy 20) When a bacterial cell is infected with a nonlytic bacteriophage with Tn5 on its chromosome, the frequency of Tn5 transposition is dramatically reduced if Tn5 is already maintained on the bacterial chromosome. Why? a) Phage integration occurs via Tn5 homology disrupting the transposase. b) Tn5 transposition is inhibited by phage gene products. c) Tn5 on the bacterial chromosome expresses a repressor of transposition. d) Transposase is not expressed from Tn5 elements integrated within the bacterial chromosome. e) None of these Answer: c Section: 21.2 Transposable Elements in Bacteria Difficulty: Medium

21) Which of the following statements is true about Tn3 elements? a) Tn3 form a cointegrate structure mediated by the tnpA gene product. b) The cointegrate structure is required for replication. c) Transposition is repressed by the dual nature of the tnpA gene product. d) Tn3 elements are highly mobile. e) All of these Answer: a Section: 21.2 Transposable Elements in Bacteria Difficulty: Medium 22) Which of the following are encoded by the genes in Tn3 elements? a) Transposase b) Beta lacatamase c) Resolvase d) Transposase and Beta lacatamase e) All of these Answer: e Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy 23) How are Tn3 elements classified? a) Cut and paste transposon b) Replicative transposon c) Retrotransposon d) Retrovirus-like transposon e) All of these Answer: b Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy 24) Which of the following has accelerated the spread of multiple drug resistance among bacterial infections? a) Conjugative R plasmids b) F plasmids c) Col plasmids d) Tn3 elements e) All of these Answer: a Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy

25) Which part of the conjugative R plasmid contains the genes for antibiotic resistance? a) RTF b) R determinant c) ORI d) Promoter e) Enhancer Answer: b Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy 26) Plasmids that can move transposons from one bacterial cell to another are known as a) conjugative R plasmids. b) F plasmids. c) Col plasmids. d) Tn3 elements. e) All of these Answer: a Section: 21.2 Transposable Elements in Bacteria Difficulty: Easy 27) Which of the following is true regarding cut and paste transposons in eukaryotes? a) These elements have inverted repeats at their termini. b) These elements create target site duplications when they insert into DNA molecules. c) Some encode a transposase that catalyzes the movement of the element from one position to another. d) All of these e) None of these Answer: d Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Medium 28) In maize, the Ac factor is not associated with a) structural identity between different Ac elements. b) the activator of chromosomal breakage in regions containing aberrant Ds. c) Cis-acting in its ability to transpose Ds. d) inverted terminal and direct repeats. e) None of these Answer: c Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Easy

29) What is another name for Ac and Ds transposable elements, which emphasizes the effect on gene expression? a) Controsons b) Controlling elements c) Cosmids d) Controsons and controlling elements e) All of these Answer: b Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Easy 30) Which of the following is true regarding the Ac element in maize?

1. Ac elements consist of 4563 nucleotide pairs.

2. Ac elements are bounded by inverted repeats that are 11 nucleotide pairs long.

3. Each Ac element is flanked by direct repeats 8 nucleotide pairs long. a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Medium 31) Which of the following is true regarding the Ds element in maize?

1. Ds elements are structurally heterogeneous 2. Some Ds elements appear to have been derived from Ac elements by the loss of internal sequences 3. They all possess the same inverted terminal repeats as Ac elements a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Medium

32) Ds elements that contain non-Ac DNA between their inverted terminal repeats are known as a) aberrant Ds elements. b) unusual Ds elements. c) double Ds elements. d) aberrant Ds elements and Unusual Ds elements. e) All of these Answer: a Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Easy 33) Ds elements are characterized by a peculiar piggybacking arrangement in which one Ds element is inserted into another, but in an inverted orientation. These are known as a) aberrant Ds elements. b) unusual Ds elements. c) double Ds elements. d) aberrant Ds elements and unusual Ds elements. e) All of these Answer: c Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Easy 34) What is the hobo element found in Drosophila? a) Ac/Ds element b) Replicative transposon c) Retrotransposon d) Tn3 e) IS Answer: a Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Easy 35) Which term is used to denote the syndrome produced from Drosophila crosses that gave rise to hybrids with an assortment of aberrant traits, including frequent mutation, chromosome breakage? a) Hybrid dysgenesis b) Hybrid displasia c) Hybrid polymorphism d) Hybrid dysgenesis and hybrid displasia e) All of these Answer: a Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Easy

36) Why do Drosophila, from a P male × M female cross, generate viable offspring? a) P element transposase is repressed in germ cells, at the level of RNA splicing. b) P element transposase is repressed in germ cells, at the level of transcription. c) P element transposase is repressed in somatic cells, at the level of RNA splicing. d) P element transposase is repressed in somatic cells, at the level of transcription. e) P male × M female crosses do not generate viable offspring.

Answer: c Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Easy 37) Some mariner elements show a high degree of similarity between organisms that are separated by millions of years of evolutionary time. What is the current explanation for this? a) Mariner is highly conserved. b) Mariner was horizontally transferred by bacterium. c) Mariner was horizontally transferred by virus. d) Mariner elements in divergent species have arisen through inter-species breeding events. e) None of these Answer: c Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Medium 38) Which of the following pairs of DNA sequences could qualify as terminal repeats of IS elements? (read all sequences as 5' to 3') a) AATTCCGTG and AATTCCGTG b) TTCGTT and TTCGTT c) TGTGGCTTT and TTTCGGTGT d) TCTTGG and GGTTCT e) TATATA and ATATAT Answer: c Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Medium 39) Which of the following crosses would result in the movement of P elements and subsequent hybrid dysgenesis? a) P cytotype females × P cytotype males b) M cytotype females × M cytotype males c) P cytotype females × M cytotype males d) M cytotype females × P cytotype males e) Hybrid dysgenesis is not induced by cytotype.

Answer: d Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Medium

40) Mutations generated by transposons can be located from a large, heterogeneous mixture of DNA by the known sequence of the transposon. What is this called? a) Transposon localization b) Transposon tagging c) Transposon linking d) Transposon marking e) Transposon identification Answer: b Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Easy 41) Retroviruses and retrovirus-like elements share all of the following genes except a) Gag. b) Pol. c) Env. d) Gag and Pol. e) Pol and Env. Answer: c Section: 21.4 Retroviruses and Retrotransposons Difficulty: Easy 42) Which transposon(s) is responsible for leaving delta-sequences in the chromosome after transposition? a) Tn5 b) Ty elements c) Mariner d) P elements e) Retroposons Answer: b Section: 21.4 Retroviruses and Retrotransposons Difficulty: Easy 43) Which of the following elements exhibit homogeneous A:T sequences at one end? a) Tn5 b) Ty elements c) Mariner d) P elements e) Retroposons Answer: e Section: 21.4 Retroviruses and Retrotransposons Difficulty: Easy

44) Which of the following statements is not true about L1 elements? a) L1 is the only transposable element known to be active in humans. b) L1 is unique to humans alone. c) L1 elements are heterogeneous in size and maintained in single copy in the human genome. d) L1 not is a retroposon. e) None of these Answer: a Section: 21.5 Transposable Elements in Humans Difficulty: Medium 45) If two transposons, in the same chromosome and in opposite orientation, pair and cross over, the segment between them will be a) inverted. b) deleted. c) inactive. d) unchanged. e) None of these Answer: a Section: 21.5 Transposable Elements in Humans Difficulty: Medium 46) Which of the following is not involved in HIV infection? a) Docks with target cell by binding the CD4 receptor via its gp120 protein b) Polymerizes 1st DNA strand synthesis by the pol gene product c) First strand synthesis is primed by snRNA. d) DNA:RNA hybrid is degraded by RNase H. e) The tRNA primer is prepackaged in the viral particle along with reverse transcriptase.

Answer: c Section: 21.5 Transposable Elements in Humans Difficulty: Medium 47) Which of the following is a transposable element found in the human genome? a) LINES b) SINES c) Retrovirus-like elements d) All of these e) None of these Answer: d Section: 21.5 Transposable Elements in Humans Difficulty: Easy

48) L1 transposition is known to cause disease when it is inserted into the gene for a) Factor VII. b) Huntingtin. c) Dystrophin. d) –globin. e) None of these Answer: c Section: 21.5 Transposable Elements in Humans Difficulty: Easy

Question Type: Essay 49) How does transposase act to catalyze transposition in IS elements?

Answer: At least some IS elements encode a protein that is needed for transposition. This protein, called transposase, seems to bind at or near the ends of the element, where it cuts both strands of the DNA. Cleavage of the DNA at these sites excises the element from the chromosome or plasmid, so that it can be inserted at a new position in the same or a different DNA molecule.

Section: 21.2 Transposable Elements in Bacteria Difficulty: Medium 50) How does the transposition of a Tn3 element occur?

Answer: The transposition of Tn3 occurs in two stages (Figure 18.6). First, the transposase mediates the fusion of two circular molecules, one carrying Tn3 and the other not. The resulting structure is called a cointegrate. During this process, the transposon is replicated, and one copy is inserted at each junction in the cointegrate; the two Tn3 elements in the cointegrate are oriented in the same direction. In the second stage of transposition, the tnpR-encoded resolvase mediates a site-specific recombination event between the two Tn3 elements. This event occurs at a sequence in Tn3 called res, the resolution site, generating two molecules, each with a copy of the transposon. Tn3 elements therefore transpose by a replicative mechanism that involves the formation of the transitional cointegrate structure. They are classified as replicative transposons.

The tnpR gene product of Tn3 has yet another function—to repress the synthesis of both the transposase and resolvase proteins. Repression occurs because the res site is located between the tnpA and tnpR genes. By binding to this site, the tnpR protein interferes with the transcription of both genes, leaving their products in chronic short supply. As a result, the Tn3 element tends to remain immobile.

Section: 21.2 Transposable Elements in Bacteria Difficulty: Medium

51) How do IS elements of E. coli contribute to horizontal gene transfer?

Answer: IS elements are contained not only in the E. coli chromosome but in the F-plasmid as well. Homologous recombination between IS elements of the chromosome and F-plasmid leads to Hfr formation. These Hfr's are capable of transmission of chromosomal markers during conjugation.

Section: 21.2 Transposable Elements in Bacteria Difficulty: Medium 52) Explain how transposable elements are responsible for chromosomal breakage in the maize Ac/Ds system as shown by McClintock.

Answer: The mechanism that McClintock proposed to explain the loss of the CI allele is diagrammed in Figure 18.9. A break at the site labeled by the arrow detaches a segment of the chromosome from its centromere, creating an acentric fragment. Such a fragment tends to be lost during cell division; thus, all the descendants of this cell will lack part of the paternally derived chromosome. Because the lost fragment carries the CI allele, none of the cells in this clone is inhibited from forming pigment. If any of them produces a part of the aleurone, a patch of purple tissue will appear, creating a mosaic kernel similar to the one shown in Figure 18.8. McClintock found that the breakage responsible for these mosaic kernels occurred at a particular site on chromosome 9. She named the factor that produced these breaks Ds, for Dissociation. However, by itself, this factor was unable to induce chromosome breakage. In fact, McClintock found that Ds had to be stimulated by another factor, called Ac, for Activator. The Ac factor was present in some maize stocks but absent in others. When different stocks were crossed, Ac could be combined with Ds to create the condition that led to chromosome breakage. This two-factor Ac/Ds system provided an explanation for the genetic instability that McClintock had observed on chromosome 9. Additional experiments demonstrated that this was only one of many instabilities present in the maize genome. McClintock found other instances of breakage at different sites on chromosome 9 and also on other chromosomes. Because breakage at these sites depended on activation by Ac, she concluded that Ds factors were also involved. To explain all these observations, McClintock proposed that Ds could exist at many different sites in the genome and that it could move from one site to another.

Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Medium 53) In your search for mutants of "your favorite gene" (yfg) you identify a mutant which you believe to be a transposon insertion. At your disposal you have a point mutant of yfg. How can you distinguish this potential yfg transposon disruption from a non-transposon mutation?

Answer: Identify the reversion frequency of your potential transposon. If the yfg disruption is caused by a transposon, revertants will be at a much higher frequency (due to transposon hopping) than if the mutation is not caused by a transposon. The yfg point mutant can be used as a standard to generate a reversion frequency for non-transposon mutations. Comparing this frequency to your "potential" transposon mutant yfg allows you to identify if your reversion rate is significantly higher than the point mutation reversion rate.

Section: 21.3 Cut-and-Paster Transposons in Eukaryotes Difficulty: Medium

54) Would you expect genes encoded by retrotransposons to contain introns? Why? Answer: No. The RNA intermediate is reverse transcribed into DNA, all of the RNA processing (splicing) that occurs in the RNA intermediate should be carried over into the DNA copy.

Section: 21.4 Retroviruses and Retrotransposons Difficulty: Medium 55) If the P element of Drosophila contained no introns, would this affect the viability of P element carriers? If so, how?

Answer: Yes. P elements are restricted to hopping only in germ line cells. P element movement is repressed in somatic cells by an alteration in RNA splicing, which leaves an intron in the coding region of the transposase gene, generating a premature stop codon. If this level of regulation didn't occur, P element hopping would lead to a high degree of somatic mutations affecting the fitness of the P element carrying individual.

Section: 21.4 Retroviruses and Retrotransposons Difficulty: Medium

Chapter 22 Test Bank Question Type: Multiple Choice 1) Which of the following organisms are considered to be premier model organisms for the genetic analysis of animal development? a) Drosophila b) C. elegans c) Chicken d) Drosophila and C. elegans e) All of these are correct.

Answer: e Section: 22.1 A Genetic Perspective on Development Difficulty: Easy 2) Why were frogs and sea urchins typically used to study development in classical anatomy and embryology?

1. The eggs developed outside of the mother.

2. The eggs were easily manipulated.

3. The animals were easy to breed. a) 1 b) 2 c) 3 d) 1 and 2 e) All of these are correct.

Answer: d Section: 22.1 A Genetic Perspective on Development Difficulty: Easy 3) Which of the following methods can be used to identify genes that are involved in important developmental events? a) Collect mutations and test for allelism b) RT-PCR c) Fluorescent labeling d) RNA and protein blotting e) All of these are correct.

Answer: e Section: 22.1 A Genetic Perspective on Development Difficulty: Easy

4) During Drosophila development, each egg is surrounded by a tough shell-like structure that is made of materials synthesized by somatic cells in the ovary. This structure is called the a) micropyle. b) chorion. c) syncytium. d) cellular blastoderm. e) imaginal disc.

Answer: b Section: 22.1 A Genetic Perspective on Development Difficulty: Easy 5) The term "syncytium" refers to which of the following stages of Drosophila development? a) A single layer of cells on the embryo's surface b) A worm-like structure that hatches by chewing through the egg shell c) A single cell with many identical nuclei d) An immobile structure with a hard skin e) An adult insect Answer: c Section: 22.1 A Genetic Perspective on Development Difficulty: Easy 6) In Drosophila, a single layer of cells on the embryo's surface which will give rise to all somatic tissues in the organism is known as a) micropyle. b) chorion. c) syncytium. d) cellular blastoderm. e) imaginal disc.

Answer: d Section: 22.1 A Genetic Perspective on Development Difficulty: Easy 7) Which of the following types of cells gives rise to germ line cells in Drosophila? a) Pole cells b) Cellular blastoderm cells c) Chorion cells d) Micropyle cells e) Imaginal discs Answer: a Section: 22.1 A Genetic Perspective on Development Difficulty: Easy

8) Structures in adult Drosophila develop from packets of cells called a) Pole cells b) Cellular blastoderm cells c) Chorion cells d) Micropyle cells e) Imaginal discs Answer: e Section: 22.1 A Genetic Perspective on Development Difficulty: Easy 9) Which of the following can be used to study developmental pathways?

1. Identifying genes whose products are involved in the differentiation of specific phenotypes 2. Identifying genes whose products are involved in the dedifferentiation of genotypes 3. Identifying proteins whose products are not involved in the differentiation of specific phenotypes a) 1 b) 2 c) 3 d) 1 and 3 e) All of these are correct.

Answer: a Section: 22.1 A Genetic Perspective on Development Difficulty: Easy 10) Which of the following is a gene product that can participate in a developmental pathway? a) Signal molecules b) Signal receptors c) Signal transducers d) Transcription factors e) All of these are correct.

Answer: e Section: 22.1 A Genetic Perspective on Development Difficulty: Easy 11) Mutations in genes that contribute to the formation of healthy eggs may have no effect on the viability or appearance of the female making those eggs. These are known as a) maternal effect mutations. b) paternal effect mutations. c) maternal gene mutations. d) maternal offspring mutations. e) embryo mutations.

Answer: a Section: 22.2 Maternal Gene Activity in Development Difficulty: Easy

12) Which of the following is true of a dorsal gene mutation? a) The embryo develops as if it has two ventral surfaces. b) The mutation can be rescued if the wild-type dorsal allele inherited from the father. c) Expression is limited to the maternal germ cells. d) All of the above e) None of the above Answer: c Section: 22.2 Maternal Gene Activity in Development Difficulty: Medium 13) Which of the following genes is a maternal-effect gene in Drosophila? a) Dorsal b) Bicoid c) Nanos d) All of these are correct. e) None of these are correct.

Answer: d Section: 22.2 Maternal Gene Activity in Development Difficulty: Easy 14) Maternal-effect gene products are involved in: 1. Dorsal-ventral axes determination in Drosophila 2. Anterior-posterior axes determination in Drosophila 3. Dorsal-ventral axes determination in C. elegans a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: d Section: 22.2 Maternal Gene Activity in Development Difficulty: Easy 15) Recessive mutations in maternal-effect genes are expressed only in embryos produced by a) females homozygous for the mutation. b) females heterozygous for the mutation. c) males homozygous for the mutation. d) males heterozygous for the mutation. e) None of these are correct.

Answer: a Section: 22.2 Maternal Gene Activity in Development Difficulty: Easy

16) The genes twist and snail are induced by a) repression of zerknüllt and decapentaplegic. b) expression of zerknüllt and decapentaplegic. c) expression of the transcription factor, dorsal. d) expression of the protein, Toll, on the embryo surface e) None of the above Answer: c Section: 22.2 Maternal Gene Activity in Development Difficulty: Medium 17) Where is the spätzle protein cleaved? a) In the perivitelline space on the dorsal side of the embryo. b) In the perivitelline space on the ventral side of the embryo. c) In the plasma membrane on the dorsal side of the embryo. d) In the plasma membrane on the ventral side of the embryo. e) In the nucleus of all cells in the embryo.

Answer: b Section: 22.2 Maternal Gene Activity in Development Difficulty: Medium 18) Which gene(s) are expressed in the cells that develop into the mesoderm? a) zerknüllt and decapentaplegic. b) snail and twist c) dorsal d) Both a and c e) Both b and c Answer: e Section: 22.2 Maternal Gene Activity in Development Difficulty: Medium 19) Which gene(s) are expressed in the cells that develop into the epidermis? a) zerknüllt and decapentaplegic. b) snail and twist c) dorsal d) Both a and c e) Both b and c Answer: a Section: 22.2 Maternal Gene Activity in Development Difficulty: Medium

20) The _________ transcription factors controls the formation of the anterior-posterior axis? a) hunchback and toll b) caudal and decapentaplegic c) dorsal and hunchback d) caudal and dorsal e) caudal and hunchback Section: 22.2 Maternal Gene Activity in Development Difficulty: Easy 21) Translation of the ________ RNA is limited to the anterior part of the embryo. a) caudal b) hunchback c) bicoid d) nanos e) all of the above Answer: b Section: 22.2 Maternal Gene Activity in Development Difficulty: Easy 22) Translation of the ________ RNA is limited to the posterior part of the embryo. a) caudal b) hunchback c) bicoid d) nanos e) all of the above Answer: a Section: 22.2 Maternal Gene Activity in Development Difficulty: Easy 23) Which RNA is synthesized in the nurse cells, and then transported into the oocyte to become anchored into the posterior end of the embryo? a) caudal b) hunchback c) bicoid d) nanos e) all of the above Answer: d Section: 22.2 Maternal Gene Activity in Development Difficulty: Easy

24) Which RNA is synthesized in the nurse cells, and then transported into the oocyte to become anchored into the anterior end of the embryo? a) caudal b) hunchback c) bicoid d) nanos e) all of the above Answer: c Section: 22.2 Maternal Gene Activity in Development Difficulty: Easy 25) Which transcription factor stimulates expression of the hunchback gene? a) caudal b) dorsal c) bicoid d) nanos e) none of the above Answer: c Section: 22.2 Maternal Gene Activity in Development Difficulty: Easy 26) Which protein would you expect to find bound to the 3′ UTR of the caudal RNA in the anterior of the embryo? a) caudal b) hunchback c) bicoid d) nanos e) none of the above Answer: c Section: 22.2 Maternal Gene Activity in Development Difficulty: Medium 27) Which protein would you expect to find bound to the 3′ UTR of the caudal RNA in the posterior of the embryo? a) caudal b) hunchback c) bicoid d) nanos e) none of the above Answer: e Section: 22.2 Maternal Gene Activity in Development Difficulty: Medium

28) Which protein would you expect to find bound to the 3′ UTR of the hunchback RNA in the posterior of the embryo? a) caudal b) hunchback c) bicoid d) nanos e) none of the above Answer: d Section: 22.2 Maternal Gene Activity in Development Difficulty: Medium 29) Which of the following statements best characterizes a homeotic gene? a) They define segmental regions in the embryo. b) They define the anterior and posterior compartments of individual segments. c) Mutations in these genes cause one body part to look like another. d) They are responsible for sex determination. e) They are involved in gene dosage compensation.

Answer: c Section: 22.3 Zygotic Gene Activity in Development Difficulty: Easy 30) Which of the following is another name for selector genes? a) Homeotic genes b) Segmentation genes c) Gap genes d) Pair-rule genes e) Segment-polarity genes Answer: a Section: 22.3 Zygotic Gene Activity in Development Difficulty: Easy 31) A mutation that causes one body part to look like another is known as a(n) a) automutation. b) homeotic mutation. c) heteromic mutation. d) xenotic mutation. e) bithorax mutation.

Answer: b Section: 22.3 Zygotic Gene Activity in Development Difficulty: Easy

32) Which of the following is a large cluster of homeotic genes?

1. Bithorax complex 2. Antennapedia complex 3. Fermurtibia complex a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: d Section: 22.3 Zygotic Gene Activity in Development Difficulty: Easy 33) Which of the following is a segmentation gene discovered along the anterior-posterior axis? a) Gap genes b) Pair-rule genes c) Segment polarity genes d) Gap genes and Pair-rule genes e) All of these are correct.

Answer: e Section: 22.3 Zygotic Gene Activity in Development Difficulty: Easy 34) Which of the following mutations were used to study the genetic control of anatomically correct organ formation? a) Eyeless b) Wingless c) Hunchback d) Nanos e) Bithorax Answer: a Section: 22.3 Zygotic Gene Activity in Development Difficulty: Easy 35) The process of determining the fate of an undifferentiated cell using a signal from a differentiated cell is known as a) deduction. b) induction. c) repression. d) inhibition. e) None of these are correct.

Answer: b Section: 22.3 Zygotic Gene Activity in Development Difficulty: Easy

36) The identity of each body segment in Drosophila is determined by the products of genes in which of the following complexes? a) Bithorax b) Antennapedia c) Femurtibia d) Bithorax and antennapedia e) Antennapedia and femurtibia Answer: d Section: 22.3 Zygotic Gene Activity in Development Difficulty: Easy 37) In Drosophila, the eyeless gene controls eye development. What is the vertebrate equivalent? a) boss b) Pax6 c) fushi tarazu d) engrailed e) sev Answer: b Section: 22.3 Zygotic Gene Activity in Development Difficulty: Easy 38) Geneticists can study development in vertebrates by 1. Applying knowledge gained from the study of model invertebrates 2. Analyzing mutations and phenocopies of mutant genes in model vertebrates such as mice and zebrafish 3. Examining the differentiation of stem cells a) 1 b) 2 c) 3 d) 1 and 2 e) All of these are correct.

Answer: e Section: 22.4 Genetic Analysis of Development in Vertebrates Difficulty: Medium 39) How is a chimeric mouse characterized? a) A combination of XX and XY cells b) An intersex phenotype c) A mouse with abnormal body morphology as a result of gene mutations d) A mouse with a homozygous recessive mutation in a gene e) A mouse with a mixture of its own and cultured ES cells Answer: e Section: 22.4 Genetic Analysis of Development in Vertebrates Difficulty: Easy

40) Which of the following is an example of colinearity? a) Replacement of a normal allele by a mutated transgene b) Migration of cells during embryogenesis to form three primary germ layers c) A process in which cells are assigned developmental fates d) The physical order of genes corresponding to their expression along the anterior-posterior axis e) Genetic dissection of a developmental pathway by analyzing gene mutations Answer: d Section: 22.4 Genetic Analysis of Development in Vertebrates Difficulty: Easy 41) Which statement is false concerning zebra fish? a) Morpholinos are used to block the translation of specific transcripts. b) Morpholinos do not act over a very long period of time after injection into fertilized eggs. c) Its genome is 1.5 billion bp in length and contains 22,000 known and predicted genes. d) Zebra fish have become a model genetic organism for vertebrates. e) None of these Answer: b Section: 22.4 Genetic Analysis of Development in Vertebrates Difficulty: Medium 42) Therapeutic cloning a) involves producing ES cells by injecting a somatic cell nucleus into an enucleated egg. b) is the same as reproductive cloning. c) involves producing individuals from ES cell nuclei injected into an enucleated egg. d) All of these are correct. e) None of these are correct.

Answer: a Section: 22.4 Genetic Analysis of Development in Vertebrates Difficulty: Easy 43) Which of the following is a vertebrate homologue of the homeotic genes in Drosophila?

A) Hox genes B) Tra genes C) Dsx genes D) Sxl genes E) None of these are correct.

Answer: a Section: 22.4 Genetic Analysis of Development in Vertebrates Difficulty: Easy

44) When genes are transcribed in the same direction, with expression proceeding from one end of a cluster to the other end, both spatially (anterior to posterior in the embryo) and temporally (early to late in development) it is known as a) bilinearity. b) colinearity. c) homologous transcription. d) heterolinearity. e) homolinearity.

Answer: b Section: 22.4 Genetic Analysis of Development in Vertebrates Difficulty: Easy 45) An insertion that is specifically targeted to a gene, disrupting the integrity of that gene is known as a) knockdown. b) knockout. c) cutout. d) positive deletion. e) negative deletion.

Answer: b Section: 22.4 Genetic Analysis of Development in Vertebrates Difficulty: Easy Question Type: Essay 46) Describe how the dorsal protein directs the formation of the dorsal-ventral axis in Drosophila.

Answer: Differentiation of a Drosophila embryo along the dorsal-ventral axis hinges on the action of the transcription factor encoded by the dorsal gene. This protein is synthesized maternally and stored in the cytoplasm of the egg. At the time of blastoderm formation, the dorsal protein enters the nuclei on the ventral side of the embryo, inducing the transcription of two genes called twist and snail (whimsically named for their mutant phenotypes). In these same nuclei, it represses the genes zerknüllt (from the German for “crumpled”) and decapentaplegic (from the Greek words for “15” and “stroke”). The selective induction and repression of these genes cause the ventral cells to differentiate into a primitive embryonic layer of tissue called the mesoderm. On the opposite side of the embryo, where the dorsal protein is excluded from the nuclei, twist and snail are not induced and zerknüllt and decapentaplegic are not repressed. Consequently, these cells differentiate into a different primitive tissue, the embryonic epidermis. The entrance of the dorsal transcription factor into the ventral nuclei and its exclusion from the dorsal nuclei therefore initiate differentiation along the dorsal-ventral axis.

Section: 22.2 Maternal Gene Activity in Development Difficulty: Medium

47) Describe how the hunchback and caudal genes direct the formation of the anterior-posterior axis in Drosophila, and the role of bicoid and nanos proteins in this process.

Answer: The anterior-posterior axis in Drosophila is created by the regional synthesis of transcription factors encoded by the hunchback and caudal genes. These two genes are transcribed in the nurse cells of the maternal germ line. These special cells support the growth and development of the oocyte. The maternal transcripts of the hunchback and caudal genes are then carried from the nurse cells into the oocyte where they become uniformly distributed in the cytoplasm. However, both types of transcripts are translated in different parts of the embryo. The hunchback RNA is translated only in the anterior part, and the caudal RNA is translated only in the posterior part. This differential translation produces concentration gradients of the proteins encoded by these two genes; hunchback protein is concentrated in the anterior part of the embryo, and caudal protein is concentrated in the posterior part. These two proteins then function to activate or repress transcription of the genes whose products are involved in the differentiation of the embryo along its anterior-posterior axis. Two maternally supplied RNAs, bicoid and nanos, are synthesized in the nurse cells of the maternal germ line and are then transported into the oocyte.

The bicoid RNA becomes anchored at the anterior end of the developing oocyte, and the nanos RNA becomes anchored at the posterior end. After fertilization, each type of RNA is translated locally, and the resulting protein products diffuse through the embryo to form concentration gradients; bicoid protein is concentrated at the anterior end, and nanos protein is concentrated at the posterior end. Bicoid acts as a transcription factor to stimulate the synthesis of RNAs from several genes, including hunchback. These RNAs are then translated into proteins that control the formation of the anterior structures of the embryo. Bicoid also prevents the translation of caudal RNA by binding to sequences in the 3′ untranslated region of that RNA. Nanos protein is concentrated in the posterior of the embryo, and there it binds to the 3′untranslated region of hunchback RNA and causes the degradation of that RNA. Consequently, hunchback protein is not synthesized in the posterior of the embryo. Instead, its synthesis is restricted to the anterior of the embryo where it acts as a transcription factor to regulate the expression of genes involved in anterior-posterior differentiation. Wherever hunchback protein is synthesized, the embryo develops anterior structures.

Section: 22.2 Maternal Gene Activity in Development Difficulty: Hard 48) What are maternal-effect genes and how can they be identified?

Answer: Mutations in genes that contribute to the formation of healthy eggs have no effect on the viability or appearance of the female making those eggs. Instead, their effects may be seen only in the next generation. Such mutations are called maternal-effect mutations because the mutant phenotype in the offspring is caused by a mutant genotype in its mother. Genes identified by such mutations are called maternal-effect genes. Mating between flies homozygous for recessive mutations in this gene produce inviable progeny. This lethal effect is strictly maternal. A cross between homozygous mutant females and homozygous wild-type males produces inviable progeny, but the reciprocal cross produces viable progeny. The lethal effect of the dorsal mutation is therefore manifested only if females are homozygous for it. The male genotype is irrelevant.

Section: 22.2 Maternal Gene Activity in Development Difficulty: Medium

49) What are the basic steps in constructing a knockout mutation in mice?

Answer: Knockout mutagenesis is possible with a gene that has already been cloned and can help researchers determine the role the normal gene plays during development. To create a knockout mutation, the sequence of the cloned gene must be altered in vitro and then introduced into cultured embryonic stem cells. At a low frequency, the mutated transgene will replace its normal allele on the chromosome by homologous recombination. This process is called gene targeting. ES cells that contain a targeted knockout mutation can be used to create chimeras, which can then be bred to produce transgenic strains that carry the knockout mutation. Two heterozygotes can then be crossed to make the knockout mutation homozygous and determine what effect it has on development.

Section: 22.4 Genetic Analysis of Development in Vertebrates Difficulty: Medium 50) Discuss some of the ethical implications that surround the study stem cells in the genetic analysis of development in vertebrates.

Answer: No matter if they are derived from embryonic or adult tissue, stem cells provide an opportunity to study the mechanisms involved in the differentiation of special cell types. Stem cells can be obtained from a variety of mammals, including mice, monkeys, and humans. They can be cultured in vitro and examined for differentiation while growing there or after being transplanted into a host organism. While in culture, stem cells can be treated in various ways to ascertain what triggers their development in a specific direction. Molecular techniques, including gene-chip technologies, allow researchers to determine which genes the cells are expressing as their developmental programs unfold. Because embryonic stem (ES) cells have the greatest developmental potential, they are ideally suited for this kind of analysis. These cells are usually derived from the inner cell mass of embryos that had been created by in vitro fertilization. The issue of procuring and analyzing human ES cells is, of course, controversial. The human ES cell lines now in use were derived from embryos that were donated by people who had sought medical help to have children through in vitro fertilization. Typically, many more embryos are created through this process than are eventually used to produce children. A couple may then decide to donate its unused embryos for research purposes. The derivation of ES cells from such embryos necessarily requires that the embryos be destroyed. Some people view the destruction of early embryos as an acceptable practice; others consider it immoral. The controversy surrounding this practice has caused many governments to withhold or restrict financial support for research on human embryonic stem cells. In the United States, for example, federal government support is provided only to projects using human stem cell lines established before August 9, 2001. Funds for projects using lines established after that date must come from other sources. The debate on funding for human embryonic stem cell research has been intensified by the prospect of using human ES cells to cure diseases that result from the loss of specific cell types, such as diabetes mellitus (in which the pancreatic islet cells have been lost) and Parkinson's disease (in which certain types of neurons in a particular region of the brain have been lost). ES cell therapy has also been proposed to treat disabilities such as those resulting from spinal cord damage. The idea is to transplant cells derived from ES cells into the diseased or injured tissue and allow these cells to regenerate the lost or damaged parts of the tissue. Experiments with mice and rats suggest that this strategy might work in humans. However, many technical problems have yet to be solved. For instance, it is not yet possible to obtain pure cultures of a particular differentiated cell type. When human ES cells develop in culture, they differentiate into many kinds of cells; isolating one

kind—say, for example, cardiac cells—is a formidable technical challenge. The proponents of human stem cell therapy also have to solve other kinds of problems. Cells derived from an in vitro culture might divide uncontrollably and form tumors upon being transplanted into a host, or they might be wiped out by the host's immune system. To circumvent the latter problem, researchers have proposed transplanting cells that are genetically identical to the host's cells. Such genetically identical cells could be created by using one of the host's somatic cells to generate the ES cell population. A somatic cell from the host could be fused with an enucleated egg cell obtained from a female donor (not necessarily the host). If the genetically altered egg, which is diploid, divides to form an embryo, cells could be isolated from that embryo to establish an ES cell line, which could then provide genetically identical material for transplantation back into the host. The production of ES cells by transferring the nucleus of a somatic cell into an enucleated egg is called therapeutic cloning. Stem cells might also be obtained by inducing somatic cells to revert to an undifferentiated state. Recent experiments in the United States and Japan indicate that this approach might be feasible. Differentiated skin cells were induced to become pluripotent cells by genetically transforming them with a mixture of four cloned genes. However, some of the genes that were used in these experiments are associated with tumor formation when they are expressed inappropriately. Thus, more research is needed before induced pluripotent cells can be used in stem cell therapy.

Section: 22.4 Genetic Analysis of Development in Vertebrates Difficulty: Medium

Chapter 23 Prelecture Question Type: Multiple Choice 1) Which of the following properties is not characteristic of a malignant cell? a) It no longer undergoes cell cycle regulation and growth arrest. b) It has undergone mutations in a c-onc and in other genes. c) It cannot leave its site of growth in the tumor. d) It may have undergone a chromosome rearrangement. e) All of these are correct.

Answer: c Section: 23.1 Cancer: A Genetic Disease Difficulty: Medium 2) Which of the following causes cancer? a) Mutations in genes that control cell growth and division b) Mutations in genes that control pigmentation in skin c) Mutations in genes that control the differentiation of cell types d) All of these are correct. e) None of these are correct.

Answer: a Section: 23.1 Cancer: A Genetic Disease Difficulty: Easy 3) What classifies a tumor as being malignant? a) When cells do not invade surrounding tissue b) When the tumor is under 10mm in diameter c) When cells detach from the tumor and invade surrounding tissue d) When the tumor is over 10mm in diameter e) None of these are correct.

Answer: c Section: 23.1 Cancer: A Genetic Disease Difficulty: Easy 4) Which of the following is not true regarding cancer? a) It is a group of diseases. b) Every type of cancer can be stopped with current medical treatments. c) Some cancers grow aggressively while others grow slowly. d) Cancer can originate in different tissues of the body. e) All statements are true.

Answer: b Section: 23.1 Cancer: A Genetic Disease Difficulty: Medium

5) Which of the following cancers is most prevalent in the United States? a) Breast cancer b) Prostate cancer c) Lung cancer d) Pancreatic cancer e) Leukemia Answer: c Section: 23.1 Cancer: A Genetic Disease Difficulty: Easy 6) An agent that can irreversibly transform normal cells into cancerous cells is most correctly known as a) tetragen b) aneugen c) mutagen d) carcinogen e) None of these are correct.

Answer: d Section: 23.1 Cancer: A Genetic Disease Difficulty: Easy 7) Which of the following is the abiding characteristic of all cancer cells? a) Caused by the presence of radiation b) Uncontrolled growth c) Multinucleated d) Form a monolayer in culture e) All of these are correct.

Answer: b Section: 23.1 Cancer: A Genetic Disease Difficulty: Easy 8) Which statement is not true about checkpoints in the cell cycle? a) They can halt cell cycle progression in response to environmental influences. b) They are a complex machinery that uses cyclins and cyclin-dependent kinases to regulate cell cycling. c) They occur only in the middle of the G1 phase of the cell cycle. d) All of these are correct. e) None of these are correct.

Answer: c Section: 23.1 Cancer: A Genetic Disease Difficulty: Medium

9) CDKs a) phosphorylate cell cycle regulating proteins independently. b) are inactivated by binding to a cyclin. c) by controlling the progression through cell cycle checkpoints. d) All of these are correct. e) None of these are correct.

Answer: c Section: 23.1 Cancer: A Genetic Disease Difficulty: Easy 10) The START checkpoint a) controls entry into the S phase. b) controls exit from the S phase. c) controls entry into the G2 phase. d) only occurs in yeast cells. e) functions at mitosis.

Answer: a Section: 23.1 Cancer: A Genetic Disease Difficulty: Easy 11) Programmed cell death is called a) cell lysis b) apoptosis c) cell popping d) hemolysis e) None of these are correct.

Answer: b Section: 23.1 Cancer: A Genetic Disease Difficulty: Easy 12) Which enzyme family plays a crucial role in apoptosis? a) Caspases b) Lactases c) Maltases d) Kinases e) Polymerases Answer: a Section: 23.1 Cancer: A Genetic Disease Difficulty: Easy

13) The genetic basis for cancer was suspected long before the scientific evidence was uncovered.

The suspicion was due to a) the cancerous property of tumor cells is clonally inherited. b) tumors can be induced by mutagenic chemicals and ionizing radiation. c) some forms of cancers tend to run in families. d) chromosomal arrangements were often associated with certain kinds of tumors. e) All of these are correct.

Answer: e Section: 23.1 Cancer: A Genetic Disease Difficulty: Easy 14) Mutations in _______ can actively promote cell division. a) Tumor suppressor genes b) Oncogenes c) Operator genes d) Promoter genes e) Silencer genes Answer: b Section: 23.2 Oncogenes Difficulty: Easy 15) Mutations in __________ lead to a failure to repress cell division. a) Tumor suppressor genes b) Oncogenes c) Operator genes d) Promoter genes e) Silencer genes Answer: a Section: 23.2 Oncogenes Difficulty: Easy 16) Which group of genes was first discovered in RNA virus genomes and is known to induce tumors in vertebrate hosts? a) Tumor suppressor genes b) Oncogenes c) Operator genes d) Promoter genes e) Silencer genes Answer: b Section: 23.2 Oncogenes Difficulty: Easy

17) Which property is not exhibited by tumor-causing retroviruses? a) It contains a gene encoding a reverse transcriptase. b) The gag gene encodes a viral surface protein. c) The v-onc gene causes the tumor. d) The viral oncogene usually has a cellular homolog. e) The env gene encodes the capsid protein of the virion.

Answer: b Section: 23.2 Oncogenes Difficulty: Medium 18) Of the four genes found in the Rous sarcoma virus, which provides tumor suppression ability? a) Gag b) Env c) Pol d) V-src e) Gag and Env Answer: d Section: 23.2 Oncogenes Difficulty: Easy 19) Which of the following genes would be considered an oncogene? a) Gag b) Env c) V-src d) Gag and Env e) All of these are correct.

Answer: c Section: 23.2 Oncogenes Difficulty: Easy 20) What are cellular homologues of viral oncogenes called? a) Non-cellular oncogenes b) Adaptive oncogenes c) Proto-oncogenes d) Pre-oncogenes e) Post-oncogenes Answer: c Section: 23.2 Oncogenes Difficulty: Easy

21) Which of the following is true regarding tumor-inducing retroviruses? a) Retroviruses are double-stranded DNA viruses. b) The enzyme reverse transcriptase is essential for transcribing the viral DNA into RNA. c) Cancer-causing genes encoded by the viral genome are called proto-oncogenes. d) Each type of viral gene that can cause cancer can potentially regulate the expression of cellular genes. e) Viral genes that can cause cancer are different from other viral genes because they possess introns.

Answer: d Section: 23.2 Oncogenes Difficulty: Easy 22) Which of the following was found to be a difference between v-src and c-src? a) C-src contained eleven introns whereas v-src had none. b) V-src contained eleven introns whereas c-src had none. c) C-src caused the formation of tumors in mice, whereas v-src caused tumor formation in chickens. d) V-src caused the formation of tumors in mice, whereas c-src caused tumor formation in chickens. e) None of these are correct.

Answer: a Section: 23.2 Oncogenes Difficulty: Easy 23) Why do v-oncs cause cancer whereas normal c-oncs do not? a) More than one v-onc occurs in the tumor cell. b) C-oncs are never expressed except when infected by a retrovirus. c) Expression of both the v-onc and the c-onc is enough to cause transformation. d) The v-oncs are expressed at much higher levels from the strong retroviral promoter than are c-oncs. e) All of these are correct.

Answer: d Section: 23.2 Oncogenes Difficulty: Easy

24) Weinberg was the first to identify a link between a c-onc and cancer by a) using a transfection test with liver DNA that transformed normal cells to hepatomas. b) cloning a DNA fragment from human colon cancer and showing that it could transform normal cells to malignant cells. c) serially transfecting DNA from bladder cells first into nontransformed cells and then into transformed cells. d) cloning the transforming DNA fragment from the transformed cells and showing it transformed normal cells. e) None of these Answer: d Section: 23.2 Oncogenes Difficulty: Easy 25) Mutations in which a single mutant allele is dominant in its ability to induce cancer are known as a) recessive activators. b) dominant activators. c) heterozygous activators. d) homozygous activators. e) penetrant activators.

Answer: b Section: 23.2 Oncogenes Difficulty: Easy 26) Burkitt’s lymphoma and chronic myelogenous leukemia are associated with which of the following? a) Lack of tumor suppressor genes b) Formation of thymidine dimers due to UV light exposure c) Mutations induced by exposure to carcinogens in cigarette smoke d) Reciprocal translocations on chromosomes e) None of these Answer: d Section: 23.2 Oncogenes Difficulty: Easy 27) The normal alleles of genes such as c-ras and c-myc produce proteins that regulate the cell cycle. When these genes are overexpressed, or when they produce proteins that function as dominant activators, the cell is a) cancerous. b) pre-disposed to become cancerous. c) immune from becoming cancerous. d) cancerous and Pre-disposed to become cancerous. e) None of these Answer: b Section: 23.2 Oncogenes Difficulty: Easy

28) Which of the following is the best example of a tumor-suppressor gene? a) The RB gene involved in retinoblastoma b) Philadelphia chromosome in chronic myelogenous leukemia c) C- ras involved in human bladder cancer d) C-myc involved in Burkitt's lymphoma. e) Platelet-derived growth factor (PDGF) Answer: a Section: 23.3 Tumor Suppressor Genes Difficulty: Easy 29) Which of the following is not a tumor suppressor gene? a) p16 b) c-myc c) RB d) NF2 e) TP53 Answer: b Section: 23.3 Tumor Suppressor Genes Difficulty: Easy 30) Tumor suppressor genes a) were discovered by studies of rare cancers exhibiting a dominant inheritance pattern. b) were suggested by Alfred Knudson's findings in his 1971 study of retinoblastoma. c) function in more than one cellular process. d) of particular classes function by binding to and inhibiting the activity of transcription factors controlling cell cycle progression. e) All of these Answer: e Section: 23.3 Tumor Suppressor Genes Difficulty: Easy 31) Which of the following is not a component of Knudson's “two hit” hypothesis? a) In the inherited cases of retinoblastoma, one of the inactivating mutations has been transmitted through the germ line. b) Two mutational “hits” are required to knock out a gene that normally functions to suppress tumor formation. c) A cancer develops only if a second mutation occurs in the somatic cells and if this mutation knocks out the function of the wild-type allele of the tumor suppressor gene. d) All of these are true. e) None of these are true.

Answer: d Section: 23.3 Tumor Suppressor Genes Difficulty: Medium

32) Which of the following is true regarding the RB gene in relation to cancer and the cell cycle?

1. In many types of cancer both copies of the RB gene have been inactivated.

2. Inactivation of both copies of the RB gene impair the ability of the RB protein to bind E2F transcription factors.

3. pRB is phosphorylated through the action of cyclin-dependent kinases. a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 23.3 Tumor Suppressor Genes Difficulty: Medium 33) Which of the following is a key step in carcinogenesis? a) Loss of p53 function b) Gain of pRB function c) Gain of p53 function d) Loss of p53 function and Gain of pRB function e) All of these Answer: a Section: 23.3 Tumor Suppressor Genes Difficulty: Easy 34) Where are most of the mutations that inactivate p53 located? a) N-terminal transcription-activation domain b) Central DNA-binding core domain c) C-terminal homo-oligomerization domain d) D-terminal binding domain e) None of these Answer: b Section: 23.3 Tumor Suppressor Genes Difficulty: Easy 35) What type of mutations are typically found in the DBD of p53? a) Recessive loss of function mutations b) Recessive gain of function mutations c) Dominant loss of function mutations d) Dominant gain of function mutations e) None of these Answer: a Section: 23.3 Tumor Suppressor Genes Difficulty: Easy

36) How does p53 respond to cell stress?

1. The level of p53 increases dramatically.

2. p53 stimulates the transcription of genes whose products arrest the cell cycle.

3. p53 activates another set of genes whose products ultimately cause the damaged cell to die. a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 23.3 Tumor Suppressor Genes Difficulty: Medium 37) Which statement is not true about the tumor suppressor gene pAPC? a) Is involved in adenomatous polyposis coli, an inherited condition leading to colorectal cancer b) Only regulates cell proliferation c) Causes cancer by binding -catenin, a regulator of apoptosis d) Causes cancer in a heritable manner only when the FAP gene is mutated as well. e) None of these Answer: b Section: 23.3 Tumor Suppressor Genes Difficulty: Medium 38) Which of the following genes have been implicated in hereditary forms of breast cancer? a) BRCA1 b) BRCA2 c) RB d) phMSH2 e) BRCA1 and BRCA2 Answer: e Section: 23.3 Tumor Suppressor Genes Difficulty: Easy 39) Mutations in HPC1 a) cause glioblastomas in the brain. b) lead to prostate cancer when mutations in TP53, RB, and NF2 also occur. c) is the gene for the inherited form of prostate cancer. d) require mutations in several other tumor suppressor genes to cause polyposis carcinomas. e) None of these Answer: c Section: 23.3 Tumor Suppressor Genes Difficulty: Easy

40) How do cancers develop? a) Through the activation of a singular proto-oncogene b) Through the activation of a singular tumor suppressor gene c) Through the accumulation of somatic mutations in proto-oncogenes and tumor suppressor genes. d) Through the inactivation of a singular tumor suppressor gene e) Through the inactivation of a singular proto-oncogene Answer: c Section: 23.3 Tumor Suppressor Genes Difficulty: Easy 41) How does the inactivation of mutations in the APC gene initiate tumor formation? a) By causing the development of abnormal tissues within the intestinal epithelium b) By causing the development of abnormal tissues within the breast tissue c) By causing the development of abnormal tissues within the connective tissue d) By causing the development of abnormal tissues within the nervous tissue e) By causing the development of abnormal tissues within the muscular tissue Answer: a Section: 23.3 Tumor Suppressor Genes Difficulty: Easy 42) Which of the following is a proto-oncogene implicated in the formation of glioblastomas? a) EGFR b) PDGF c) AST d) EGFR and PDGF e) All of these Answer: d Section: 23.3 Tumor Suppressor Genes Difficulty: Easy 43) Which of the following is not a hallmark of the malignant cancer pathway? a) Cancer cells acquire self-sufficiency in the signalling processes that stimulate division and growth. b) Cancer cells are normally sensitive to signals that inhibit growth. c) Cancer cells can evade programmed cell death. d) Cancer cells acquire limitless replicative potential. e) Cancer cells develop ways to nourish themselves.

Answer: b Section: 23.3 Tumor Suppressor Genes Difficulty: Easy

44) Blood vessels are induced to grow among cancer cells through a process known as a) angiogenesis. b) cardiogenesis. c) vasculargenesis. d) arterialgenesis. e) None of these Answer: a Section: 23.3 Tumor Suppressor Genes Difficulty: Easy 45) Which of the following human behaviors contribute to the risk of cancer? a) Cigarette smoking b) Exposure to UV light c) Consumption of fatty foods d) All of these e) None of these Answer: d Section: 23.3 Tumor Suppressor Genes Difficulty: Easy Question Type: Essay 46) How does the START checkpoint regulates cell cycle growth, and what happens when this checkpoint is deregulated? Explain your answer by discussing the appropriate cyclins and CDK molecules and their functions.

Answer: One of the most important cell-cycle checkpoints, called START, is in mid-G1. The cell receives both external and internal signals at this checkpoint to determine when it is appropriate to move into the S phase. This checkpoint is regulated by D-type cyclins in conjunction with CDK4.

If a cell is driven past the START checkpoint by the cyclin D/CDK4 complex, it becomes committed to another round of DNA replication. Inhibitory proteins with the capability of sensing problems in the late G1 phase, such as low levels of nutrients or DNA damage, can put a brake on the cyclin/CDK complex and prevent the cell from entering the S phase. In the absence of such problems, the cyclin D/CDK4 complex drives the cell through the end of the G1 phase and into the S phase, thereby initiating the DNA replication that is a prelude to cell division. Cells in which the START checkpoint is dysfunctional are especially prone to become cancerous. The START checkpoint controls entry into the S phase of the cell cycle. If DNA within a cell has been damaged, it is important that entry into the S phase be delayed to allow for the damaged DNA to be repaired. Otherwise, the damaged DNA will be replicated and transmitted to all the cell's descendants. Normal cells are programmed to pause at the START checkpoint to ensure that repair is completed before DNA replication commences. By contrast, cells in which the START checkpoint is dysfunctional move into S phase without repairing their damaged DNA. Over a series of cell cycles, mutations that result from the replication of unrepaired DNA may accumulate and cause further deregulation of the cell cycle. A clone of cells with a dysfunctional START checkpoint may therefore become aggressively cancerous.

Section: 23.1 Cancer: A Genetic Disease Difficulty: Medium

47) Outline the killing events that take place during the process of apoptosis and discuss how this process can relate to the formation of cancer.

Answer: A family of proteolytic enzymes called caspases plays a crucial role in the cell death phenomenon. The caspases remove small parts of other proteins by cleaving peptide bonds.

Through this enzymatic trimming, the target proteins are inactivated. The caspases attack many different kinds of proteins, including the lamins, which make up the inner lining of the nuclear envelope, and several components of the cytoskeleton. The collective impact of this proteolytic cleavage is that cells in which it occurs lose their integrity; their chromatin becomes fragmented, blebs of cytoplasm form at their surfaces, and they begin to shrink. Cells undergoing this kind of disintegration are usually engulfed by phagocytes, which are scavenger cells of the immune system, and are then destroyed. If the apoptotic mechanism has been impaired or inactivated, a cell that should otherwise be killed can survive and proliferate. Such a cell has the potential to form a clone that could become cancerous if it acquires the ability to divide uncontrollably.

Section: 23.1 Cancer: A Genetic Disease Difficulty: Medium 48) Why do c-oncs have introns whereas v-oncs do not; and how would this benefit a retrovirus?

Answer: The most plausible answer is that v-oncs were derived from c-oncs by the insertion of a fully processed c-onc mRNA into the genome of a retrovirus. A virion that packaged such a recombinant molecule would then be able to transduce the c-onc gene whenever it infected another cell. During infection, the recombinant RNA would be reverse-transcribed into DNA and then integrated into the cell's chromosomes. What could be of greater value to a virus than to have a new gene that stimulates increased growth of its host, while its integrated genome goes along for the ride?

Section: 23.2 Oncogenes Difficulty: Medium 49) How do c-ras mutations, like the c-H ras gene, cause cancer?

Answer: Unlike viral oncogenes, the mutant c-H-ras gene does not synthesize abnormally large amounts of protein. Instead, the valine-for-glycine substitution at position 12 impairs the ability of the mutant c-H-ras protein to hydrolyze one of its substrates, guanosine triphosphate (GTP).

Because of this impairment, the mutant protein is kept in an active signaling mode, transmitting information that ultimately stimulates the cells to divide in an uncontrolled way (Figure 22.5).Mutant versions of the c-ras oncogenes have now been found in a large number of different human tumors, including lung, colon, mammary, prostate, and bladder tumors, as well as neuroblastomas (nerve cell cancers), fibrosarcomas (cancers of the connective tissues), and teratocarcinomas (cancers that contain different embryonic cell types). In all cases, the mutations involve amino acid changes in one of three positions—12, 59, or 61. Each of these amino acid changes impairs the ability of the mutant Ras protein to switch out of its active signaling mode.

These types of mutations therefore stimulate cells to grow and divide.

Section: 23.2 Oncogenes Difficulty: Medium

50) What is the role of the tumor suppressor gene in the regulation of the cell cycle? Include in your answer discussion of at least one specific tumor suppressor gene such as pRB. Answer: The proteins encoded by tumor suppressor genes function in a diverse array of cellular processes, including division, differentiation, programmed cell death, and DNA repair.

Molecular and biochemical analyses have elucidated the role of pRB in cell-cycle regulation.

Early in the G1 phase of the cell cycle, pRB binds to the E2F proteins, a family of transcription factors that control the expression of several genes whose products move the cell through its cycle.

When E2F transcription factors are bound to pRB, they cannot bind to specific enhancer sequences in their target genes. Consequently, the cell-cycle factors encoded by these genes are not produced, and the machinery for DNA synthesis and cell division remains quiescent. Later in G1, pRB is phosphorylated through the action of cyclin-dependent kinases. In this changed state, pRB releases the E2F transcription factors that have bound to it. These released transcription factors are then free to activate their target genes, which encode proteins that induce the cell to progress through S phase and into mitosis. After mitosis, pRB is dephosphorylated, and each of the daughter cells enters the quiescent phase of a new cell cycle.

Section: 23.3 Tumor Suppressor Genes Difficulty: Medium

Chapter 24 Test Bank Question Type: Multiple Choice 1) Who is credited with proposing the original theory of evolution? a) Charles Darwin b) James Watson c) Francis Crick d) Alfred Wallace e) None of these Answer: a Section: 24.1 The Emergence of Evolutionary Theory Difficulty: Easy 2) Darwin proposed that a species changes as a result of a) generations of malnutrition. b) generations of competition among individuals. c) lack of intelligence. d) genetic abnormalities. e) None of these Answer: b Section: 24.1 The Emergence of Evolutionary Theory Difficulty: Easy 3) According to Darwin’s theory of natural selection: 1. Individuals within a species vary with regards to heritable traits.

2. Individuals who have certain characteristics are more able to survive and reproduce.

3. Individuals who pass their traits down to offspring will eventually change the characteristics of the species. a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 24.1 The Emergence of Evolutionary Theory Difficulty: Medium 4) What did Darwin propose to be the driving force of evolution in nature? a) Selection b) Migration c) Emigration d) Inbreeding e) Random mating Answer: a Section: 24.1 The Emergence of Evolutionary Theory Difficulty: Easy

5) Which publication by Charles Darwin outlines his theory for evolution? a) The Origin of Evolution b) The Origin of the Galapagos c) The Origin of Species d) Voyage to the Galapagos e) The Beagle's Voyage Answer: c Section: 24.1 The Emergence of Evolutionary Theory Difficulty: Easy 6) Which of the following were not explained in Darwin’s theory of evolution?

1. The origin of variation among individuals 2. How particular variants are inherited 3. How individuals are selected in nature to pass on their traits a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: d Section: 24.1 The Emergence of Evolutionary Theory Difficulty: Medium 7) Which of the following principles grounded Darwin’s work and allowed for an explanation of how variations are inherited?

1. Wallace’s theory of inheritance 2. Mendel’s principles of inheritance 3. Morgan’s principles of inheritance a) 1 b) 2 c) 3 d) 1 and 2 e) 1 and 3 Answer: b Section: 24.1 The Emergence of Evolutionary Theory Difficulty: Easy

8) Different colored bands on the shells of land snails would be classified as what type of variation? a) Molecular variation b) Chromosomal variation c) Phenotypic variation d) Molecular variation and chromosomal variation e) Chromosomal variation and phenotypic variation Answer: c Section: 24.2 Genetic Variation in Natural Populations Difficulty: Easy 9) Which type of variation was first to be analyzed by naturalists and geneticists? a) Molecular variation b) Chromosomal variation c) Phenotypic variation d) Molecular variation and chromosomal variation e) Chromosomal variation and phenotypic variation Answer: c Section: 24.2 Genetic Variation in Natural Populations Difficulty: Easy 10) Which of the following is not considered a polymorphism? a) Light and dark forms of peppered moths b) White or blue flowered Lianthus parryae c) Blood types d) White eyes in Drosophila e) All of these Answer: d Section: 24.2 Genetic Variation in Natural Populations Difficulty: Easy 11) The status of the Duffy polymorphism for blood type varies among a) individual members of the same family. b) human ethnic groups. c) human religious groups. d) inbred family groups. e) None of these Answer: b Section: 24.2 Genetic Variation in Natural Populations Difficulty: Easy

12) Which of the following would be the best example of polymorphisms changing in various environments? a) Yellow eyes and white wings in Drosophila b) The Duffy polymorphism in humans c) The light and dark forms of the Peppered moth d) The white and blue flowers of Lianthus parryae e) All of these Answer: c Section: 24.2 Genetic Variation in Natural Populations Difficulty: Medium 13) Which structure was examined by Dobzhansky and his collaborators, enabling the detection of variability underlying a phenotype? a) Polytene chromosomes in Drosophila b) Polytene chromosomes in C. elegans c) Plasmids in E. coli d) Transposons in humans e) None of these Answer: a Section: 24.2 Genetic Variation in Natural Populations Difficulty: Easy 14) What type of mutation created the different arrangements of banding patterns in the polytene chromosomes studied by Dobzhansky? a) Transversion b) Transition c) Inversion d) Point mutation e) Frameshift mutation Answer: c Section: 24.2 Genetic Variation in Natural Populations Difficulty: Easy 15) In the study of polytene chromosomes among Drosophila it was determined that: 1. Different arrangements predominated different geographical regions.

2. Frequencies of the arrangements changed seasonally.

3. Long-term changes in the frequencies of arrangements occur in some populations. a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 24.2 Genetic Variation in Natural Populations Difficulty: Medium

16) In the study of polytene chromosomes among Drosophila it was determined that __________ selection plays an important role in maintaining chromosomal polymorphisms in nature.

A) Artificial selection B) Balancing selection C) Disruptive selection D) Dispersive selection E) None of these Answer: b Section: 24.2 Genetic Variation in Natural Populations Difficulty: Medium 17) Different forms of an enzyme encoded by different alleles of a gene are known as a) isoalleles. b) isoenzymes. c) allozymes. d) alloleles. e) isozymes.

Answer: c Section: 24.3 Molecular Evolution Difficulty: Easy 18) Allozymes differ from each other by a) one amino acid in their central sequence. b) one or more amino acids in their overall sequences. c) less than one amino acid in their central sequences. d) the tertiary structure. e) None of these Answer: b Section: 24.3 Molecular Evolution Difficulty: Easy 19) Proteins that exhibit electrophoretic variation with at least two of the variants having frequencies greater than 1 percent in the population are known as a) polymorphic b) monomorphic c) amorphic d) aneuplodic e) None of these Answer: a Section: 24.3 Molecular Evolution Difficulty: Easy

20) Which of the following is a disadvantage of protein gel electrophoresis?

1. Nonsoluble, hydrophobic proteins cannot readily be analyzed.

2. It focuses on gene products rather than on the genes themselves.

3. It tells nothing about variation in the nongenic portion of a genome. a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 24.3 Molecular Evolution Difficulty: Medium 21) Nucelotide differences that have no effect on the amino acid sequence of the polypeptide are known as a) null polymorphisms. b) lethal polymorphisms. c) isopolymorphisms. d) silent polymorphisms. e) missense polymorphisms.

Answer: d Section: 24.3 Molecular Evolution Difficulty: Easy 22) How are silent polymorphisms possible?

1. The degeneracy of the genetic code 2. The commaless nature of the genetic code 3. The high rate of mutations in species a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: a Section: 24.3 Molecular Evolution Difficulty: Medium 23) Polymorphisms in DNA structure have been detected by a) sequencing cloned DNA. b) sequencing PCR amplified DNA. c) diagnostic gene chips. d) sequencing cloned DNA and Sequencing PCR amplified DNA. e) All of these Answer: e Section: 24.3 Molecular Evolution Difficulty: Easy

24) In The Origin of Species, Darwin repeatedly referred to evolution as a process of a) descent with variation. b) descent with modification. c) descent with change. d) descent of speciation. e) None of these Answer: b Section: 24.3 Molecular Evolution Difficulty: Easy 25) The molecular source of variation that underlies the process of evolution is a) mutation b) genetic drift c) artificial selection d) migration e) emigration Answer: a Section: 24.3 Molecular Evolution Difficulty: Easy 26) Which type of evidence was used by Darwin to propose that species evolve? a) DNA b) Protein structure c) Nucleotide sequence d) Fossils e) All of these Answer: d Section: 24.3 Molecular Evolution Difficulty: Easy 27) Which of the following is considered a “document of evolutionary history” today? a) Fossils b) DNA c) Proteins d) Fossils and DNA e) All of these Answer: e Section: 24.3 Molecular Evolution Difficulty: Easy

28) Which of the following is not an advantage of studying evolution through the use of DNA and proteins? a) DNA and protein sequences follow simple rules of heredity. b) Molecular sequence data are easy to obtain. c) Molecular sequence data are amenable to quantitative analyses framed in the context of evolutionary genetics theory. d) Researchers usually cannot obtain DNA or protein sequence data from extinct organisms. e) Molecular sequence data allow researchers to investigate evolutionary relationships among organisms that are phenotypically very dissimilar.

Answer: d Section: 24.3 Molecular Evolution Difficulty: Medium 29) Evolutionary relationships among organisms are summarized in diagrams called a) pedigree charts. b) phylogenetic trees. c) family trees. d) pedigree charts and phylogenetic trees. e) pedigree charts and family trees.

Answer: b Section: 24.3 Molecular Evolution Difficulty: Easy 30) Which of the following can be shown using a phylogenetic tree?

1. Relationships among organisms 2. How each organism in the tree evolved over time 3. Chromosomal makeup of each organism a) 1 b) 2 c) 3 d) 1 and 2 e) 1 and 3 Answer: d Section: 24.3 Molecular Evolution Difficulty: Medium 31) Each bifurcation in a phylogenetic tree represents a) a different ancestor. b) a common ancestor. c) a common descendent. d) a different descendent. e) a mutation Answer: b Section: 24.3 Molecular Evolution Difficulty: Easy

32) Even if they have diverged significantly from the ancestor and are different from each other descendants of an ancestral DNA or protein sequence are said to be a) homologous. b) heterologous. c) analogous. d) heterozygous. e) homozygous.

Answer: a Section: 24.3 Molecular Evolution Difficulty: Easy 33) Two molecular sequences that come to resemble each other even though they are derived from entirely different ancestral sequences are said to be a) homologous. b) heterologous. c) analogous. d) heterozygous. e) homozygous.

Answer: c Section: 24.3 Molecular Evolution Difficulty: Easy 34) Which type of molecular sequence would is best for constructing a phylogenetic tree? a) Homologous b) Heterologous c) Analogous d) Heterozygous e) Homozygous Answer: a Section: 24.3 Molecular Evolution Difficulty: Easy 35) Which of the following is not a common feature shared by the methods used to construct a phylogenetic tree? a) Aligning the sequences to allow comparisons among them b) Ascertaining the amount of similarity (or difference) between any two sequences c) Grouping the sequences on the basis of similarity d) Placing the sequences at the tips of a tree e) Placing a chromosome map at the base of each tree Answer: e Section: 24.3 Molecular Evolution Difficulty: Medium

36) A distinct basic structure in a phylogenetic tree is known as a) heterology. b) topography. c) topology. d) parsimony. e) homology.

Answer: c Section: 24.3 Molecular Evolution Difficulty: Easy 37) The best phylogenetic tree is one that requires the fewest mutational changes to explain the evolution of all the tree's sequences from a common ancestor. This is known as a) the principle of partrology. b) the principle of topology. c) the principle of parsimony. d) the principle of topography. e) the principle of homology.

Answer: c Section: 24.3 Molecular Evolution Difficulty: Easy 38) How can the rate of evolution be determined using a phylogenetic tree?

1. Link the branch points of a tree to specific times in the evolutionary history of the sequences 2. Link the common ancestor to a specific evolutionary era in history 3. Link the analogous sequences to specific times in evolutionary history a) 1 b) 2 c) 3 d) 1 and 2 e) None of these Answer: a Section: 24.3 Molecular Evolution Difficulty: Medium 39) Which sequence was used to determine an evolutionary rate? a) -globin b) Cro c) RecA d) -globin and RecA e) All of these Answer: a Section: 24.3 Molecular Evolution Difficulty: Easy

40) Which type of DNA sequence exhibits the highest evolutionary rates? a) Nucleotides in the second position of a codon b) Nucleotides in the first position of a codon c) Pseudogenes d) Introns e) Rates among DNA sequences do not vary Answer: c Section: 24.3 Molecular Evolution Difficulty: Easy 41) The Neutral Theory states that: 1. For selectively neutral mutations, the rate of molecular evolution is equal to the rate at which these mutations occur in the population.

2. The rate of evolution does not depend on population size, the efficiency of selection, or peculiarities of the mating system.

3. If the neutral mutation rate is constant, then nucleotide and amino acid substitutions, which are due to mutations, should occur in clocklike fashion in all evolving lineages. a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e Section: 24.3 Molecular Evolution Difficulty: Medium 42) What can be can be determined by calculating the average number of amino acid or nucleotide changes that have occurred per site in a molecule since two or more evolving lineages diverged from a common ancestor?

1. The amount of phenotypic evolution 2. The rate of molecular evolution 3. The future of phenotypic evolution a) 1 b) 2 c) 3 d) 1 and 3 e) All of these Answer: b Section: 24.3 Molecular Evolution Difficulty: Medium

43) In evolutionary genetics a group of populations that share a common gene pool is known as a/an a) class. b) order. c) genus. d) species. e) family.

Answer: d Section: 24.4 Speciation Difficulty: Easy 44) Which of the following is a key event in the speciation process?

1. Artificial selection 2. Genetic drift 3. Development of reproductive isolation between populations a) 1 b) 2 c) 3 d) 1 and 2 e) 1 and 3 Answer: c Section: 24.4 Speciation Difficulty: Easy 45) The process whereby subpopulations evolve reproductive isolation while they are geographically separated is called a) allopatric speciation. b) sympatric speciation. c) synonymous speciation. d) analogous speciation. e) homologous speciation.

Answer: a Section: 24.4 Speciation Difficulty: Easy 46) The process of evolving reproductive isolation between subpopulations that exist in the same territory is called a) allopatric speciation. b) sympatric speciation. c) synonymous speciation. d) analogous speciation. e) homologous speciation.

Answer: b Section: 24.4 Speciation Difficulty: Easy

47) Fossil evidence indicates that the remote ancestors of human beings evolved in Africa, beginning about a) 3,000 years ago. b) 50,000-100,000 years ago. c) 4-5 million years ago. d) 100 million years ago. e) 60,000 years ago.

Answer: c Section: 24.5 Human Evolution Difficulty: Easy 48) Genetic evidence indicates that modern human populations may have emerged from Africa and subsequently spread to other continents how many years ago? a) 50,000-100,000 years ago b) 100,000-200,000 years ago c) 4-5 million years ago d) 100 million years ago e) 60,000 years ago Answer: b Section: 24.5 Human Evolution Difficulty: Easy

Question Type: Essay 49) Briefly explain the procedure of gel electrophoresis and how proteins can be separated and studied using this procedure.

Answer: Gel electrophoresis is a sieving technique that separates macromolecules on the basis of size and charge. The sieving agent is a thin, rectangular gel made from a polymer such as starch or polyacrylamide. Samples containing the macromolecules of interest are deposited in wells formed in a line across the gel. The gel is immersed in a solution of buffer formulated to conduct electricity, and the ends of the tank containing the buffer are connected to a power source. When the power is turned on, an electric field is created in the buffer tank. In this field, macromolecules migrate through the gel at a rate that depends on their size and charge. Smaller, more highly charged molecules migrate faster than larger, less highly charged molecules. Proteins that differ in size and charge can therefore be separated from each other by moving them through the gel. After sufficient time has elapsed, the power is turned off and the gel is treated to reveal how far each protein has migrated. The treatment may involve a reagent that stains the proteins, or, if the proteins are enzymes, it may involve a substrate whose chemical change is coupled to the production of a characteristic color. Protein gel electrophoresis therefore allows a researcher to detect variation at the level of gene products—that is, as amino acid differences in polypeptide chains. This technique can be applied to proteins extracted from almost any kind of organism, including those that are not amenable to genetic analysis in the laboratory. In addition, because protein extracts are easy to obtain, it provides a way to survey genetic variation in large samples of individuals from different populations. Protein gel electrophoresis therefore allows researchers to investigate the spatial and temporal dimensions of genetic variation in nature.

Section: 24.2 Genetic Variation in Natural Populations Difficulty: Medium 50) Briefly explain the various forms of DNA technologies and how they are used to study genetic variability in various species.

Answer: DNA sequencing provides the ultimate data on genetic variation. Any sequence—coding, noncoding, genic, nongenic—can be analyzed. The first efforts to study genetic variation by DNA sequencing used material that had been cloned from the genomes of different individuals.

Each clone was obtained by virtue of its ability to hybridize with a specific DNA probe. The clones were then sequenced, and the sequences were compared to identify differences along their lengths.

Today, obtaining DNA sequence data to study naturally occurring genetic variation is not nearly as difficult as it used to be. Particular regions of the genome can be amplified by PCR, and the resulting DNA products can be sequenced by machine. Sophisticated computer programs can then be used to analyze the sequence data and identify variation among individuals. This technique permits researchers to assess the level of variation in functionally different regions of DNA—for instance, in exons compared to introns. Gene chip technologies provide another means of documenting variation at the DNA level. These technologies allow researchers to screen genomic DNA for single-nucleotide polymorphisms (SNPs), which are found every 1-2 kb. Many different genomic DNA samples can be analyzed in parallel, and a great many SNPs can be detected on a single chip.

Section: 24.2 Genetic Variation in Natural Populations Difficulty: Medium

51) What are the advantages of studying evolution using DNA and protein sequences as compared to more traditional methods such as comparative anatomy and physiology?

Answer: The analysis of DNA and protein sequences has several advantages over more traditional methods of studying evolution based on comparative anatomy, physiology, and embryology. First, DNA and protein sequences follow simple rules of heredity. By contrast, anatomical, physiological, and embryological traits are subject to all the vicissitudes of complex heredity.

Second, molecular sequence data are easy to obtain, and they are also amenable to quantitative analyses framed in the context of evolutionary genetics theory. The interpretation of these analyses is usually much more straightforward than the interpretation of analyses based on morphological data. Third, molecular sequence data allow researchers to investigate evolutionary relationships among organisms that are phenotypically very dissimilar. For instance, DNA and protein sequences from bacteria, yeast, protozoa, and humans can be compared to study the evolutionary relationships among them.

Section: 24.3 Molecular Evolution Difficulty: Medium 52) A researcher has sequenced a gene from four populations of an organism, denoted 1, 2, 3, and 4. This gene consists of two exons separated by an intron; a 5 untranslated region (UTR) is included in the first exon, and a 3 UTR is included in the second exon. When we align the four sequences of this gene, we find that two of them have a transposable element insertion in the 3 UTR, and three of them have lost a short sequence within the intron. We also see that each of the four sequences has at least one feature that uniquely distinguishes it from all the other sequences—a G:C base pair near the start of the coding sequence in exon I in sequence 1, an A:T base pair near the start of exon II in sequence 2, a C:G base pair in the 5 UTR and a T:A base pair near the end of the coding sequence in exon II in sequence 3, and an A:T base pair in the 5 UTR, a C:G base pair in the middle of exon II, and a G:C base pair in the 3 UTR in sequence 4. Based on these similarities and differences, draw a phylogenetic tree to show how the four sequences are related.

Answer: see Figures 25.9b and Figure 25.9c-- Sequences 1 and 2 are the most similar—they both have the transposon insertion in the 3 UTR, and they are both missing the A:T base pair within the intron. Sequence 3 also lacks the A:T base pair within the intron. Because of these similarities, we could place sequences 1 and 2 close together in the tree—on branches diverging from a common point—and we could place sequence 3 on another branch nearby. Sequence 4 has none of the features found in the other sequences. Thus, because it is the most different in the sample—the outlier—we could place it on a branch that diverged from the other lineages at the root of the tree.

Section: 24.3 Molecular Evolution Difficulty: Medium

53) How do geneticists explain the variation in evolution rates?

Answer: Geneticists hypothesize that in more rapidly evolving proteins, the exact amino acid sequence is not as important as it is in more slowly evolving proteins. They speculate that in some proteins, amino acid changes can occur with relative impunity, whereas in others, they are rigorously selected against. According to this view, the rate of evolution depends on the degree to which the amino acid sequence of a protein is constrained by selection to preserve that protein's function. Slowly evolving proteins are more constrained than rapidly evolving proteins. Variation in evolutionary rates is therefore explained by the amount of functional constraint on the amino acid sequence. This idea also applies to parts of proteins. For example, the specific amino acids at or near the active sites of enzymes might be expected to be more rigorously constrained by selection than amino acids that simply take up space, such as those in the bridge segment of preproinsulin, which is discarded during the formation of the active insulin molecule. Thus, functionally more important proteins, or parts of proteins, evolve more slowly than functionally less important ones.

Section: 24.3 Molecular Evolution Difficulty: Medium

Answers to All Questions and Problems Chapter 1 1.1 In a few sentences, what were Mendel’s key ideas about inheritance?

ANS: Mendel postulated transmissible factors—genes—to explain the inheritance of traits. He discovered that genes exist in different forms, which we now call alleles.

Each organism carries two copies of each gene. During reproduction, one of the gene copies is randomly incorporated into each gamete. When the male and female gametes unite at fertilization, the gene copy number is restored to two. Different alleles may coexist in an organism. During the production of gametes, they separate from each other without having been altered by coexistence.

ANS: There are 3 × 141 = 423 nucleotides in the gene’s coding sequence. Its polypeptide product will contain 141 amino acids.

1.7 The template strand of a gene being transcribed is CTTGCCAGT. What will be the sequence of the RNA made from this template?

ANS: GAACGGUCT 1.8 What is the difference between transcription and translation?

ANS: Transcription is the production of an RNA chain using a DNA chain as a template. Translation is the production of a chain of amino acids—that is, a polypeptide—using an RNA chain as a template.

1.2 Both DNA and RNA are composed of nucleotides. What molecules combine to form a nucleotide?

1.9 RNA is synthesized using DNA as a template. Is DNA ever synthesized using RNA as a template? Explain.

ANS: Each nucleotide consists of a sugar, a nitrogen-containing base, and a phosphate.

ANS: Sometimes, DNA is synthesized from RNA in a process called reverse transcription. This process plays an important role in the life cycles of some viruses.

1.3 Which bases are present in DNA? Which bases are present in RNA? Which sugars are present in each of these nucleic acids?

ANS: The bases present in DNA are adenine, thymine, guanine, and cytosine; the bases present in RNA are adenine, uracil, guanine, and cytosine. The sugar in DNA is deoxyribose; the sugar in RNA is ribose.

1.4 What is a genome?

ANS: A genome is the set of all the DNA molecules that are characteristic of an organism. Each DNA molecule forms one chromosome in a cell of the organism.

1.5 The sequence of a strand of DNA is ATTGCCGTC. If this strand serves as the template for DNA synthesis, what will be the sequence of the newly synthesized strand?

ANS: TAACGGCAG 1.6 A gene contains 141 codons. How many nucleotides are present in the gene’s coding sequence? How many amino acids are expected to be present in the polypeptide encoded by this gene?

1.10 The gene for a-globin is present in all vertebrate species.

Over millions of years, the DNA sequence of this gene has changed in the lineage of each species. Consequently, the amino acid sequence of a-globin has also changed in these lineages. Among the 141 amino acid positions in this polypeptide, human a-globin differs from shark a-globin in 79 positions; it differs from carp a-globin in 68 and from cow a-globin in 17. Do these data suggest an evolutionary phylogeny for these vertebrate species?

ANS: The human and cow a-globins are least different; therefore, on the assumption that differences in a-globin reflect the degree of phylogenetic relationship, the human and the cow are the most closely related organisms among those mentioned. The next closest “relative” of humans is the carp, and the most distant relative is the shark.

1.11 Sickle-cell anemia is caused by a mutation in one of the codons in the gene for b-globin; because of this mutation, the sixth amino acid in the b-globin polypeptide is a valine instead of a glutamic acid. A less severe type of anemia is caused by a mutation that changes this same codon

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2-WC Answers to All Questions and Problems to one specifying lysine as the sixth amino acid in the b-globin polypeptide. What word is used to describe the two mutant forms of this gene? Do you think that an individual carrying these two mutant forms of the b-globin gene would suffer from anemia? Explain.

ANS: The two mutant forms of the b-globin gene are properly described as alleles. Because neither of the mutant alleles can specify a “normal” polypeptide, an individual who carries each of them would probably suffer from anemia.

1.12 Hemophilia is an inherited disorder in which the bloodclotting mechanism is defective. Because of this defect, people with hemophilia may die from cuts or bruises, especially if internal organs such as the liver, lungs, or kidneys have been damaged. One method of treatment involves injecting a blood-clotting factor that has been purified from blood donations. This factor is a protein encoded by a human gene. Suggest a way in which modern genetic technology could be used to produce this factor on an industrial scale. Is there a way in which the inborn error of hemophilia could be corrected by human gene therapy?

ANS: The gene for the human clotting factor could be isolated from the human genome and transferred into bacteria, which could then be grown in vats to produce large amounts of the gene’s protein product. This product could be isolated from the bacteria, purified, and then injected into patients to treat hemophilia. Another approach would be to transfer a normal copy of the clotting factor gene into the cells of people who have hemophilia. If expressed properly, the transferred normal gene might be able to compensate for the mutant allele these people naturally carry. For this approach to succeed, the normal clotting factor gene would have to be transferred into the cells that produce clotting factor, or into their precursors.

Chapter 2 2.1 Carbohydrates and proteins are linear polymers. What types of molecules combine to form these polymers?

ANS: Sugars combine to form carbohydrates; amino acids combine to form proteins.

2.2 All cells are surrounded by a membrane; some cells are surrounded by a wall. What are the differences between cell membranes and cell walls?

ANS: Cell membranes are made of lipids and proteins; they have a fluid structure. Cell walls are made of more rigid materials such as cellulose.

Eukaryotic cells usually possess a well-developed internal system of membranes and they also have membranebounded subcellular organelles such as mitochondria and chloroplasts; prokaryotic cells do not typically have a system of internal membranes (although some do), nor do they possess membrane-bounded organelles.

2.4 Distinguish between the haploid and diploid states.

What types of cells are haploid? What types of cells are diploid?

ANS: In the haploid state, each chromosome is represented once; in the diploid state, each chromosome is represented twice. Among multicellular eukaryotes, gametes are haploid and somatic cells are diploid.

2.5 Compare the sizes and structures of prokaryotic and eukaryotic chromosomes.

ANS: Prokaryotic chromosomes are typically (but not always) smaller than eukaryotic chromosomes; in addition, prokaryotic chromosomes are circular, whereas eukaryotic chromosomes are linear. For example, the circular chromosome of E. coli, a prokaryote, is about 1.4 mm in circumference. By contrast, a linear human chromosome may be 10–30 cm long. Prokaryotic chromosomes also have a comparatively simple composition: DNA, some RNA, and some protein. Eukaryotic chromosomes are more complex: DNA, some RNA, and a lot of protein.

2.6 With a focus on the chromosomes, what are the key events during interphase and M phase in the eukaryotic cell cycle?

ANS: During interphase, the chromosomes duplicate. During M phase (mitosis), the duplicated chromosomes, each consisting of two identical sister chromatids, condense (a feature of prophase), migrate to the equatorial plane of the cell (a feature of metaphase), and then split so that their constituent sister chromatids are separated into different daughter cells (a feature of anaphase); this last process is called sister chromatid disjunction.

2.7 Which typically lasts longer, interphase or M phase? Can you explain why one of these phases lasts longer than the other?

ANS: Interphase typically lasts longer than M phase. During interphase, DNA must be synthesized to replicate all the chromosomes. Other materials must also be synthesized to prepare for the upcoming cell division.

2.8 In what way do the microtubule organizing centers of plant and animal cells differ?

2.3 What are the principal differences between prokaryotic and eukaryotic cells?

ANS: The microtubule organizing centers of animal cells have distinct centrosomes, whereas the microtubule organizing centers of plant cells do not.

ANS: In a eukaryotic cell, the many chromosomes are contained within a membrane-bounded structure called the nucleus; the chromosomes of prokaryotic cells are not contained within a special subcellular compartment.

2.9 Match the stages of mitosis with the events they encompass: Stages: (1) anaphase, (2) metaphase, (3) prophase, and (4) telophase. Events: (a) reformation of the nucleolus, (b) disappearance of the nuclear membrane,

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Answers to All Questions and Problems WC-3

(c) condensation of the chromosomes, (d) formation of the mitotic spindle, (e) movement of chromosomes to the equatorial plane, (f) movement of chromosomes to the poles, (g) decondensation of the chromosomes, (h) splitting of the centromere, and (i) attachment of microtubules to the kinetochore.

ANS: (1) Anaphase: (f), (h); (2) metaphase: (e), (i); (3) prophase: (b), (c), (d); (4) telophase: (a), (g).

2.10 Arrange the following events in the correct temporal sequence during eukaryotic cell division, starting with the earliest: (a) condensation of the chromosomes, (b) movement of chromosomes to the poles, (c) duplication of the chromosomes, (d) formation of the nuclear membrane, (e) attachment of microtubules to the kinetochores, and (f) migration of centrosomes to positions on opposite sides of the nucleus.

ANS: (c), (f), (a), (e), (b), (d).

2.11 In human beings, the gene for b-globin is located on chromosome 11, and the gene for a-globin, which is another component of the hemoglobin protein, is located on chromosome 16. Would these two chromosomes be expected to pair with each other during meiosis? Explain your answer.

ANS: Chromosomes 11 and 16 would not be expected to pair with each other during meiosis; these chromosomes are heterologues, not homologues.

2.12 A sperm cell from the fruit fly Drosophila melanogaster contains four chromosomes. How many chromosomes would be present in a spermatogonial cell about to enter meiosis? How many chromatids would be present in a spermatogonial cell at metaphase I of meiosis? How many would be present at metaphase II? ANS: There are eight chromosomes in a Drosophila spermatogonial cell about to enter meiosis. There are 16 chromatids in a Drosophilia spermatogonial cell at metaphase I of meiosis. There are eight chromatids in a Drosophilia cell at metaphase II of meiosis.

2.13 Does crossing over occur before or after chromosome duplication in cells going through meiosis?

ANS: Crossing over occurs after chromosomes have duplicated in cells going through meiosis.

2.14 What visible characteristics of chromosomes indicate that they have undergone crossing over during meiosis?

ANS: The chiasmata, which are visible late in prophase I of meiosis, indicate that chromosomes have crossed over.

2.15 During meiosis, when does chromosome disjunction occur? When does chromatid disjunction occur?

How many are present in the male gametophyte? Are these nuclei haploid or diploid?

ANS: Leaf tissue is diploid. The female gametophyte contains eight identical haploid nuclei. The male gametophyte contains three identical haploid nuclei.

2.17 From the information given in Table 2.1 in this chapter, is there a relationship between genome size (measured in base pairs of DNA) and gene number? Explain.

ANS: Among eukaryotes, there does not seem to be a clear relationship between genome size and gene number. For example, humans, with 3.2 billion base pairs of genomic DNA, have about 20,500 genes, and Arabidopsis plants, with about 150 million base pairs of genomic DNA, have roughly the same number of genes as humans. However, among prokaryotes, gene number is rather tightly correlated with genome size, probably because there is so little nongenic DNA. 2.18 Are the synergid cells in an Arabidopsis female gametophyte genetically identical to the egg cell nestled between them?

ANS: Yes. 2.19 A cell of the bacterium Escherichia coli, a prokaryote, contains one chromosome with about 4.6 million base pairs of DNA comprising 4288 protein-encoding genes. A cell of the yeast Saccharomyces cerevisiae, a eukaryote, contains about 12 million base pairs of DNA comprising 6268 genes, and this DNA is distributed over 16 distinct chromosomes. Are you surprised that the chromosome of a prokaryote is larger than some of the chromosomes of a eukaryote? Explain your answer.

ANS: It is a bit surprising that yeast chromosomes are, on average, smaller than E. coli chromosomes because, as a rule, eukaryotic chromosomes are larger than prokaryotic chromosomes. Yeast is an exception because its genome— not quite three times the size of the E. coli genome—is distributed over 16 separate chromosomes.

2.20 Given the way that chromosomes behave during meiosis, is there any advantage for an organism to have an even number of chromosome pairs (such as Drosophila does), as opposed to an odd number of chromosome pairs (such as human beings do)?

ANS: No, there is no advantage associated with an even number of chromosomes. As long as the chromosomes come in pairs, they will be able to synapse during prophase I and then disjoin during anaphase I to distribute the genetic material properly to the two daughter cells.

ANS: Chromosome disjunction occurs during anaphase I. Chromatid disjunction occurs during anaphase II.

2.21 In flowering plants, two nuclei from the pollen grain participate in the events of fertilization. With which nuclei from the female gametophyte do these nuclei combine?

What tissues are formed from the fertilization events?

2.16 In Arabidopsis, is leaf tissue haploid or diploid? How many nuclei are present in the female gametophyte?

ANS: One of the pollen nuclei fuses with the egg nucleus in the female gametophyte to form the zygote, which then

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4-WC Answers to All Questions and Problems develops into an embryo and ultimately into a sporophyte. The other genetically functional pollen nucleus fuses with two nuclei in the female gametophyte to form a triploid nucleus, which then develops into a triploid tissue, the endosperm; this tissue nourishes the developing plant embryo.

2.22 The mouse haploid genome contains about 2.9 × 109 nucleotide pairs of DNA. How many nucleotide pairs of DNA are present in each of the following mouse cells: (a) somatic cell, (b) sperm cell, (c) fertilized egg, (d) primary oocyte, (e) first polar body, and (f) secondary spermatocyte?

ANS: (a) 5.8 × 109 nucleotide pairs (np); (b) 2.9 × 109 np; (c) 5.8 × 109 np; (d) 11.6 × 109 np; (e) 5.8 × 109 np; and (f) 5.8 × 109 np 2.23 Arabidopsis plants have 10 chromosomes (five pairs) in their somatic cells. How many chromosomes are present in each of the following: (a) egg cell nucleus in the female gametophyte, (b) generative cell nucleus in a pollen grain, (c) fertilized endosperm nucleus, and (d) fertilized egg nucleus?

ANS: (a) 5, (b) 5, (c) 15, (d) 10.

Chapter 3 3.1 On the basis of Mendel’s observations, predict the results from the following crosses with peas: (a) a tall (dominant and homozygous) variety crossed with a dwarf variety; (b) the progeny of (a) self-fertilized; (c) the progeny from (a) crossed with the original tall parent; (d) the progeny of (a) crossed with the original dwarf parent.

ANS: (a) All tall; (b) 3/4 tall, 1/4 dwarf; (c) all tall; (d) 1/2 tall, 1/2 dwarf.

3.2 Mendel crossed pea plants that produced round seeds with those that produced wrinkled seeds and self-fertilized the progeny. In the F2, he observed 5474 round seeds and 1850 wrinkled seeds. Using the letters W and w for the seed texture alleles, diagram Mendel’s crosses, showing the genotypes of the plants in each generation.

Are the results consistent with the Principle of Segregation?

ANS: Round (WW ) × wrinkled (ww) → F1 round (Ww); F1 self-fertilized → F2 3/4 round (2 WW; 1 Ww), 1/4 wrinkled (ww). The expected results in the F2 are 5493 round, 1831 wrinkled. To compare the observed and expected results, compute c2 with one degree of freedom; (5474 − 5493)2/5493 = (1850 − 1831)2/1831 = 0.263, which is not significant at the 5% level. Thus, the results are consistent with the Principle of Segregation.

3.3 A geneticist crossed wild, gray-colored mice with white (albino) mice. All the progeny were gray. These progeny were intercrossed to produce an F2, which consisted of 198 gray and 72 white mice. Propose a hypothesis to

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explain these results, diagram the crosses, and compare the results with the predictions of the hypothesis.

ANS: The data suggest that coat color is controlled by a single gene with two alleles, C (gray) and c (albino), and that C is dominant over c. On this hypothesis, the crosses are gray (CC) × albino (cc) → F1 gray (Cc); F1 × F1 → 3/4 gray (2 CC: 1 Cc), 1/4 albino (cc). The expected results in the F2 are 203 gray and 67 albino. To compare the observed and expected results, compute c2 with one degree of freedom: (198 − 203)2/203 + (67 − 72)2/72 = 0.470, which is not significant at the 5% level. Thus, the results are consistent with the hypothesis.

3.4 A woman has a rare abnormality of the eyelids called ptosis, which prevents her from opening her eyes completely. This condition is caused by a dominant allele, P. The woman’s father had ptosis, but her mother had normal eyelids. Her father’s mother had normal eyelids. (a) What are the genotypes of the woman, her father, and her mother? (b) What proportion of the woman’s children will have ptosis if she marries a man with normal eyelids?

ANS: (a) Woman’s genotype Pp, father’s genotype Pp, mother’s genotype pp; (b) ½ 3.5 In pigeons, a dominant allele C causes a checkered pattern in the feathers; its recessive allele c produces a plain pattern. Feather coloration is controlled by an independently assorting gene; the dominant allele B produces red feathers, and the recessive allele b produces brown feathers. Birds from a true-breeding checkered, red variety are crossed to birds from a true-breeding plain, brown variety. (a) Predict the phenotype of their progeny. (b) If these progeny are intercrossed, what phenotypes will appear in the F2 and in what proportions?

ANS: (a) Checkered, red (CC BB) × plain, brown (cc bb) → F1 all checkered, red (Cc Bb); (b) F2 progeny: 9/16 checkered, red (C- B-), 3/16 plain, red (cc B-), 3/16 checkered, brown (C- bb), 1/16 plain, brown (cc bb).

3.6 In mice, the allele C for colored fur is dominant over the allele c for white fur, and the allele V for normal behavior is dominant over the allele v for waltzing behavior, a form of dis-coordination. Given the genotypes of the parents in each of the following crosses: (a) Colored, normal mice mated with white, normal mice produced 29 colored, normal, and 10 colored, waltzing progeny (b) Colored, normal mice mated with colored, normal mice produced 38 colored, normal, 15 colored, waltzing, 11 white, normal, and 4 white, waltzing progeny

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Answers to All Questions and Problems WC-5

(c) Colored, normal mice mated with white, waltzing mice produced 8 colored, normal, 7 colored, waltzing, 9 white, normal, and 6 white, waltzing progeny.

ANS: (a) colored, normal (CC Vv) × white, normal (cc Vv) (b) colored, normal (Cc Vv) × colored, normal (Cc Vv); (c) colored, normal (Cc Vv) × white, waltzing (cc vv).

3.10 A researcher studied six independently assorting genes in a plant. Each gene has a dominant and a recessive allele: R black stem, r red stem; D tall plant, d dwarf plant; C full pods, c constricted pods; O round fruit, o oval fruit; H hairless leaves, h hairy leaves; W purple flower, w white flower. From the cross (P1) Rr Dd cc Oo Hh Ww × (P2) Rr dd Cc oo Hh ww,

3.7 In rabbits, the dominant allele B causes black fur and the recessive allele b causes brown fur; for an independently assorting gene, the dominant allele R causes long fur and the recessive allele r (for rex) causes short fur. A homozygous rabbit with long, black fur is crossed with a rabbit with short, brown fur, and the offspring are intercrossed.

In the F2, what proportion of the rabbits with long, black fur will be homozygous for both genes?

(a) How many kinds of gametes can be formed by P1?

ANS: Among the F2 progeny with long, black fur, the genotypic ratio is 1 BB RR: 2 BB Rr: 2 Bb RR: 4 Bb Rr; thus, 1/9 of the rabbits with long, black fur are homozygous for both genes.

(e) What is the probability of obtaining a black, dwarf, constricted, oval, hairy, purple phenotype in the progeny?

3.8 In shorthorn cattle, the genotype RR causes a red coat, the genotype rr causes a white coat, and the genotype Rr causes a roan coat. A breeder has red, white, and roan cows and bulls. What phenotypes might be expected from the following matings and in what proportions? (a) Red × red (b) Red × roan

(b) How many genotypes are possible among the progeny of this cross? (c) How many phenotypes are possible among the progeny? (d) What is the probability of obtaining the Rr Dd cc Oo hh ww genotype in the progeny?

ANS: (a) 2 × 2 × 1 × 2 × 2 × 2 = 32; (b) 3 × 2 × 2 × 2 × 3 × 2 = 144; (c) 2 × 2 × 2 × 2 × 2 × 2 = 64; (d) (1/2) × (1/2) × (1/2) × (1/2) × (1/4) × (1/2) = 1/128; (e) (3/4) × (1/2) × (1/2) × (1/2) × (1/4) × (1/2) = 3/256.

3.11 For each of the following situations, determine the degrees of freedom associated with the c2 statistic and decide whether or not the observed c2 value warrants acceptance or rejection of the hypothesized genetic ratio.

Hypothesized Ratio

(c) Red × white (d) Roan × roan.

ANS: (a) All red; (b) 1/2 red, 1/2 roan; (c) all roan; (d) 1/4 red, 1/2 roan, 1/4 white 3.9 How many different kinds of F1 gametes, F2 genotypes, and F2 phenotypes would be expected from the following crosses: (a) AA × aa; (b) AA BB × aa bb; (c) AA BB CC × aa bb cc? (d) What general formulas are suggested by these answers?

ANS: F1 Gametes

F2 Genotypes

F2 Phenotypes

(a) 2

3

2

(b) 2 × 2 = 4

3×3=9

2×2=4

(c) 2 × 2 × 2 = 8

3 × 3 × 3 = 27

2×2×2=8

(d) 2n 3n

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2n, where n is the number of genes

Observed−χ2

(a) 3:1

7.0

(b) 1:2:1

7.0

(c) 1:1:1:1

7.0

(d) 9:3:3:1

5.0

ANS: (a) 1, reject; (b) 2, reject; (c) 3, accept; (d) 3, accept.

3.12 Mendel testcrossed pea plants grown from yellow, round F1 seeds to plants grown from green, wrinkled seeds and obtained the following results: 31 yellow, round; 26 green, round; 27 yellow, wrinkled; and 26 green, wrinkled. Are these results consistent with the hypothesis that seed color and seed texture are controlled by independently assorting genes, each segregating two alleles?

ANS: On the hypothesis, the expected number in each class is 27.5; c2 with three degrees of freedom is calculated as (31 − 27.5)2/27.5 + (26 − 27.5)2/27.5 + (27 − 27.5)2/27.5 + (26 − 27.5)2/27.5 = 0.618, which is not significant at the 5% level. Thus, the results are consistent with the hypothesis of two independently assorting genes, each segregating two alleles.

3.13 Perform a chi-square test to determine if an observed ratio of 30 tall to 20 dwarf pea plants is consistent with an expected ratio of 1:1 from the cross Dd × dd.

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6-WC Answers to All Questions and Problems ANS: c2 = (30 − 25)2/25 + (20 − 25)2/25 = 2, which is less than 3.84, the 5 percent critical value for a chi-square statistic with one degree of freedom; consequently, the observed segregation ratio is consistent with the expected ratio of 1:1.

3.14 Seed capsules of the Shepherd’s purse are either triangular or ovoid. A cross between a plant with triangular seed capsules and a plant with ovoid seed capsules yielded F1 hybrids that all had triangular seed capsules. When these F1 hybrids were intercrossed, they produced 80 F2 plants, 72 of which had triangular seed capsules and 8 of which had ovoid seed capsules. Are these results consistent with the hypothesis that capsule shape is determined by a single gene with two alleles?

ANS: If capsule shape is determined by a single gene with two alleles, the F2 plants should segregate in a 3:1 ratio. To test for agreement between the observed segregation data and the expected ratio, compute the expected number of plants with either triangular or ovoid seed capsules: (3/4) × 80 = 60 triangular and (1/4) × 80 = 20 ovoid; then compute a c2 statistic with one degree of freedom: c2 = (72 − 60)2/60 + (8 − 20)2/20 = 9.6, which exceeds the critical value of 3.84. Consequently, the data are inconsistent with the hypothesis that capsule shape is determined by a single gene with two alleles.

3.15 Albinism in humans is caused by a recessive allele a.

From marriages between people known to be carriers (Aa) and people with albinism (aa), what proportion of the children would be expected to have albinism? Among three children, what is the chance of one without albinism and two with albinism?

ANS: Man (Cc ff ) × woman (cc Ff ). (a) cc ff, (1/2) × (1/2) = 1/4; (b) Cc ff, (1/2) × (1/2) = 1/4; (c) cc Ff, (1/2) × (1/2) = 1/4; (d) Cc Ff, (1/2) × (1/2) = 1/4.

3.18 In generation V in the pedigree in Figure 3.15, what is the probability of observing seven children without the cancer-causing mutation and two children with this mutation among a total of nine children?

ANS: 9!/(7! 2!) × (1/2)7 × (1/2)2 = 0.07 3.19 If a man and a woman are heterozygous for a gene, and if they have three children, what is the chance that all three will also be heterozygous?

ANS: (1/2)3 = 1/8 3.20 If four babies are born on a given day: (a) What is the chance that two will be boys and two will be girls? (b) What is the chance that all four will be girls? (c) What combination of boys and girls among four babies is most likely? (d) What is the chance that at least one baby will be a girl?

ANS: (a) 4 × (1/2)2 × (1/2)2 = 4/16; (b) (1/2)4 = 1/16; (c) 2 boys, girls; (d) 1 − probability that all four are boys = 1 − (1/2)4 = 15/16.

3.21 In a family of six children, what is the chance that at least three are girls?

ANS: (20/64) + (15/64) + (6/64) + (1/64) = 42/64 3.22 The following pedigree shows the inheritance of a dominant trait. What is the chance that the offspring of the following matings will show the trait: (a) III-1 × III-3; (b) III-2 × III-4?

ANS: Half the children from Aa × aa matings would have ­albinism. In a family of three children, the chance that one will be unaffected and two affected is 3 × (1/2)1 × (1/2)2 = 3/8.

3.16 If both husband and wife are known to be carriers of the allele for albinism, what is the chance of the following combinations in a family of four children: (a) all four unaffected; (b) three unaffected and one affected; (c) two unaffected and two affected; (d) one unaffected and three affected?

ANS: (a) (3/4)4 = 81/256; (b) 4 × (3/4)3 × (1/4)1 = 108/256; (c) 6 × (3/4)2 × (1/4)2 = 54/256; (d) 4 × (3/4)1 × (1/4)3 = 12/256.

3.17 In humans, cataracts in the eyes and fragility of the bones are caused by dominant alleles that assort independently.

A man with cataracts and normal bones marries a woman without cataracts but with fragile bones. The man’s father had normal eyes, and the woman’s father had normal bones. What is the probability that the first child of this couple will (a) be free from both abnormalities; (b) have cataracts but not have fragile bones; (c) have fragile bones but not have cataracts; (d) have both cataracts and fragile bones?

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I 1

2

II 1

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III

ANS: (a) zero; (b) 1/2 3.23 The following pedigree shows the inheritance of a recessive trait. Unless there is evidence to the contrary, assume that the individuals who have married into the family do not carry the recessive allele. What is the chance that the offspring of the following matings will show the trait: (a) III-1 × III-12; (b) II-4 × III-14; (c) III-6 × III-13; (d) IV-1 × IV-2?

I 1

II III

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ANS: (a) (1/2) × (1/4) = 1/8; (b) (1/2) × (1/2) × (1/4) = 1/16; (c) (2/3) × (1/4) = 1/6; (d) (2/3) × (1/2) × (1/2) × (1/4) = 1/24

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3.24 In the following pedigrees, determine whether the trait is more likely to be due to a dominant or a recessive allele.

Assume the trait is rare in the population.

I 1

2

II 1

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(a) I II 1

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III 1 IV

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2

ANS: 1/2 3.28 A geneticist crosses tall pea plants with short pea plants.

All the F1 plants are tall. The F1 plants are then allowed to self-fertilize, and the F2 plants are classified by height: 62 tall and 26 short. From these results, the geneticist concludes that shortness in peas is due to a recessive allele (s) and that tallness is due to a dominant allele (S).

On this hypothesis, 2/3 of the tall F2 plants should be heterozygous Ss. To test this prediction, the geneticist uses pollen from each of the 62 tall plants to fertilize the ovules of emasculated flowers on short pea plants. The next year, three seeds from each of the 62 crosses are sown in the garden and the resulting plants are grown to maturity. If none of the three plants from a cross is short, the male parent is classified as having been homozygous SS; if at least one of the three plants from a cross is short, the male parent is classified as having been heterozygous Ss. Using this system of progeny testing, the geneticist concludes that 29 of the 62 tall F2 plants were homozygous SS and that 33 of these plants were heterozygous Ss.

3

(a) Using the chi-square procedure, evaluate these results for goodness of fit to the prediction that 2/3 of the tall F2 plants should be heterozygous.

(b)

ANS: (a) Recessive; (b) dominant.

(b) Informed by what you read in A Milestone in Genetics: Mendel’s 1866 Paper, which you can find in the Student Companion Site, explain why the geneticist’s procedure for classifying tall F2 plants by genotype is not definitive.

3.25 In pedigree (b) of Problem 3.24, what is the chance that the couple III-1 and III-2 will have an affected child?

What is the chance that the couple IV-2 and IV-3 will have an affected child?

ANS: For III-1 × III-2, the chance of an affected child is 1/2.

For IV-2 × IV-3, the chance is zero.

3.26 Peas heterozygous for three independently assorting genes were intercrossed.

(c) Adjust for the uncertainty in the geneticist’s classification procedure and calculate the expected frequencies of homozygotes and heterozygotes among the tall F2 plants.

(a) What proportion of the offspring will be homozygous for all three recessive alleles?

(d) Evaluate the predictions obtained in (c) using the chisquare procedure.

(b) What proportion of the offspring will be homozygous for all three genes?

ANS: (a) The observed numbers, expected numbers, and chisquare calculation are laid out in the following table:

(c) What proportion of the offspring will be homozygous for one gene and heterozygous for the other two? (d) What proportion of the offspring will be homozygous for the recessive allele of at least one gene?

ANS: (a) (1/4)3 = 1/64; (b) (1/2)3 = 1/8; (c) 3 × (1/2)1 × (1/2)2 = 3/8; (d) 1 − probability that the offspring is not homozygous for the recessive allele of any gene = 1 − (3/4)3 = 37/64.

3.27 The following pedigree shows the inheritance of a recessive trait. What is the chance that the couple III-3 and III-4 will have an affected child?

I 1

2

II 1

2

2

3

III 1

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4

Observed

Expected

(Obs − Exp)2 /Exp

Dominant homozygotes (SS)

29

62 × 1/3 = 20.7

3.33

Heterozygotes (Ss)

33

62 × 2/3 = 41.3

1.67

Total

62

62

5.00

The total chi-square value is greater than the critical value for a chi-square statistic with one degree of freedom (3.84). Therefore, we reject the hypothesis that the expected proportions are 1/3 and 2/3. (b) The problem with the geneticist’s classification procedure is that it allows for a heterozygote to be

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8-WC Answers to All Questions and Problems misclassified as a homozygote if none of its three progeny shows the recessive (short) phenotype. The probability of this event is 1/2 for any one offspring—therefore (1/2)3 = 1/8 for all three offspring. (c) The predicted frequencies must take into account the probability of misclassifying a heterozygote as a homozygote. The frequency of heterozygotes expected a priori (2/3) must be decreased by the probability of misclassification (1/8); thus, the predicted frequency of heterozygotes is 62 × (2/3) × (1 − 1/8) = 62 × (7/12) = 36.2.

The predicted frequency of homozygotes is obtained by subtraction: 62 – 36.2 = 25.8. (d) The chi-square calculation is (29 − 25.8)2/25.8 + (33 − 36.2)2/36.2 = 0.68, which is much less than the critical value for a chi-square statistic with one degree of freedom. Therefore, we tentatively accept the idea that adjusting for the probability of misclassification explains the observed data.

3.29 A researcher who has been studying albinism has identified a large group of families with four children in which at least one child shows albinism. None of the parents in this group of families shows albinism. Among the children, the ratio of those without albinism to those with albinism is 1.7:1. The researcher is surprised by this result because he thought that a 3:1 ratio would be expected on the basis of Mendel’s Principle of Segregation. Can you explain the apparently non-Mendelian segregation ratio in the researcher’s data?

ANS: The researcher has obtained what appears to be a nonMendelian ratio because he has been studying only families in which at least one child shows albinism. In these families, both parents are heterozygous for the mutant allele that causes albinism. However, other couples in the population might also be heterozygous for this allele but, simply due to chance, have failed to produce a child with albinism. If a man and a woman are both heterozygous carriers of the mutant allele, the chance that a child they produce will not have albinism is 3/4. The chance that four children they produce will not have albinism is therefore (3/4)4 = 0.316. In the entire population of families in which two heterozygous parents have produced a total of four children, the average number of affected children is 1. Among families in which two heterozygous parents have produced at least one affected child among a total of four children, the average must be greater than 1.

To calculate this conditional average, let us denote the number of children with albinism by x, and the probability that exactly x of the four children have albinism by P(x). The average number of affected children among families in which at least one of the four children is affected—that is, the conditional average—is therefore SxP(x)/(1 − P(0)), where the sum starts at x = 1 and ends at x = 4. We start the sum at x = 1 because we must exclude those cases in which none of the four children is affected. The divisor (1 − P(0)) is the probability that the couple has had at least one affected child among their

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four children. Now P(0) = 0.316 and SxP(x) = 1. Therefore, the average we seek is simply 1/(1 − 0.316) = 1.46.

If, in the subset of families with at least one affected child, the average number of affected children is 1.46, then the average number of unaffected children is 4 – 1.46 = 2.54.

Thus, the expected ratio of unaffected to affected children in these families is 2.54:1.46, or 1.74:1, which is what the researcher has observed.

Chapter 4 4.1 What blood types could be observed in children born to a woman who has blood type M and a man who has blood type MN? ANS: M and MN. 4.2 In rabbits, coloration of the fur depends on alleles of the gene c. From information given in the chapter, what phenotypes and proportions would be expected from the following crosses: (a) c+c+ × cc; (b) c+c × c+c; (c) c+c h × c+c ch; (d) cc ch × cc; (e) c+c h × c+c; (f) c hc × cc?

ANS: (a) All wild-type; (b) 3/4 wild-type, 1/4 albino; (c) 3/4 wild-type, 1/4 chinchilla; (d) 1/2 chinchilla, 1/2 albino; (e) 3/4 wild-type, 1/4 Himalayan; (f) 1/2 Himalayan, 1/2 albino.

4.3 In mice, a series of five alleles determines fur color. In order of dominance, these alleles are as follows: AY, yellow fur but homozygous lethal; AL, agouti with light belly; A+, agouti (wild-type); at, black and tan; and a, black. For each of the following crosses, give the coat color of the parents and the phenotypic ratios expected among the progeny: (a) AYAL × AYAL; (b) AYa × ALat; (c) ata × AYa; (d) ALat × ALAL; (e) ALAL × AYA+; (f) A+at × ata; (g) ata × aa; (h) AYAL × A+at; and (i) AYaL × AYA+.

ANS: Parents (a)

Offspring

Yellow × yellow

2 yellow: 1 light belly

(b) Yellow × light belly

2 yellow: 1 light belly: 1 black and tan

(c) Black and tan × yellow

2 yellow: 1 black and tan: 1 black

(d)

Light belly × light belly

All light belly

(e)

Light belly × yellow

1 yellow: 1 light belly

(f)

Agouti × black and tan

1 agouti: 1 black and tan

(g)

Black and tan × black

1 black and tan: 1 black

(h)

Yellow × agouti

1 yellow: 1 light belly

(i)

Yellow × yellow

2 yellow: 1 light belly

4.4 In several plants, such as tobacco, primrose, and red clover, combinations of alleles in eggs and pollen have been found to influence the reproductive compatibility of the plants. Homozygous combinations, such as S1 S1, do not

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Answers to All Questions and Problems WC-9

develop because S1 pollen is not effective on S1 stigmas.

However, S1 pollen is effective on S2 S3 stigmas. What progeny might be expected from the following crosses (seed parent written first): (a) S1 S2 × S2 S3; (b) S1 S2 × S3 S4; (c) S4 S5 × S4 S5; and (d) S3 S4 × S5 S6? .

ANS: (a) S S , S S , S S ; (b) S S , S S , S S , S S ; (c) S S ; (d) S3 S5, S3 S6, S4 S5, S4 S6.

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4

5

4.5 From information in the chapter about the ABO blood types, what phenotypes and ratios are expected from the following matings: (a) IA IA × IB IB; (b) IA IB × ii; (c) IA i × IB i; and (d) IA i × ii; ANS: (a) All AB; (b) 1 A: 1 B; (c) 1 A: 1 B: 1 AB: 1 O; (d) 1 A: 1 O. 4.6 A woman with type O blood gave birth to a baby, also with type O blood. The woman stated that a man with type AB blood was the father of the baby. Is there any merit to her statement?

ANS: No. The woman must be ii; if her mate is IA IB; they could not have an ii child.

4.7 A woman with type AB blood gave birth to a baby with type B blood. Two different men claim to be the father.

One has type A blood, the other has type B blood. Can the genetic evidence decide in favor of either?

ANS: No. The woman is IAIB. One man could be either IAIA or IAi; the other could be either IBIB or IBi. Given the uncertainty in the genotype of each man, either could be the father of the child.

4.8 The flower colors of plants in a particular population may be blue, purple, turquoise, light blue, or white. A series of crosses between different members of the population produced the following results: Cross

Parents

Progeny

1

Purple × blue

All purple

2

Purple × purple

76 purple, 25 turquoise

3

Blue × blue

86 blue, 29 turquoise

4

Purple × turquoise

49 purple, 52 turquoise

5

Purple × purple

69 purple, 22 blue

6

Purple × blue

50 purple, 51 blue

7

Purple × blue

54 purple, 26 blue, 25 turquoise

8

Turquoise × turquoise

All turquoise

9

Purple × blue

49 purple, 25 blue, 23 light blue

10

Light blue × light blue

60 light blue, 29 turquoise, 31 white

11

Turquoise × white

All light blue

12

White × white

All white

13

Purple × white

All purple

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How many genes and alleles are involved in the inheritance of flower color? Indicate all possible genotypes for the following phenotypes: (a) purple; (b) blue; (c) turquoise; (d) light blue; (e) white.

ANS: One gene with four alleles. (a) purple: cp cp, cp cb; cp ct, cp cw; (b) blue: cb cb, cb ct, cb cw; (c) turquoise: ct ct, ct cw; (d) light blue: ct cw; (e) white: cw cw.

4.9 A woman who has blood type O and blood type M marries a man who has blood type AB and blood type MN. If we assume that the genes for the A-B-O and M-N bloodtyping systems assort independently, what blood types might the children of this couple have, and in what proportions?

ANS: The woman is ii LMLM; the man is IAIB LMLN; the blood types of the children will be A and M, A and MN, B and M, and B and MN, all equally likely.

4.10 A Japanese strain of mice has a peculiar, uncoordinated gait called waltzing, which is due to a recessive allele, v.

The dominant allele V causes mice to move in a coordinated manner. A mouse geneticist has recently isolated another recessive mutation that causes uncoordinated movement. This mutation, called tango, could be an allele of the waltzing gene, or it could be a mutation in an entirely different gene. Propose a test to determine whether the waltzing and tango mutations are alleles, and if they are, propose symbols to denote them.

ANS: Cross homozygous waltzing with homozygous tango. If the mutations are alleles, all the offspring will have an uncoordinated gait; if they are not alleles, all the offspring will be wild-type. If the two mutations are alleles, they could be denoted with the symbols v (waltzing) and vt (tango).

4.11 Congenital deafness in human beings is inherited as a recessive condition. In the following pedigree, two deaf individuals, each presumably homozygous for a recessive mutation, have married and produced four children with normal hearing. Propose an explanation.

I II III IV

ANS: The individuals III-4 and III-5 must be homozygous for recessive mutations in different genes; that is, one is aa BB and the other is AA bb; none of their children is deaf because all of them are heterozygous for both genes (Aa Bb).

4.12 In the fruit fly, recessive mutations in either of two independently assorting genes, brown and purple, prevent the synthesis of red pigment in the eyes. Thus, homozygotes for either of these mutations have brownish-purple eyes.

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10-WC Answers to All Questions and Problems However, heterozygotes for both of these mutations have dark red, that is, wild-type eyes. If such double heterozygotes are intercrossed, what kinds of progeny will be produced, and in what proportions?

ANS: 9/16 dark red, 7/16 brownish purple.

4.13 The dominant mutation Plum in the fruit fly also causes brownish-purple eyes. Is it possible to determine by genetic experiments whether Plum is an allele of the brown or purple genes?

(a) RR Pp × rr Pp; (b) rr PP × Rr Pp; (c) Rr Pp × Rr pp; (d) Rr pp × rr pp.

ANS: (a) 3/4 walnut, 1/4 rose; (b) 1/2 walnut, 1/2 pea; (c) 3/8 walnut, 3/8 rose, 1/8 pea, 1/8 single; (d) 1/2 rose, 1/2 single.

ANS: No. The test for allelism cannot be performed with dominant mutations.

4.18 Rose-comb chickens mated with walnut-comb chickens produced 15 walnut-, 14 rose-, 5 pea-, and 6 single-comb chicks. Determine the genotypes of the parents.

4.14 From information given in the chapter, explain why mice with yellow coat color are not true-breeding.

ANS: Rr pp × Rr Pp.

ANS: The allele for yellow fur is homozygous lethal.

4.15 A couple has four children. Neither the father nor the mother is bald; one of the two sons is bald, but neither of the daughters is bald. (a) If one of the daughters marries a nonbald man and they have a son, what is the chance that the son will become bald as an adult? (b) If the couple has a daughter, what is the chance that she will become bald as an adult?

ANS: The mother is Bb and the father is bb. The chance that a daughter is Bb is 1/2. (a) The chance that the daughter will have a bald son is (1/2) × (1/2) = 1/4. (b) The chance that the daughter will have a bald daughter is zero.

4.16 The following pedigree shows the inheritance of ataxia, a rare neurological disorder characterized by uncoordinated movements. Is ataxia caused by a dominant or a recessive allele? Explain.

I II III IV

ANS: Dominant. The condition appears in every generation and nearly every affected individual has an affected parent. The exception, IV-2, had a father who carried the ataxia allele but did not manifest the trait—an example of incomplete penetrance.

4.17 Chickens that carry both the alleles for rose comb (R) and pea comb (P) have walnut combs, whereas chickens that lack both of these alleles (i.e., they are genotypically rr pp) have single combs. From the information about interactions between these two genes given in the chapter, determine the phenotypes and proportions expected from the following crosses:

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4.19 Summer squash plants with the dominant allele C bear white fruit, whereas plants homozygous for the recessive allele c bear colored fruit. When the fruit is colored, the dominant allele G causes it to be yellow; in the absence of this allele (i.e., with genotype gg), the fruit color is green. What are the F2 phenotypes and proportions expected from intercrossing the progeny of CC GG and cc gg plants? Assume that the C and G genes assort independently.

ANS: 12/16 white, 3/16 yellow, 1/16 green.

4.20 The white Leghorn breed of chickens is homozygous for the dominant allele C, which produces colored feathers.

However, this breed is also homozygous for the dominant allele I of an independently assorting gene that inhibits coloration of the feathers. Consequently, Leghorn chickens have white feathers. The white Wyandotte breed of chickens has neither the allele for color nor the inhibitor of color; it is therefore genotypically cc ii. What are the F2 phenotypes and proportions expected from intercrossing the progeny of a white Leghorn hen and a white Wyandotte rooster?

ANS: 13/16 white, 3/16 colored.

4.21 Fruit flies homozygous for the recessive mutation scarlet have bright red eyes because they cannot synthesize brown pigment. Fruit flies homozygous for the recessive mutation brown have brownish-purple eyes because they cannot synthesize red pigment. Fruit flies homozygous for both of these mutations have white eyes because they cannot synthesize either type of pigment. The brown and scarlet mutations assort independently. If fruit flies that are heterozygous for both of these mutations are intercrossed, what kinds of progeny will they produce, and in what proportions?

ANS: 9/16 dark red (wild-type), 3/16 brownish purple, 3/16 bright red, 1/16 white.

4.22 Consider the following hypothetical scheme of determination of coat color in a mammal. Gene A controls the conversion of a white pigment P0 into a gray pigment

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Answers to All Questions and Problems WC-11

P1; the dominant allele A produces the enzyme necessary for this conversion, and the recessive allele a produces an enzyme without biochemical activity. Gene B controls the conversion of the gray pigment P1 into a black pigment P2; the dominant allele B produces the active enzyme for this conversion, and the recessive allele b produces an enzyme without activity. The dominant allele C of a third gene produces a polypeptide that completely inhibits the activity of the enzyme produced by gene A; that is, it prevents the reaction P0→P1. Allele c of this gene produces a defective polypeptide that does not inhibit the reaction P0→P1. Genes A, B, and C assort independently, and no other genes are involved. In the F2 of the cross AA bb CC × aa BB cc, what is the expected phenotypic segregation ratio?

Progeny genotypes:

4.23 What F2 phenotypic segregation ratio would be expected for the cross described in the preceding problem if the dominant allele, C, of the third gene produced a product that completely inhibited the activity of the enzyme produced by gene B—that is, prevented the reaction P1→P2, rather than inhibiting the activity of the enzyme produced by gene A? ANS: 9 black: 39 gray: 16 white.

4.24 The Micronesian Kingfisher, Halcyon cinnamomina, has a cinnamon-colored face. In some birds, the color continues onto the chest, producing one of three patterns: a circle, a shield, or a triangle; in other birds, there is no color on the chest. A male with a colored triangle was crossed with a female that had no color on her chest, and all their offspring had a colored shield on the chest.

When these offspring were intercrossed, they produced an F2with a phenotypic ratio of 3 circle: 6 shield: 3 triangle: 4 no color. (a) Determine the mode of inheritance for this trait and indicate the genotypes of the birds in all three generations. (b) If a male without color on his chest is mated to a female with a colored shield on her chest and the F1 segregate in the ratio of 1 circle: 2 shield: 1 triangle, what are the genotypes of the parents and their progeny? (a) The simplest explanation for the inheritance of the trait is recessive epistasis combined with incomplete dominance, summarized in the following table: Genotype

Phenotype

Frequency in F2

AA B-

Circle

3/16

Aa B-

Shield

6/16

aa B-

Triangular

3/16

A- bb

No color

3/16

aa bb

No color

1/16

Circle

Shield

Triangle

AA Bb

Aa Bb

aa Bb

4.25 In a species of tree, seed color is determined by four independently assorting genes: A, B, C, and D. The recessive alleles of each of these genes (a, b, c, and d) produce abnormal enzymes that cannot catalyze a reaction in the biosynthetic pathway for seed pigment. This pathway is diagrammed as follows:

White precursor

ANS: 9 black: 3 gray: 52 white.

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(b) Father’s genotype: Aa bb; mother’s genotype: Aa BB

A

Yellow

B

C

Orange D

Red Blue

When both red and blue pigments are present, the seeds are purple. Trees with the genotypes Aa Bb Cc Dd and Aa Bb Cc dd were crossed. (a) What color are the seeds in these two parental genotypes? (b) What proportion of the offspring from the cross will have white seeds? (c) Determine the relative proportions of red, white, and blue offspring from the cross.

ANS: (a) Purple × red; (b) proportion white (aa) = 1/4; (c) proportion red (A- B- C- dd) = (3/4)(3/4)(3/4)(1/2) = 27/128, proportion white (aa) = 1/4 = 32/128, proportion blue (A- B- cc Dd) = (3/4)(3/4)(1/4)(1/2) = 9/128.

4.26 Multiple crosses were made between true-breeding lines of black and yellow Labrador retrievers. All the F1 progeny were black. When these progeny were intercrossed, they produced an F2 consisting of 91 black, 39 yellow, and 30 chocolate. (a) Propose an explanation for the inheritance of coat color in Labrador retrievers. (b) Propose a biochemical pathway for coat color determination and indicate how the relevant genes control coat coloration.

ANS: (a) Because the F2 segregation is approximately 9 black: 3 chocolate: 4 yellow, coat color is determined by epistasis between two independently assorting genes: black = B- E-; chocolate = bb E-; yellow = B- ee or bb ee. (b) Yellow pigment—E brown pigment—B black pigment.

4.27 Two plants with white flowers, each from true-breeding strains, were crossed. All the F1 plants had red flowers.

When these F1 plants were intercrossed, they produced an F2 consisting of 177 plants with red flowers and

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12-WC Answers to All Questions and Problems 142 with white flowers. (a) Propose an explanation for the inheritance of flower color in this plant species. (b) Propose a biochemical pathway for flower pigmentation and indicate which genes control which steps in this pathway.

ANS: (a) Because the F2 segregation is approximately 9 red: 7 white, flower color is due to epistasis between two independently assorting genes: red = A- B- and white = aa B-, A- bb, or aa bb. (b) Colorless precursor—A colorless product—B red pigment.

4.28 Consider the following genetically controlled biosynthetic pathway for pigments in the flowers of a hypothetical plant:

P0

Gene A

Gene B

Gene C

Enzyme A

Enzyme B

Enzyme C

P1

P2

B Offspring of first cousins once removed

P3

Assume that gene A controls the conversion of a white pigment, P0, into another white pigment, P1; the dominant allele A specifies an enzyme necessary for this conversion, and the recessive allele a specifies a defective enzyme without biochemical function. Gene B controls the conversion of the white pigment, P1, into a pink pigment, P2; the dominant allele, B, produces the enzyme necessary for this conversion, and the recessive allele, b, produces a defective enzyme.

The dominant allele, C, of the third gene specifies an enzyme that converts the pink pigment, P2, into a red pigment, P3; its recessive allele, c, produces an altered enzyme that cannot carry out this conversion. The dominant allele, D, of a fourth gene produces a polypeptide that completely inhibits the function of enzyme C; that is, it blocks the reaction P2→P3. Its recessive allele, d, produces a defective polypeptide that does not block this reaction. Assume that flower color is determined solely by these four genes and that they assort independently. In the F2 of a cross between plants of the genotype AA bb CC DD and plants of the genotype aa BB cc dd, what proportion of the plants will have (a) red flowers? (b) pink flowers? (c) white flowers?

ANS: (a) Proportion red = (3/4)3 × (1/4) = 27/256; (b) proportion pink = (3/4)4 + [(3/4)2 × (1/4)] = 117/256; (c) proportion white = 1 − 144/256 = 112/256.

4.29 In the following pedigrees, what are the inbreeding coefficients of A, B, and C?

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A Offspring of half–first cousins

C Offspring of second cousins

ANS: FA = (1/2)5 = 1/32; FB = 2 × (1/2)6 = 1/32; FC = 2 × (1/2)7 = 1/64.

4.30 A, B, and C are inbred strains of mice, assumed to be completely homozygous. A is mated to B and B to C. Then the A × B hybrids are mated to C, and the offspring of this mating are mated to the B × C hybrids.

What is the inbreeding coefficient of the offspring of this last mating?

ANS: From the following pedigree, the inbreeding coefficient is (1/2)3 (1 + FC) + (1/2)4(1 + FB) = 3/8 because FB = FC = 1.

A

B

C

A×B

B×C

(A × B) × C

4.31 Mabel and Frank are half siblings, as are Tina and Tim. However, these two pairs of half siblings do not have any common ancestors. If Mabel marries Tim and Frank marries Tina and each couple has a child, what fraction of their genes will these children share by virtue of

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Answers to All Questions and Problems WC-13

common ancestry? Will the children be more or less closely related than first cousins?

ANS: The pedigree is shown below:

Frank

Mabel Tina

Tim

Chapter 5 5.1 What are the genetic differences between male- and female-determining sperm in animals with heterogametic males?

ANS: The male-determining sperm carries a Y chro­ mo­ some; the female-determining sperm carries an X chromosome.

5.2 A male with singed bristles appeared in a culture of Drosophila. How would you determine if this unusual phenotype was due to an X-linked mutation?

The coefficient of relationship between the offspring of the two couples is obtained by calculating the inbreeding coefficient of the imaginary child from a mating between these offspring and multiplying by 2: [(1/2)5 × 2] × 2 = 1/8. This is the same degree of relatedness as first cousins.

4.32 Suppose that the inbreeding coefficient of I in the following pedigree is 0.25. What is the inbreeding coefficient of I’s common ancestor, C? C

I

ANS: FI = (1/2)3(1 + FC) = 0.25; thus, FC = 1.

4.33 A randomly pollinated strain of maize produces ears that are 24 cm long, on average. After one generation of selffertilization, the ear length is reduced to 20 cm. Predict the ear length if self-fertilization is continued for one more generation.

ANS: The mean ear length for randomly mated maize is 24 cm and that for maize from one generation of selffertilization is 20 cm. The inbreeding coefficient of the offspring of one generation of self-fertilization is 1/2, and the inbreeding coefficient of the offspring of two generations of self-fertilization is (1/2)(1+1/2) = 3/4.

Mean ear length (Y ) is expected to decline linearly with inbreeding according to the equation Y = 24 − b F1 where b is the slope of the line. The value of b can be determined from the two values of Y that are given. The difference between these two values (4 cm) corresponds to an increase in F from 0 to 1/2. Thus, b = 4/(1/2) = 8 cm, and for F = 3/4, the predicted mean ear length is Y = 24 − 8 × (3/4) = 18 cm.

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ANS: Cross the singed male to wild-type females and then intercross the offspring. If the singed bristle phenotype is due to an X-linked mutation, approximately half the F2 males, but none of the F2 females, will show it.

5.3 In grasshoppers, rosy body color is caused by a recessive mutation; the wild-type body color is green. If the gene for body color is on the X chromosome, what kind of progeny would be obtained from a mating between a homozygous rosy female and a hemizygous wild-type male? (In grasshoppers, females are XX and males are XO.) ANS: All the daughters will be green and all the sons will be rosy.

5.4 In the mosquito Anopheles culicifacies, golden body (go) is a recessive X-linked mutation, and brown eyes (bw) is a recessive autosomal mutation. A homozygous XX female with golden body is mated to a homozygous XY male with brown eyes. Predict the phenotypes of their F1 offspring. If the F1 progeny are intercrossed, what kinds of progeny will appear in the F2 and in what proportions?

ANS: The cross is go/go +/+ female × +/Y bw/bw male → F1: go/+ bw/+ females (wild-type eyes and body) and go/Y bw/+ males (golden body, wild-type eyes). An intercross of the F1 offspring yields the following F2 phenotypes in both sexes.

Body

Eyes

Genotype

Proportion

Golden

Brown

go/go or Y

bw/bw

(1/2) × (1/4) = 1/8

Golden

Wildtype

go/go or Y

+/bw or +

(1/2) × (3/4) = 3/8

Wildtype

Brown

+/go or Y

bw/bw

(1/2) × (1/4) = 1/8

Wildtype

Wildtype

+/go or Y

+/bw or +

(1/2) × (3/4) = 3/8

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14-WC Answers to All Questions and Problems 5.5 What are the sexual phenotypes of the following genotypes in Drosophila: XX, XY, XXY, XXX, XO?

Color blindness is determined by an X-linked gene, and blood type is determined by an autosomal gene.

ANS: XX is female, XY is male, XXY is female, XXX is female (but barely viable), XO is male (but sterile).

(a) What are the genotypes of the man and the woman?

5.6 In human beings, a recessive X-linked mutation, g, causes green-defective color vision; the wild-type allele, G, causes normal color vision. A man (a) and a woman (b), both with normal vision, have three children, all married to people with normal vision: a color-defective son (c), who has a daughter with normal vision (f); a daughter with normal vision (d), who has one color-defective son (g) and two normal sons (h); and a daughter with normal vision (e), who has six normal sons (i). Give the most likely genotypes for the individuals (a–i) in this family. (a)

(b)

G/Y

G/g

(c)

(d)

(e)

g/Y

g/G

G/G

(f) G/g

g/Y

(i) 6

(h)

(g) G/Y

G/Y

ANS: (a) XGY; (b) XGXg; (c) XgY; (d) XGXg; (e) XGXG; (f) XGXg; (g) XgY; (h) XGY; (i) XGY 5.7 If both father and son have defective color vision, is it likely that the son inherited the trait from his father?

ANS: No. Defective color vision is caused by an X-linked mutation. The son’s X chromosome came from his mother, not his father.

5.8 A normal woman, whose father had hemophilia, marries a normal man. What is the chance that their first child will have hemophilia?

ANS: The risk for the child is P(woman transmits mutant allele) × P(child is male) = (1/2) × (1/2) = 1/4.

5.9 A man with X-linked color blindness marries a woman with no history of color blindness in her family. The daughter of this couple marries a normal man, and their daughter also marries a normal man. What is the chance that this last couple will have a child with color blindness?

If this couple has already had a child with color blindness, what is the chance that their next child will be color blind?

ANS: The risk for the child is P(mother is C/c) × P(mother transmits c) × P(child is male) = (1/2) × (1/2) × (1/2) = 1/8; if the couple has already had a child with color blindness, P(mother is C/c) = 1, and the risk for each subsequent child is 1/4.

5.10 A man who has color blindness and type O blood has children with a woman who has normal color vision and type AB blood. The woman’s father had color blindness.

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(b) What proportion of their children will have color blindness and type B blood? (c) What proportion of their children will have color blindness and type A blood? (d) What proportion of their children will be color blind and have type AB blood?

ANS: (a) The man is X c Y ii; the woman is X+ Xc IA IB. (b) Probability color blind = 1/2; probability type B blood = 1/2; combined probability = (1/2) × (1/2) = 1/4. (c) Probability color blind = 1/2; probability type A blood = 1/2; combined probability (1/2) × (1/2) = 1/4. (d) 0.

5.11 A Drosophila female homozygous for a recessive X-linked mutation that causes vermilion eyes is mated to a wildtype male with red eyes. Among their progeny, all the sons have vermilion eyes, and nearly all the daughters have red eyes; however, a few daughters have vermilion eyes. Explain the origin of these vermilion-eyed daughters.

ANS: Each of the rare vermilion daughters must have resulted from the union of an X(v) X(v) egg with a Y-bearing sperm. The diplo-X eggs must have originated through nondisjunction of the X chromosomes during oogenesis in the mother. However, we cannot determine if the nondisjunction occurred in the first or the second meiotic division.

5.12 In Drosophila, vermilion eye color is due to a recessive allele (v) located on the X chromosome. Curved wings are due to a recessive allele (cu) located on one autosome, and ebony body is due to a recessive allele (e) located on another autosome. A vermilion male is mated to a curved, ebony female, and the F1 males are phenotypically wildtype. If these males were backcrossed to curved, ebony females, what proportion of the F2 offspring will be wildtype males?

ANS: P(male) = 1/2; P(male transmits first wild-type autosome) = 1/2; P(male transmits other wild-type autosome) = 1/2; therefore, combined proportion, P(wild-type male) = 1/8 5.13 A Drosophila female heterozygous for the recessive X-linked mutation w (for white eyes) and its wild-type allele w 1 is mated to a wild-type male with red eyes.

Among the sons, half have white eyes and half have red eyes. Among the daughters, nearly all have red eyes; however, a few have white eyes. Explain the origin of these white-eyed daughters.

ANS: Each of the rare white-eyed daughters must have resulted from the union of an X(w) X(w) egg with a Y-bearing sperm. The rare diplo-X eggs must have originated through nondisjunction of the X chromosomes during the second meiotic division in the mother.

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Answers to All Questions and Problems WC-15

5.14 In Drosophila, a recessive mutation called chocolate (c) causes the eyes to be darkly pigmented. The mutant phenotype is indistinguishable from that of an autosomal recessive mutation called brown (bw). A cross of chocolate-eyed females to homozygous brown males yielded wild-type F1 females and darkly pigmented F1 males. If the F1 flies are intercrossed, what types of progeny are expected, and in what proportions? (Assume that the double mutant combination has the same phenotype as either of the single mutants alone.)

ANS: 3/8 wild-type (red), 5/8 brown for both male and female F2 progeny.

5.15 Suppose that a mutation occurred in the SRY gene on the human Y chromosome, knocking out its ability to produce the testis-determining factor. Predict the phenotype of an individual who carried this mutation and a normal X chromosome.

ANS: Female.

5.16 A woman carries the androgen-insensitivity mutation (ar) on one of her X chromosomes; the other X carries the wild-type allele (AR). If the woman marries a normal man, what fraction of her children will be phenotypically female? Of these, what fraction will be fertile?

ANS: Three-fourths will be phenotypically female (genotypically ar/AR, AR/AR, or ar/Y). Among the females, 2/3 (ar/AR and AR/AR) will be fertile; the ar/Y females will be sterile.

5.17 Would a human with two X chromosomes and a Y chromosome be male or female?

ANS: Male.

5.18 In Drosophila, the gene for bobbed bristles (recessive allele bb, bobbed bristles; wild-type allele+, normal bristles) is located on the X chromosome and on a homologous segment of the Y chromosome. Give the genotypes and phenotypes of the offspring from the following crosses:

(c) 2X 3A; (d) 1X 3A; (e) 2X 2A; (f) 1X 2A. ANS: (a) Female; (b) intersex; (c) intersex; (d) male: (e) female; (f) male.

5.20 In chickens, the absence of barred feathers is due to a recessive allele. A barred rooster was mated with a nonbarred hen, and all the offspring were barred. These F1 chickens were intercrossed to produce F2 progeny, among which all the males were barred; half the females were barred and half were nonbarred. Are these results consistent with the hypothesis that the gene for barred feathers is located on one of the sex chromosomes?

ANS: Yes. The gene for feather patterning is on the Z chromosome. If we denote the allele for barred feathers as B and the allele for nonbarred feathers as b, the crosses are as follows: B/B (barred) male × b/W (nonbarred) female → F1: B/b (barred) males and B/W (barred) females. Intercrossing the F1 produces B/B (barred) males, B/b (barred) males, B/W (barred) females, and b/W (nonbarred) females, all in equal proportions.

5.21 A Drosophila male carrying a recessive X-linked mutation for yellow body is mated to a homozygous wild-type female with gray body. The daughters of this mating all have uniformly gray bodies. Why are not their bodies a mosaic of yellow and gray patches?

ANS: Drosophila does not achieve dosage compensation by inactivating one of the X chromosomes in females.

5.22 What is the maximum number of Barr bodies in the nuclei of human cells with the following chromosome compositions: (a) XY; (b) XX;

+

(a) X X × X Y ;

(c) XXY;

(b) Xbb Xbb × Xbb Y+;

(d) XXX;

(c) X X × X Y ;

(e) XXXX;

(d) X+ Xbb × Xbb Y+.

(f) XYY?

bb

+

bb

bb

bb

+

bb

ANS: (a) 1/2 Xbb Xbb bobbed females, 1/2 Xbb Y+ wild-type males; (b) 1/2 X+ Xbb wild-type females, 1/2 Xbb Ybb bobbed males; (c) 1/4 X+ X+ wild-type females, 1/4 X+ Xbb wild-type females, 1/4 X+ Ybb wild-type males, 1/4 Xbb Ybb bobbed males; (d) 1/4 X+ Xbb wild-type females, 1/4 Xbb Xbb bobbed females, 1/4 X+ Y+ wild-type males, 1/4 Xbb Y+ wild-type males.

5.19 Predict the sex of Drosophila with the following chromosome compositions (A = haploid set of autosomes): (a) 4X 4A; (b) 3X 4A;

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 15

ANS: a) Zero; (b) one; (c) one; (d) two; (e) three; (f) zero.

5.23 Males in a certain species of deer have two nonhomologous X chromosomes, denoted X1 and X2, and a Y chromosome. Each X chromosome is about half as large as the Y chromosome, and its centromere is located near one of the ends; the centromere of the Y chromosome is located in the middle. Females in this species have two copies of each of the X chromosomes and lack a Y chromosome. How would you predict the X and Y chromosomes to pair and disjoin during spermatogenesis to produce equal numbers of male- and female-determining sperm?

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16-WC Answers to All Questions and Problems ANS:

Cross 2 Metaphase

Anaphase

P F1

X2

Since the centromere is at the end of each small X chromosome but in the middle of the larger Y, both X1 and X2 pair at the centromere of the Y chromosome during metaphase so that the two X chromosomes disjoin together and segregate from the Y chromosome during anaphase.

5.24 A breeder of sun conures (a type of bird) has obtained two true-breeding strains, A and B, which have red eyes instead of the normal brown found in natural populations. In Cross 1, a male from strain A was mated to a female from strain B, and the male and female offspring all had brown eyes. In Cross 2, a female from strain A was mated to a male from strain B, and the male offspring had brown eyes and the female offspring had red eyes.

When the F1 birds from each cross were mated brother to sister, the breeder obtained the following results: Phenotype

Proportion in F2 of Cross 1

Proportion in F2 of Cross 2

Brown male

6/16

3/16

Red male

3/16

5/16

Brown female

3/16

3/16

Red female

5/16

5/16

Provide a genetic explanation for these results.

ANS: Color is determined by an autosomal gene (alleles A and a) and a sex-linked gene (alleles B and b) on the Z chromosome (females are ZW and males are ZZ) and the recessive alleles are mutually epistatic—that is, aa, bb, or bW birds have red eyes, and A- B- or A- BW birds have brown eyes.

Cross 1 P

Strain A red male

×

aa BB F1 F2

×

aa BW

Y

X1

Strain A red female

Aa Bb

Strain B red female AA bW

×

Aa BW

Brown males

Brown females

A- Bb

A- BW

Brown males (6/16)

Brown females (3/16)

aa Bb

A- bW

Red males (2/16)

Red females (4/16) aa BW Red females (1/16)

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F2

Aa Bb

Strain B red male AA bb

×

Aa bW

Brown males

Red females

A- Bb

A- bW

Brown males (3/16)

Brown females (3/16)

A- bb

A- bW

Red males (3/16)

Red females (3/16)

aa b-

aa -W

Red males 2/16

Red females (2/16)

5.25 In 1908, F. M. Durham and D. C. E. Marryat reported the results of breeding experiments with canaries. Cinnamon canaries have pink eyes when they first hatch, whereas green canaries have black eyes. Durham and Marryat crossed cinnamon females with green males and observed that all the F1 progeny had black eyes, just like those of the green strain. When the F1 males were crossed to green females, all the male progeny had black eyes, whereas all the female progeny had either black or pink eyes, in about equal proportions.

When the F1 males were crossed to cinnamon females, four classes of progeny were obtained: females with black eyes, females with pink eyes, males with black eyes, and males with pink eyes—all in approximately equal proportions. Propose an explanation for these findings.

ANS: Eye color in canaries is due to a gene on the Z chromosome, which is present in two copies in males and one copy in females. The allele for pink color at hatching (p) is recessive to the allele for black color at hatching (P). There is no eye color gene on the other sex chromosome (W), which is present in one copy in females and absent in males. The parental birds were genotypically p/W (cinnamon females) and P/P (green males).

Their F1 sons were genotypically p/P (with black eyes at hatching). When these sons were crossed to green females (genotype P/W), they produced F2 progeny that sorted into three categories: males with black eyes at hatching (P/-, half the total progeny), females with black eyes at hatching (P/W, a fourth of the total progeny), and females with pink eyes at hatching (p/W, a fourth of the total progeny). When these sons were crossed to cinnamon females (genotype p/W), they produced F2 progeny that sorted into four equally frequent categories: males with black eyes at hatching (genotype P/p), males with pink eyes at hatching (genotype p/p), females with black eyes at hatching (genotype P/W), and females with pink eyes at hatching (genotype p/W).

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Answers to All Questions and Problems WC-17

Chapter 6 6.1 In the human karyotype, the X chromosome is approximately the same size as seven of the autosomes (the so-called C group of chromosomes). What procedure could be used to distinguish the X chromosome from the other members of this group?

ANS: Use one of the banding techniques.

6.2 In humans, a cytologically abnormal chromosome 22, called the “Philadelphia” chromosome because of the city in which it was discovered, is associated with chronic leukemia. This chromosome is missing part of its long arm. How would you denote the karyotype of an ­individual who had 46 chromosomes in his somatic cells, including one normal 22 and one Philadelphia chromosome?

ANS: 46, XX, del(22)(q) or 46, XY, del(22)(q), depending on the sex chromosome constitution.

6.3 During meiosis, why do some tetraploids behave more regularly than triploids?

ANS: In allotetraploids, each member of the different sets of chromosomes can pair with a homologous partner during prophase I and then disjoin during anaphase I. In triploids, disjunction is irregular because homol­o­ gous chromosomes associate during prophase I either by forming bivalents and univalents or by forming trivalents.

6.4 The following table presents chromosome data on four species of plants and their F1 hybrids: Meiosis I Metaphase Root Tip Chromosome Number

Number of Bivalents

Number of Univalents

A

20

10

0

B

20

10

0

C

10

5

0

D

10

5

0

A×B

20

0

20

A×C

15

5

5

A×D

15

5

5

C×D

10

0

10

Species or F1 Hybrid

(a) Deduce the chromosomal origin of species A. (b) How many bivalents and univalents would you expect to observe at meiotic metaphase I in a hybrid between species C and species B? (c) How many bivalents and univalents would you expect to observe at meiotic metaphase I in a hybrid between species D and species B?

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ANS: (a) Species A is an allotetraploid with a genome from each of species C and species D; (b) 0 bivalents and 15 univalents; (c) 0 bivalents and 15 univalents.

6.5 A plant species A, which has seven chromosomes in its gametes, was crossed with a related species B, which has nine. The hybrids were sterile, and microscopic observation of their pollen mother cells showed no chromosome pairing. A section from one of the hybrids that grew vigorously was propagated vegetatively, producing a plant with 32 chromosomes in its somatic cells. This plant was fertile. Explain.

ANS: The fertile plant is an allotetraploid with 7 pairs of chromosomes from species A and 9 pairs of chromosomes from species B; the total number of chromosomes is (2 × 7) + (2 × 9) = 32.

6.6 A plant species X with n = 5 was crossed with a related species Y with n = 7. The F1 hybrid produced only a few pollen grains, which were used to fertilize the ovules of species Y. A few plants were produced from this cross, and all had 19 chromosomes. Following self-fertilization, the F1 hybrids produced a few F2 plants, each with 24 chromosomes. These plants were phenotypically different from either of the original species and were highly fertile. Explain the sequence of events that produced these fertile F2 hybrids.

ANS: The F1 hybrid had 5 chromosomes from species X and 7 chromosomes from species Y, for a total of 12. When this hybrid was backcrossed to species Y, the few progeny that were produced had 5 + 7 = 12 chromosomes from the hybrid and 7 from species Y, for a total of 19. This hybrid was therefore a triploid. Upon self-fertilization, a few F2 plants were formed, each with 24 chromosomes.

Presumably the chromosomes in these plants consisted of 2 × 5 = 10 from species X and 2 × 7 = 14 from species Y. These vigorous and fertile F2 plants were therefore allotetraploids.

6.7 Identify the sexual phenotypes of the following genotypes in human beings: XX, XY, XO, XXX, XXY, XYY. ANS: XX is female, XY is male, XO is female (but sterile), XXX is female, XXY is male (but sterile), and XYY is male.

6.8 If nondisjunction of chromosome 21 occurs in the division of a secondary oocyte in a human female, what is the chance that a mature egg derived from this division will receive two number 21 chromosomes?

ANS: 1/2 6.9 A Drosophila female homozygous for a recessive X-linked mutation causing yellow body was crossed to a wild-type male. Among the progeny, one fly had sectors of yellow pigment in an otherwise gray body. These yellow sectors were distinctly male, whereas the gray areas were female.

Explain the peculiar phenotype of this fly.

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18-WC Answers to All Questions and Problems ANS: The fly is a gynandromorph, that is, a sexual mosaic. The yellow tissue is X(y)/O and the gray tissue is X(y)/X(+).

This mosaicism must have arisen through loss of the X chromosome that carried the wild-type allele, pre­ sumably during one of the early embryonic cleavage divisions.

chromosomes from her heterozygous mother through nondisjunction during the second meiotic division in the mother’s germline. Furthermore, because the daughter did not receive a sex chromosome from her father, sex chromosome nondisjunction must have occurred during meiosis in his germline too.

6.10 The Drosophila fourth chromosome is so small that flies monosomic or trisomic for it survive and are fertile. Several genes, including eyeless (ey), have been located on this chromosome. If a cytologically normal fly homozygous for a recessive eyeless mutation is crossed to a fly monosomic for a wild-type fourth chromosome, what kinds of progeny will be produced, and in what proportions?

6.13 Although XYY men are phenotypically normal, would they be expected to produce more children with sex chromosome abnormalities than XY men? Explain.

ANS: Approximately half the progeny should be disomic ey/+ and half should be monosomic ey/O. The disomic progeny will be wild-type, and the monosomic progeny will be eyeless.

6.11 A woman with X-linked color blindness and Turner syndrome had a color-blind father and a normal mother. In which of her parents did nondisjunction of the sex chromosomes occur?

ANS: Nondisjunction must have occurred in the mother. The color blind woman with Turner syndrome was produced by the union of an X-bearing sperm, which carried the mutant allele for color blindness, and a nullo-X egg.

6.12 In humans, Hunter syndrome is known to be an X-linked trait with complete penetrance. In family A, two phenotypically normal parents have produced a normal son, a daughter with Hunter and Turner syndromes, and a son with Hunter syndrome. In family B, two phenotypically normal parents have produced two phenotypically normal daughters and a son with Hunter and Klinefelter syndromes. In family C, two phenotypically normal parents have produced a phenotypically normal daughter, a daughter with Hunter syndrome, and a son with Hunter syndrome. For each family, explain the origin of the child indicated in italics.

ANS: The daughter with Turner and Hunter syndromes in family A must have received her single X chromosome from her mother, who is heterozygous for the mutation causing Hunter syndrome. The daughter did not receive a sex chromosome from her father because sex chromosome nondisjunction must have occurred during meiosis in his germline. The son with Klinefelter syndrome in family B is karyotypically XXY, and both of his X chromosomes carry the mutant allele for Hunter syndrome.

This individual must have received two mutant X chromosomes from his heterozygous mother due to X chromosome nondisjunction during the second meiotic division in her germline. The daughter with Hunter syndrome in family C is karyotypically XX, and both of her X chromosomes carry the mutant allele for Hunter syndrome. This individual received the two mutant X

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ANS: XYY men would produce more children with sex chromosome abnormalities because their three sex chromosomes will disjoin irregularly during meiosis. This irregular disjunction will produce a variety of aneuploid gametes, including the XY, YY, XYY, and nullo sex chromosome constitutions.

6.14 In a Drosophila salivary chromosome, the bands have a sequence of 1 2 3 4 5 6 7 8. The homologue with which this chromosome is synapsed has a sequence of 1 2 3 6 5 4 7 8. What kind of chromosome change has occurred?

Draw the synapsed chromosomes.

ANS: The animal is heterozygous for an inversion: 5 5 6

6

4

4

1

2

3

7

8

1

2

3

7

8

6.15 Other chromosomes have sequences as follows: (a) 1 2 5 6 7 8; (b) 1 2 3 4 4 5 6 7 8; (c) 1 2 3 4 5 8 7 6. What kind of chromosome change is present in each? Illustrate how these chromosomes would pair with a chromosome whose sequence is 1 2 3 4 5 6 7 8.

ANS: (a) Deletion: 3 1

2

1

2

4

3

4

5

6

7

8

5

6

7

8

(b) Duplication: 4 1

2

3

4

1

2

3

4

4

5

6

7

8

5

6

7

8

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Answers to All Questions and Problems WC-19

(c) A terminal inversion:

not between a mutant chromosome and a wild-type chromosome, that is, T(bw; +) or T(+; st). If it were, the progeny would be either brown or scarlet, not either wild-type or white.

7 6 1

2

3

4

5

1

2

3

4

5

7 6

8

8

6.16 In plants translocation heterozygotes display about 50 percent pollen abortion. Why? ANS: In translocation heterozygotes, only alternate segregation leads to euploid gametes, and the frequency of alternate segregation is typically around 5 percent.

6.17 One chromosome in a plant has the sequence A B C D E F, and another has the sequence M N O P Q R. A reciprocal translocation between these chromosomes produced the following arrangement: A B C P Q R on one chromosome and M N O D E F on the other. Illustrate how these translocated chromosomes would pair with their normal counterparts in a heterozygous individual during meiosis.

ANS:

A

B

C

A

B

C

F

F

E

E

D

D

P

P

Q

Q

R

R

O

N

M

O

N

M

6.18 In Drosophila, the genes bw and st are located on chromosomes 2 and 3, respectively. Flies homozygous for bw mutations have brown eyes, flies homozygous for st mutations have scarlet eyes, and flies homozygous for bw and st mutations have white eyes. Doubly heterozygous males were mated individually to homozygous bw; st females. All but one of the matings produced four classes of progeny: wild-type, and brown-, scarlet- and white-eyed. The single exception produced only wildtype and white-eyed progeny. Explain the nature of this exception.

ANS: The exceptional male, whose genotype is bw/+ st/+, is heterozygous for a translocation between chromosomes 2 and 3. It is not possible to determine whether the translocation is between the two mutant chromosomes or between the two wild-type chromosomes, that is, whether it is T(bw; st) or T(+; +); however, it clearly is

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6.19 A phenotypically normal boy has 45 chromosomes, but his sister, who has Down syndrome, has 46. Suggest an explanation for this paradox.

ANS: The boy carries a translocation between chromosome 21 and another chromosome, say chromosome 14. He also carries a normal chromosome 21 and a normal chromosome 14. The boy’s sister carries the translocation, one normal chromosome 14, and two normal copies of chromosome 21.

6.20 Distinguish between a compound chromosome and a Robertsonian translocation.

ANS: A compound chromosome is composed of segments from the same pair of chromosomes, as when two X chromosomes become attached to each other. A Robertsonian translocation involves a fusion of segments from two different pairs of chromosomes. These segments fuse at or near the centromeres, usually with the loss of the short arms of each of the par­ tic­ i­ pat­ ing chromosomes.

6.21 A yellow-bodied Drosophila female with attached-X chromosomes was crossed to a white-eyed male. Both of the parental phenotypes are caused by X-linked recessive mutations. Predict the phenotypes of the progeny.

ANS: All the daughters will be yellow-bodied and all the sons will be white-eyed.

6.22 A man has attached chromosomes 21. If his wife is cytologically normal, what is the chance their first child will have Down syndrome?

ANS: Zygotes produced by this couple will be either trisomic or monosomic for chromosome 21. Thus, 100 percent of their viable children will develop Down syndrome.

6.23 Analysis of the polytene chromosomes of three populations of Drosophila has revealed three different banding sequences in a region of the second chromosome: Population

Banding Sequence

P1

1 2 3 4 5 6 7 8 9 10

P2

1 2 3 9 8 7 6 5 4 10

P3

1 2 3 9 8 5 6 7 4 10

Explain the evolutionary relationships among these populations.

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20-WC Answers to All Questions and Problems ANS: The three populations are related by a series of inversions: P1

1 2 3 4 5 6 7 8 9 10

P2

1 2 3 9 8 7 6 5 4 10

P3

1 2 3 9 8 5 6 7 4 10

6.24 Each of six populations of Drosophila in different geographic regions had a specific arrangement of bands in one of the large autosomes: (a) 12345678 (b) 12263478 (c) 15432678 (d) 14322678 (e) 16223478 (f) 154322678 Assume that arrangement (a) is the original one. In what order did the other arrangements most likely arise, and what type of chromosomal aberration is responsible for each change?

ANS: Arrangement (a) produced (c) by inversion of segment 2345; (c) produced (f) by a duplication of band 2; (f) produced (d) by a deletion of band 5; (d) produced (e) by inversion of segment 43226; (e) produced (b) by inversion of segment 622.

6.25 The following diagram shows two pairs of chromosomes in the karyotypes of a man, a woman, and their child.

The man and the woman are phenotypically normal, but the child (a boy) suffers from a syndrome of abnormalities, including poor motor control and severe mental impairment. What is the genetic basis of the child’s abnormal phenotype? Is the child hyperploid or hypoploid for a segment in one of his chromosomes?

Mother

Father

Child

ANS: The mother is heterozygous for a reciprocal translocation between the long arms of the large and small chromosomes; a piece from the long arm of the large chromosome has been broken off and attached to the long arm of the short chromosome. The child has inherited the rearranged large chromosome and the normal small chromosome from the mother. Thus, because the rearranged large chromosome is deficient for some of its genes, the child is hypoploid.

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6.26 A male mouse that is heterozygous for a reciprocal translocation between the X chromosome and an autosome is crossed to a female mouse with a normal karyotype. The autosome involved in the translocation carries a gene responsible for coloration of the fur. The allele on the male’s translocated autosome is wild-type, and the allele on its nontranslocated autosome is mutant; however, because the wild-type allele is dominant to the mutant allele, the male’s fur is wild-type (dark in color). The female mouse has light color in her fur because she is homozygous for the mutant allele of the color-determining gene. When the offspring of the cross are examined, all the males have light fur and all the females have patches of light and dark fur. Explain these peculiar results.

ANS: The phenotype in the female offspring is mosaic because one of the X chromosomes is inactivated in each of their cells. If the translocated X is inactivated, the autosome attached to it could also be partially inactivated by a spreading of the inactivation process across the translocation breakpoint. This spreading could therefore inactivate the color-determining gene on the translocated autosome and cause patches of tissue to be phenotypically mutant.

6.27 In Drosophila, the autosomal genes cinnabar (cn) and brown (bw) control the production of brown and red eye pigments, respectively. Flies homozygous for cinnabar mutations have bright red eyes, flies homozygous for brown mutations have brown eyes, and flies homozygous for mutations in both of these genes have white eyes. A male homozygous for mutations in the cn and bw genes has bright red eyes because a small duplication that carries the wild-type allele of bw (bw+) is attached to the Y chromosome. If this male is mated to a karyotypically normal female that is homozygous for the cn and bw mutations, what types of progeny will be produced?

ANS: The sons will have bright red eyes because they will inherit the Y chromosome with the bw+ allele from their father. The daughters will have white eyes because they will inherit an X chromosome from their father.

6.28 In Drosophila, vestigial wing (vg), hairy body (h), and eyeless (ey) are recessive mutations on chromosomes 2, 3, and 4, respectively. Wild-type males that had been irradiated with X rays were crossed to triply homozygous recessive females. The F1 males (all phenotypically wildtype) were then testcrossed to triply homozygous recessive females. Most of the F1 males produced eight classes of progeny in approximately equal proportions, as would be expected if the vg, h, and ey genes assort independently. However, one F1 male produced only four classes of offspring, each approximately one-fourth of the total: (1) wild-type, (2) eyeless, (3) vestigial, hairy, and (4) vestigial, hairy, eyeless. What kind of chromosome aberration did the exceptional F1 male carry, and which chromosomes were involved?

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Answers to All Questions and Problems WC-21

ANS: A reciprocal translocation between chromosomes 2 and 3. One translocated chromosome carries the wildtype alleles of vg and h on chromosomes 2 and 3, respectively, and the other carries the recessive mutant alleles of these genes. The chromosome that carries the ey gene (chromosome 4) is not involved in the rearrangement.

6.29 Cytological examination of the sex chromosomes in a man has revealed that he carries an insertional translocation. A small segment has been deleted from the Y chromosome and inserted into the short arm of the X chromosome; this segment contains the gene responsible for male differentiation (SRY). If this man marries a karyotypically normal woman, what types of progeny will the couple produce?

ANS: XX zygotes will develop into males because one of their X chromosomes carries the SRY gene that was translocated from the Y chromosome. XY zygotes will develop into females because their Y chromosome has lost the SRY gene.

Chapter 7 7.1 Mendel did not know of the existence of chromosomes.

Had he known, what change might he have made in his Principle of Independent Assortment?

ANS: If Mendel had known of the existence of chromosomes, he would have realized that the number of factors ­determining traits exceeds the number of chromosomes, and he would have concluded that some factors must be linked on the same chromosome. Thus, Mendel would have revised the Principle of Independent Assortment to say that factors on different chromosomes (or far apart on the same chromosome) are inherited independently.

7.2 From a cross between individuals with the genotypes Cc Dd Ee × cc dd ee, 1000 offspring were produced. The class that was C- D- ee included 351 individuals. Are the genes c, d, and e on the same or different chromosomes? Explain.

ANS: The class represented by 351 offspring indicates that at least two of the three genes are linked.

7.3 If a is linked to b, and b to c, and c to d, does it follow that a recombination experiment would detect linkage between a and d? Explain.

ANS: No. The genes a and d could be very far apart on the same chromosome—so far apart that they recombine freely, that is, 50 percent of the time.

7.4 Mice have 19 autosomes in their genome, each about the same size. If two autosomal genes are chosen randomly, what is the chance that they will be on the same chromosome?

ANS: 1/19 7.5 Genes on different chromosomes recombine with a frequency of 50 percent. Is it possible for two genes on the same chromosome to recombine with this frequency?

ANS: Yes, if they are very far apart.

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7.6 If two loci are 10 cM apart, what proportion of the cells in prophase of the first meiotic division will contain a single crossover in the region between them?

ANS: 20% 7.7 Genes a and b are 20 cM apart. An a+ b+/a+ b+ individual was mated with an a b/a b individual.

(a) Diagram the cross and show the gametes produced by each parent and the genotype of the F1.

(b) What gametes can the F1 produce, and in what proportions?

(c) If the F1 was crossed to a b/a b individuals, what offspring would be expected, and in what proportions?

(d) Is this an example of the coupling or repulsion linkage phase?

(e) If the F1 were intercrossed, what offspring would be expected, and in what proportions?

ANS: (a) Cross: a+ b+/a+ b+ × a b/a b. Gametes: a+ b+ from one parent, a b from the other. F1: a+ b+/a b (b) 40% a+ b+, 40% a b, 10% a+ b, 10% a b+ (c) F2 from testcross: 40% a+ b+/a b, 40% a b/a b, 10% a+ b/a b, 10% a b+/a b (d) Coupling linkage phase (e) F2 from intercross:

+

+

40% a b

Eggs

40% a b 10% a+ b +

10% a b

40% a+ b+

Sperm 40% a b 10% a+ b

10% a b+

16% + + + + a b /a b 16% a b/a+ b+ 4% a+ b/a+ b+

16% + + a b /a b 16% a b/a b 4% a+ b/a b

4% + + + a b /a b 4% a b/a+ b 1% a+ b/a+ b

4% + + + a b /a b 4% a b/a b+ 1% a+ b/a b+

4% a b+/a+ b+

4% a b+/a b

1% a b+/a+ b

1% a b+/a b+

Summary of phenotypes: a+ and b+

66%

a+ and b

9%

a and b+

9%

a and b

16%

7.8 Answer questions (a)–(e) in the preceding problem under the assumption that the original cross was a+ b/a+ b × a b+/a b+.

ANS: (a) Cross: a+ b/a+ b × a b+/a b+ Gametes: a+ b from one parent, a b+ from the other F1: a+ b/a b+ (b) 40% a+ b, 40% a b+, 10% a+ b+, 10% a b

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22-WC Answers to All Questions and Problems (c) F2 from testcross: 40% a+ b/a b, 40% a b+/a b, 10% a+ b+/a b, 10% a b/a b

b vg female b+ vg+

(d) Repulsion linkage phase (e) F2 from intercross: 40% a b

Sperm 40% a b+ 10% a+ b+

10% a b

16% a+ b/a+ b 16% + + a b /a b 4% a+ b+/a+ b 4%

16% + + a b/a b 16% + + a b /a b 4% a+ b+/a b+ 4%

4% + a b/a b 4% + a b /a b 1% a+ b+/a b 1%

+

40% a+ b

Eggs

40% a b+ + + 10% a b

10% a b

a b/a+ b

a b/a b+

4% + + + a b/a b 4% + + + a b /a b 1% a+ b+/a+ b+ 1% a b/a+ b+

a b/a b

Summary of phenotypes: a+ and b+

51%

a+ and b

24%

a and b+

24%

a and b

1%

7.9 If the recombination frequency in the previous two problems were 40 percent instead of 20 percent, what change would occur in the proportions of gametes and testcross progeny?

ANS: Coupling heterozygotes a+ b+/a b would produce the following gametes: 30% a+ b+, 30% a b, 20% a+ b, 20% a b+; repulsion heterozygotes a+ b/a b+ would produce the following gametes: 30% a+ b, 30% a b+, 20% a+ b+, 20% a b. In each case, the frequencies of the testcross progeny would correspond to the frequencies of the gametes.

7.10 A homozygous variety of maize with red leaves and normal seeds was crossed with another homozygous variety with green leaves and tassel seeds. The hybrids were then backcrossed to the green, tassel-seeded variety, and the following offspring were obtained: red, normal 124; red, tassel 126; green, normal 125; green, tassel 123. Are the genes for plant color and seed type linked? Explain.

ANS: No. The leaf color and tassel seed traits assort independently.

7.11 A phenotypically wild-type female fruit fly that was heterozygous for genes controlling body color and wing length was crossed to a homozygous mutant male with black body (allele b) and vestigial wings (allele vg). The cross produced the following progeny: gray body, normal wings 126; gray body, vestigial wings 24; black body, normal wings 26; black body, vestigial wings 124. Do these data indicate linkage between the genes for body color and wing length? What is the frequency of recombination? Diagram the cross, showing the arrangement of the genetic markers on the chromosomes.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 22

ANS: Yes. Recombination frequency = (24 + 26)/(126 + 24 + 26 + 124) = 0.167. Cross: b vg male b vg

×

b vg

b vg

b vg

b vg

b+ vg+ 126

b vg 124

b+ vg 24

b vg+ 26

7.12 Another phenotypically wild-type female fruit fly heterozygous for the two genes mentioned in the previous problem was crossed to a homozygous black, vestigial male. The cross produced the following progeny: gray body, normal wings 23; gray body, vestigial wings 127; black body, normal wings 124; black body, vestigial wings 26. Do these data indicate linkage? What is the frequency of recombination? Diagram the cross, showing the arrangement of the genetic markers on the chromosomes.

ANS: Yes. Recombination frequency = (23 + 26)/(23 + 127 + 124 + 26) = 0.163. Cross: b+ vg female b vg+

b vg male b vg

×

b vg b+ vg+ 23

b vg b+ vg 127

b vg b vg 26

b vg b vg+ 124

7.13 In rabbits, the dominant allele C is required for colored fur; the recessive allele c makes the fur colorless (albino).

In the presence of at least one C allele, another gene determines whether the fur is black (B, dominant) or brown (b, recessive). A homozygous strain of brown rabbits was crossed with a homozygous strain of albinos.

The F1 were then crossed to homozygous double recessive rabbits, yielding the following results: black 34; brown 66; albino 100. Are the genes b and c linked? What is the frequency of recombination? Diagram the crosses, showing the arrangement of the genetic markers on the chromosomes.

ANS: Yes. Recombination frequency is estimated by the frequency of black offspring among the colored offspring: 34/(66 + 34) = 0.34. Cross:

Cb cb brown 66

Cb Cb

×

cB cB

Cb cB

×

cb cb

cB cb albino

cb cb albino 100

CB cb black 34

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Answers to All Questions and Problems WC-23

7.14 In tomatoes, tall vine (D) is dominant over dwarf (d), and spherical fruit shape (P) is dominant over pear shape (p). The genes for vine height and fruit shape are linked with 20 percent recombination between them.

One tall plant (I) with spherical fruit was crossed with a dwarf, pear-fruited plant. The cross produced the following results: tall, spherical 81; dwarf, pear 79; tall, pear 22; dwarf spherical 17. Another tall plant with spherical fruit (II) was crossed with the dwarf, pear-fruited plant, and the following results were obtained: tall, pear 21; dwarf, spherical 18; tall, spherical 5; dwarf, pear 4. Diagram these two crosses, showing the genetic markers on the chromosomes. If the two tall plants with spherical fruit were crossed with each other, that is, I × II, what phenotypic classes would you expect from the cross, and in what proportions?

ANS: Plant I has the genotype D P/d p, and when crossed to a d p/d p plant produces four classes of progeny: DP dp

dp dp

Dp dp

dP dp

81

79

22

17

homozygous for e+ was mated with an ebony male homozygous for sr+. All the offspring were phenotypically wild-type (gray body and unstriped). (a) What kind of gametes will be produced by the F1 females, and in what proportions? (b) What kind of gametes will be produced by the F1 males, and in what proportions? (c) If the F1 females are mated with striped, ebony males, what offspring are expected, and in what proportions? (d) If the F1 males and females are intercrossed, what offspring would you expect from this intercross, and in what proportions?

ANS: (a) The F1 females, which are sr e+/sr+ e, produce four types of gametes: 46% sr e+, 46% sr+ e, 4% sr e, 4% sr+ e+. (b) The F1 males, which have the same genotype as the F1 females, produce two types of gametes: 50% sr e+, 50% sr+ e; remember, there is no crossing over in Drosophila males. (c) 46% striped, gray; 46% unstriped, ebony; 4% striped, ebony; 4% unstriped, gray. (d) The offspring from the intercross can be obtained from the follow­ing table.

Plant II has the genotype D p/d P, and when crossed to a d p/d p plant produces four classes of progeny: Dp dp

dP dp

DP dp

dp dp

21

18

5

4

If the two plants are crossed (D P/d p × D p/d P), the phenotypes of the offspring can be predicted from the following table.

Gametes from plant II

Dp 0.40 dP 0.40 DP 0.10 dp 0.10

Gametes from plant I dp Dp 0.40 0.10

dP 0.10

D p/D P 0.16 d P/D P 0.16 D P/D P 0.04 d p/D P 0.04

D p/d p 0.16 d P/d p 0.16 D P/d p 0.04 d p/d p 0.04

D p/d P 0.04 d P/d P 0.04 D P/d P 0.01 d p/d P 0.01

Summary of phenotypes: Tall, spherical Tall, pear

0.54 0.21

Dwarf, spherical

0.21

Dwarf, pear

0.04

7.15 In Drosophila, the genes sr (stripe thorax) and e (ebony body) are located at 62 and 70 cM, respectively, from the left end of chromosome 3. A striped female

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 23

+

sr e 0.50

0.002

0.002

sr e+ sr e+/sr e+ sr e+/sr+ e 0.46 0.23 0.23 + Eggs sr e sr+ e/sr e+ sr+ e/sr+ e 0.46 0.23 0.23 sr e sr e/sr e+ sr e/sr+ e 0.04 0.002 0.002 + + sr e sr+ e+/sr e+ sr+ e+/sr+ e 0.04

Summary of phenotypes:

Dp 0.40

D p/D p 0.04 d P/D p 0.04 D P/D p 0.01 d p/D p 0.01

Sperm +

sr e 0.50

Striped, gray

0.25

Unstriped, gray

0.50

Striped, ebony

0

Unstriped, ebony

0.25

7.16 In Drosophila, genes a and b are located at positions 22.0 and 42.0 on chromosome 2, and genes c and d are located at positions 10.0 and 25.0 on chromosome 3.

A fly homozygous for the wild-type alleles of these four genes was crossed with a fly homozygous for the recessive alleles, and the F1 daughters were backcrossed to their quadruply recessive fathers. What offspring would you expect from this backcross, and in what proportions?

ANS: Because the two chromosomes assort independently, the genetic makeup of the gametes (and, therefore, of the backcross progeny) can be obtained from the following table.

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24-WC Answers to All Questions and Problems Chromosome 3 in gametes

Chromosome 2 in gametes

+

+

+

+

cd 0.425

c d 0.425

ab 0.40 + + a b 0.40 + a b 0.10

abcd 0.17 + + a b cd 0.17 + a bcd 0.0425

abc d 0.17 + + + + a b c d 0.17 + + + + a b c d 0.0425

abcd abc d 0.03 0.03 + + + + + + a b cd a b c d 0.03 0.03 + + + + a bcd a bc d 0.0075 0.0075

ab 0.10

ab cd 0.0425

ab c d 0.0425

ab cd 0.0075

+

+

cd 0.075

+

+

+

+

+

c d 0.075 +

+

+

+

+

+

ab c d 0.0075

7.17 The Drosophila genes vg (vestigial wings) and cn (cinnabar eyes) are located at 67.0 and 57.0, respectively, on chromosome 2. A female from a homozygous strain of vestigial flies was crossed with a male from a homozygous strain of cinnabar flies. The F1 hybrids were phenotypically wild-type (long wings and dark red eyes).

5

Scarlet, spineless, ebony

0.0516

0.06

6

Wild-type

0.0516

0.06

7

Scarlet

0.0084

0

8

Spineless, ebony

0.0084

0

7.19 In maize, the genes Pl for purple leaves (dominant over Pl for green leaves), sm for salmon silk (recessive to Sm for yellow silk), and py for pigmy plant (recessive to Py for normal-size plant) are on chromosome 6, with map positions as shown:

(a) How many different kinds of gametes could the F1 females produce, and in what proportions?

7.18 In Drosophila, the genes st (scarlet eyes), ss (spineless bristles), and e (ebony body) are located on chromosome 3, with map positions as indicated: st ss e 44 58 70 Each of these mutations is recessive to its wild-type allele (st+, dark red eyes; ss+, smooth bristles; e+, gray body).

Phenotypically wild-type females with the genotype st ss e+/st+ st+ ss+ e were crossed with triply recessive males.

Predict the phenotypes of the progeny and the frequencies with which they will occur assuming (a) no interference and (b) complete interference.

ANS: In the following enumeration, classes 1 and 2 are parental types, classes 3 and 4 result from a single crossover between st and ss, classes 5 and 6 result from a single crossover between ss and e, and classes 7 and 8 result from a double crossover, with one of the exchanges between st and ss and the other between ss and e. (a) Frequency with No Interference

(b) Frequency with Complete Interference

Class

Phenotypes

1

Scarlet, spineless

0.3784

0.37

2

Ebony

0.3784

0.37

3

Scarlet, ebony

0.0616

0.07

4

Spineless

0.0616

0.07

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 24

sm

py

45

55

65

Hybrids from the cross Pl sm py / Pl sm py × pl Sm Py / pl Sm Py were testcrossed with pl sm py / pl sm py plants. Predict the phenotypes of the offspring and their frequencies assuming (a) no interference and (b) complete interference.

(b) If these females are mated with cinnabar, vestigial males, what kinds of progeny would you expect, and in what proportions?

ANS: (a) The F1 females, which are cn vg+/cn+ vg, produce four types of gametes: 45% cn vg+, 45% cn+ vg, 5% cn+ vg+, 5% cn vg. (b) 45% cinnabar eyes, normal wings; 45% reddish-brown eyes, vestigial wings; 5% reddish-brown eyes, normal wings; 5% cinnabar eyes, vestigial wings.

pl

ANS: In the enumeration below, classes 1 and 2 are parental types, classes 3 and 4 result from a single crossover between Pl and Sm, classes 5 and 6 result from a single crossover between Sm and Py, and classes 7 and 8 result from a double crossover, with one of the exchanges between Pl and Sm and the other between Sm and Py. (a) Frequency with No Interference

(b) Frequency with Complete Interference

Class

Phenotypes

1

Purple, salmon, pigmy

0.405

0.40

2

Green, yellow, normal

0.405

0.40

3

Purple, yellow, normal

0.045

0.05

4

Green, salmon, pigmy

0.045

0.05

5

Purple, salmon, normal

0.045

0.05

6

Green, yellow, pigmy

0.045

0.05

7

Purple, yellow, pigmy

0.005

0

8

Green, salmon, normal

0.005

0

7.20 In maize, the genes Tu, j2, and gl3 are located on chromosome 4 at map positions 101, 106, and 112, respectively. If plants homozygous for the recessive alleles of these genes are crossed with plants homozygous for the dominant alleles, and the F1 plants are testcrossed to triply recessive plants, what genotypes would you expect,

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Answers to All Questions and Problems WC-25

and in what proportions? Assume that interference is complete over this map interval.

ANS: In the following enumeration, classes 1 and 2 are parental types, classes 3 and 4 result from crossing over between Tu and j2, and classes 5 and 6 result from crossing over between j2 and Gl3; only the chromosome from the triply heterozygous F1 plant is shown. Because interference is complete, there are no double crossover progeny.

Class

Genotype

Frequency

1

tu j2 gl3

0.445

2

Tu J2 Gl3

0.445

3

tu J2 Gl3

0.025

4

Tu j2 gl3

0.025

5

tu j2 Gl3

0.030

6

Tu J2 gl3

0.030

Phenotype

Number

}

Yellow, bar, vermilion wild-type

546

}

Yellow bar, vermilion

244

}

Yellow, vermilion bar

160

}

Yellow, bar vermilion

50

Determine the order of these three loci on the X chromosome and estimate the distances between them.

7.21 A Drosophila geneticist made a cross between females homozygous for three X-linked recessive mutations (y, yellow body; ec, echinus eye shape; w, white eye color) and wild-type males. He then mated the F1 females to triply mutant males and obtained the following results:

ANS: The last two classes, consisting of yellow, bar flies and vermilion flies, with a total of 50 progeny, result from double crossovers. Thus, the order of the genes is y—v—B, and the F1 females had the genotype y v B/+ + +. The distance between y and v is the average number of crossovers between them: (244 + 50)/1000 = 29.4 cM; likewise, the distance between v and B is (160 + 50)/1000 = 21.0 cM. Thus, the genetic map is y—29.4 cM—v—21.0 cM—B. 7.23 Female Drosophila heterozygous for three recessive mutations e (ebony body), st (scarlet eyes), and ss (spineless bristles) were testcrossed, and the following progeny were obtained:

Females

Males

Number

+ + + / y ec w

+++

475

Phenotype

Number

y ec w / y ec w

y ec w

469

Wild-type

67

y++

8

Ebony

8 68

y + + / y ec w + ec w / y ec w

+ ec w

7

Ebony, scarlet

y + w / y ec w

y+w

18

Ebony, spineless

347

+ ec + / y ec w

+ ec +

23

Ebony, scarlet, spineless

78

+ + w / y ec w

++w

0

Scarlet

368

y ec + / y ec w

y ec +

0

Scarlet, spineless

10

Spineless

54

Determine the order of the three loci y, ec, and w, and estimate the distances between them on the linkage map of the X chromosome.

ANS: The double crossover classes, which are the two that were not observed, establish that the gene order is y—w—ec. Thus, the F1 females had the genotype y w ec/+ + +. The distance between y and w is estimated by the frequency of recombination between these two genes: (8 + 7)/1000 = 0.015; similarly, the distance between w and ec is (18 + 23)/1000 = 0.041. Thus, the genetic map for this segment of the X chromosome is y—1.5 cM—w— 4.1 cM—ec.

7.22 A Drosophila geneticist crossed females homozygous for three X-linked mutations (y, yellow body; B, bar eye shape; v, vermilion eye color) to wild-type males. The F1 females, which had gray bodies and bar eyes with dark red pigment, were then crossed to y B+ v males, yielding the following results:

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 25

(a) What indicates that the genes are linked? (b) What was the genotype of the original heterozygous females? (c) What is the order of the genes? (d) What is the map distance between e and st? (e) What is the map distance between e and ss? (f) What is the coefficient of coincidence? (g) Diagram the crosses in this experiment.

ANS: (a) Two of the classes (the parental types) vastly outnumber the other six classes (recombinant types); (b) st + +/+ ss e; (c) st—ss—e; (d) [(145 + 122) × 1 + (18) × 2]/1000 = 30.3 cM; (e) (122 + 18)/1000 = 14.0 cM; (f) (0.018)/(0.163 × 0.140) = 0.789. (g) st + +/+ ss e females × st ss e/st ss e males → two parental classes and six recombinant classes.

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26-WC Answers to All Questions and Problems 7.24 Consider a female Drosophila with the following X chromosome genotype: w dor+ w+ dor The recessive alleles w and dor cause mutant eye colors (white and deep orange, respectively). However, w is epistatic over dor; that is, the genotypes w dor / Y and w dor/ w dor have white eyes. If there is 40 percent recombination between w and dor, what proportion of the sons from this heterozygous female will show a mutant phenotype? What proportion will have either red or deep orange eyes?

ANS: The female will produce four kinds of gametes: 30% w +, 30% + dor, 20% w dor, and 20% + +; thus, 80% of the progeny will be mutant (either white or deep orange) and 50% will be pigmented (either red or deep orange).

7.25 In Drosophila, the X-linked recessive mutations prune (pn) and garnet (g) recombine with a frequency of 0.4. Both of these mutations cause the eyes to be brown instead of dark red. Females homozygous for the pn mutation were crossed to males hemizygous for the g mutation, and the F1 daughters, all with dark red eyes, were crossed with their brown-eyed brothers. Predict the frequency of sons from this last cross that will have dark red eyes.

ANS: The F1 females are genotypically pn +/+ g. Among their sons, 40 percent will be recombinant for the two X-linked genes, and half of the recombinants will have the wildtype alleles of these genes. Thus, the frequency of sons with dark red eyes will be 1/2 × 40% = 20%.

7.26 Assume that in Drosophila there are three genes x, y, and z, with each mutant allele recessive to the wild-type allele. A cross between females heterozygous for these three loci and wild-type males yielded the following progeny: Females

+++

1010

Males

+++

39

++z

430

+yz

32

x++

27

xy+

441

xyz

31 Total: 2010

Using these data, construct a linkage map of the three genes and calculate the coefficient of coincidence.

ANS: Ignore the female progeny and base the map on the male progeny. The parental types are + + z and x y +. The two missing classes (+ y + and x + z) must represent double crossovers; thus, the gene order is y—x—z. The distance between y and x is (32 + 27)/1000 = 5.9 cM and

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 26

that between x and z is (31 + 39)/1000 = 7.0 cM. Thus, the map is y—5.9 cM—x—7.0 cM—z. The coefficient of coincidence is zero.

7.27 In the nematode Caenorhabditis elegans, the linked genes dpy (dumpy body) and unc (uncoordinated behavior) recombine with a frequency P. If a repulsion heterozygote carrying recessive mutations in these genes is self-fertilized, what fraction of the offspring will be both dumpy and uncoordinated?

ANS: (P/2)2 7.28 In the following testcross, genes a and b are 20 cM apart, and genes b and c are 10 cM apart: a + c / + b + × a b c / a b c. If the coefficient of coincidence is 0.5 over this interval on the linkage map, how many triply homozygous recessive individuals are expected among 1000 progeny?

ANS: 5.

7.29 Drosophila females heterozygous for three recessive mutations, a, b, and c, were crossed to males homozygous for all three mutations. The cross yielded the following results: Phenotype +++ ++c +bc a++ ab+ abc

Number 75 348 96 110 306 65

Construct a linkage map showing the correct order of these genes and estimate the distances between them.

ANS: From the parental classes, + + c and a b +, the heterozygous females must have had the genotype + + c/a b +.

The missing classes, + b + and a + c, which would represent double crossovers, establish that the gene order is b—a—c. The distance between b and a is (96 + 110)/ 1000 = 20.6 cM and that between a and c is (65 + 75)/1000 = 14.0 cM. Thus, the genetic map is b—20.6 cM—a—14.0 cM—c.

7.30 A Drosophila second chromosome that carried a recessive lethal mutation, l(2)g14, was maintained in a stock with a balancer chromosome marked with a dominant mutation for curly wings. This latter mutation, denoted Cy, is also associated with a recessive lethal effect—but this effect is different from that of l(2)g14. Thus, l(2)g14 /Cy flies survive, and they have curly wings. Flies without the Cy mutation have straight wings. A researcher crossed l(2)g14 / Cy females to males that carried second chromosomes with different deletions (all homozygous lethal) balanced over the Cy chromosome (genotype Df /Cy).

Each cross was scored for the presence or absence of progeny with straight wings.

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Answers to All Questions and Problems WC-27 Cross

Location of deletion

12 3 4

5 6 7

Non-Curly progeny

10 8 9

11 12

1

no

2

no

3

yes

4

yes

5

no

In which band is the lethal mutation l(2)g14 located?

ANS: The lethal mutation resides in band 7.

7.31 The following pedigree, described in 1937 by C. L. Birch, shows the inheritance of X-linked color blindness and hemophilia in a family. What is the genotype of II-2?

Do any of her children provide evidence for recombination between the genes for color blindness and hemophilia?

I

Key to phenotypes: 1

2

1

2

3

1

2

3

Normal Color blind

II

Hemophilic

III

ANS: II-1 has the genotype C h/c H, that is, she is a repulsion heterozygote for the alleles for color blindness (c) and hemophilia (h). None of her children are recombinant for these alleles.

7.32 The following pedigree, described in 1938 by B. Rath, shows the inheritance of X-linked color blindness and hemophilia in a family. What are the possible genotypes of II-1? For each possible genotype, evaluate the children of II-1 for evidence of recombination between the color blindness and hemophilia genes.

Key to phenotypes:

I 1

2

Normal Color blind

II 1

2

Hemophilic Color blind and hemophilic

III 1

2

3

4

Color vision uncertain

ANS: II-1 is either (a) C h/c H or (b) c h/C H. Her four sons have the genotypes c h (1), C h (2), c H (3), and C H (4). If II-1 has the genotype C h/c H, sons 1 and 4 are recombinant and sons 2 and 3 are nonrecombinant. If II-1 has the genotype c h/C H, sons 2 and 3 are recombinant and

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 27

sons 1 and 4 are nonrecombinant. Either way, the frequency of recombination is 0.5.

7.33 A normal woman with a color-blind father married a normal man, and their first child, a boy, had hemophilia.

Both color blindness and hemophilia are due to X-linked recessive mutations, and the relevant genes are separated by 10 cM. This couple plans to have a second child. What is the probability that it will have hemophilia? Color blindness? Both hemophilia and color blindness? Neither hemophilia nor color blindness?

ANS: The woman is a repulsion heterozygote for the alleles for color blindness and hemophilia—that is, she is C h/c H. If the woman has a boy, the chance that he will have hemophilia is 0.5 and the chance that he will have color blindness is 0.5. If we specify that the boy have only one of these two conditions, then the chance that he will have color blindness is 0.45. The reason is that the boy will inherit a nonrecombinant X chromosome with a probability of 0.9, and half the nonrecombinant X chromosomes will carry the mutant allele for color blindness and the other half will carry the mutant allele for hemophilia. The chance that the boy will have both conditions is 0.05, and the chance that he will have neither condition is 0.05. The reason is that the boy will inherit a recombinant X chromosome with a probability of 0.1, and half the recombinant X chromosomes will carry both mutant alleles and the other half will carry neither mutant allele.

7.34 Two strains of maize, M1 and M2, are homozygous for four recessive mutations, a, b, c, and d, on one of the large chromosomes in the genome. Strain W1 is homozygous for the dominant alleles of these mutations. Hybrids produced by crossing M1 and W1 yield many different classes of recombinants, whereas hybrids produced by crossing M2 and W1 do not yield any recombinants at all. What is the difference between M1 and M2?

ANS: M2 carries an inversion that suppresses recombination in the chromosome.

7.35 A Drosophila geneticist has identified a strain of flies with a large inversion in the left arm of chromosome 3. This inversion includes two mutations, e (ebony body) and cd (cardinal eyes), and is flanked by two other mutations, sr (stripe thorax) on the right and ro (rough eyes) on the left.

The geneticist wishes to replace the e and cd mutations inside the inversion with their wild-type alleles; he plans to accomplish this by recombining the multiply mutant, inverted chromosome with a wild-type, inversion-free chromosome. What event is the geneticist counting on to achieve his objective? Explain.

ANS: A two-strand double crossover within the inversion; the exchange points of the double crossover must lie between the genetic markers and the inversion breakpoints.

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28-WC Answers to All Questions and Problems Chapter 8 8.1 By what criteria are viruses living? nonliving?

ANS: Viruses reproduce and transmit their genes to progeny viruses. They utilize energy provided by host cells and respond to environmental and cellular signals like other living organisms. However, viruses are obligate parasites; they can reproduce only in appropriate host cells.

8.2 How do bacteriophages differ from other viruses?

ANS: Bacteriophages reproduce in bacteria; other viruses reproduce in protists, plants, and animals.

8.3 In what ways do the life cycles of bacteriophages T4 and l differ? In what aspects are they the same?

ANS: Bacteriophage T4 is a virulent phage. When it infects a host cell, it reproduces and kills the host cell in the process. Bacteriophage lambda can reproduce and kill the host bacterium—the lytic response—just like phage T4, or it can insert its chromosome into the chromosome of the host and remain there in a dormant state—the lysogenic response.

8.4 How does the structure of the l prophage differ from the structure of the l chromosome packaged in the l head?

ANS: The mature (packaged) lambda chromosome and the lambda prophage are circular permutations of one another (see Figure 8.5).

8.5 In what way does the integration of the l chromosome into the host chromosome during a lysogenic infection differ from crossing over between homologous chromosomes?

ANS: The insertion of the phage l chromosome into the host chromosome is a site-specific recombination process catalyzed by an enzyme that recognizes specific sequences in the l and E. coli chromosomes. Crossing over between homologous chromosomes is not sequence specific. It can occur at many sites along the two chromosomes.

8.6 Geneticists have used mutations that cause altered phenotypes such as white eyes in Drosophila, white flowers and wrinkled seeds in peas, and altered coat color in rabbits to determine the locations of genes on the chromosomes of these eukaryotes. What kinds of mutant phenotypes have been used map genes in bacteria?

ANS: Three main types of bacterial mutants have been used to map genes in bacteria; these include mutants unable to utilize specific sugars as energy sources (such as lactose), mutants unable to synthesize essential metabolites (these are called auxotrophs), and mutants resistant to drugs and antibiotics. Wild-type bacteria can use almost any sugar as an energy source, can grow on minimal media, and are killed by antibiotics, whereas mutants in genes controlling these processes result in different growth characteristics. These growth phenotypes can be used to map genes in bacteria.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 28

8.7 You have identified three mutations—a, b, and c—in Streptococcus pneumoniae. All three are recessive to their wild-type alleles a+, b+, and c+. You prepare DNA from a wild-type donor strain and use it to transform a strain with genotype a b c. You observe a+b+ transformants and a+c+ transformants, but no b+c+ transformants. Are these mutations closely linked? If so, what is their order on the Streptococcus chromosome?

ANS: The a, b, and c mutations are closely linked and in the order b—a—c on the chromosome.

8.8 A nutritionally defective E. coli strain grows only on a medium containing thymine, whereas another nutritionally defective strain grows only on a medium containing leucine. When these two strains were grown together, a few progeny were able to grow on a minimal medium with neither thymine nor leucine. How can this result be explained?

ANS: There are two possible explanations. One possibility is that a spontaneous mutation caused reversion of either auxotrophic strain to the prototrophic condition. Because this requires only one mutation in one cell, this is a possibility, although rare. Another, more likely, possibility is that conjugation occurred between the E. coli parental auxotrophic strains. During conjugation, genes from the parental strains recombined. Because each parent had a wild-type gene copy for either thymine or leucine, recombinant progeny containing the wild-type copy of each gene would be able to synthesize both nutrients and grow on minimal medium.

8.9 Assume that you have just demonstrated genetic recombination (e.g., when a strain of genotype a b+ is present with a strain of genotype a+ b, some recombinant genotypes, a+ b+ and a b, are formed) in a previously unstudied species of bacteria. How would you determine whether the observed recombination resulted from transformation, conjugation, or transduction?

ANS: Perform two experiments: (1) determine whether the process is sensitive to DNase and (2) determine whether cell contact is required for the process to take place. The cell contact requirement can be tested by a U-tube experiment (see Figure 8.9). If the process is sensitive to DNase, it is similar to transformation. If cell contact is required, it is similar to conjugation. If it is neither sensitive to DNase nor requires cell contact, it is similar to transduction.

8.10 (a) What are the genotypic differences between F- cells, F+ cells, and Hfr cells? (b) What are the phenotypic differences? (c) By what mechanism are F− cells converted into F+ cells? F+ cells to Hfr cells? Hfr cells to F+ cells?

ANS: (a) F− cells, no F factor present; F+ cells, autonomous F factor; Hfr cells, integrated F factor (see Figure 8.14). (b) F+ and Hfr cells have F pili; F− cells do not. (c) F− cells are converted into F+ cells by the conjugative transfer of F factors from F+ cells. Hfr cells are formed when

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Answers to All Questions and Problems WC-29

F factors in F+ cells become integrated into the chromosomes of these cells. Hfr cells become F+ cells when the integrated F factors exit the chromosome and become autonomous (self-replicating) genetic elements.

8.11 (a) Of what use are F′ factors in genetic analysis? (b) How are F′ factors formed? (c) By what mechanism does sexduction occur?

ANS: (a) F′ factors are useful for genetic analyses where two copies of a gene must be present in the same cell, for example, in determining dominance relationships. (b) F′ factors are formed by abnormal excision of F factors from Hfr chromosomes (see Figure 8.21). (c) By the conjugative transfer of an F′ factor from a donor cell to a recipient (F−) cell.

8.12 What are the basic differences between generalized transduction and specialized transduction?

ANS: Generalized transduction: (1) transducing particles often contain only host DNA; (2) transducing particles may carry any segment of the host chromosome. Thus, all host genes are transduced. Specialized transduction: (1) transducing particles carry a recombinant chromosome, which contains both phage DNA and host DNA; (2) only host genes that are adjacent to the prophage integration site are transduced.

8.13 What roles do IS elements play in the integration of F factors?

ANS: IS elements (or insertion sequences) are short (800– 1400 nucleotide pairs) DNA sequences that are transposable—that is, capable of moving from one position in a chromosome to another position or from one chromosome to another chromosome. IS elements mediate recombination between nonhomologous DNA molecules—for example, between F factors and bacterial chromosomes.

8.14 How can bacterial genes be mapped by interrupted mating experiments?

frequency can be used as an estimate of the linkage distance. Specific cotransduction-linkage functions must be prepared for each phage–host system studied.

8.16 In E. coli, the ability to utilize lactose as a carbon source requires the presence of the enzymes b-galactosidase and b-galactoside permease. These enzymes are encoded by two closely linked genes, lacZ and lacY, respectively.

Another gene, proC, controls, in part, the ability of E. coli cells to synthesize the amino acid proline. The alleles strr and strs, respectively, control resistance and sensitivity to streptomycin. Hfr H is known to transfer the two lac genes, proC, and str, in that order, during conjugation. A cross was made between Hfr H of genotype lacZ− lacY+ proC+ strs and an F− strain of genotype lacZ+ lacY− proC− strr. After about 2 hours, the mixture was diluted and plated out on medium containing streptomycin but no proline. When the resulting proC+ strr recombinant colonies were checked for their ability to grow on medium containing lactose as the sole carbon source, very few of them were capable of fermenting lactose. When the reciprocal cross (Hfr H lacZ+ lacY− proC+ strs X F− lacZ− lacY+ proC− strr) was done, many of the proC+ strr recombinants were able to grow on medium containing lactose as the sole carbon source. What is the order of the lacZ and lacY genes relative to proC? ANS: lacY—lacZ—proC. 8.17 An F+ strain, marked at 10 loci, gives rise spontaneously to Hfr progeny whenever the F factor becomes incorporated into the chromosome of the F+ strain. The F factor can integrate into the circular chromosome at many points, so that the resulting Hfr strains transfer the genetic markers in different orders. For any Hfr strain, the order of markers entering a recipient cell can be determined by interrupted mating experiments. From the following data for several Hfr strains derived from the same F+, determine the order of markers in the F+ strain.

ANS: By interrupting conjugation at various times after the donor and recipient cells are mixed (using a blender or other form of agitation), one can determine the length of time required to transfer a given genetic marker from an Hfr cell to an F−.

8.15 What does the term cotransduction mean? How can cotransduction frequencies be used to map genetic markers?

ANS: Cotransduction refers to the simultaneous transduction of two different genetic markers to a single recipient cell. Since bacteriophage particles can package only 1/100 to 1/50 of the total bacterial chromosome, only markers that are relatively closely linked can be cotransduced. The frequency of cotransduction of any two markers will be an inverse function of the distance between them on the chromosome. As such, this

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 29

Hfr Strain

Markers Donated in order

1

—Z–H–E–R→

2

—O–K–S–R→

3

—K–O–W–I→

4

—Z–T–I–W→

5

—H–Z–T–I→

ANS: T

Z

H

I

E

W

R O

K

S

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30-WC Answers to All Questions and Problems 8.18 The data in the following table were obtained from three-point transduction tests made to determine the order of mutant sites in the A gene encoding the a subunit of tryptophan synthetase in E. coli. Anth is a linked, unselected marker. In each cross, trp+ recombinants were selected and then scored for the anth marker (anth+ or anth-). What is the linear order of anth and the three mutant alleles of the A gene indicated by the data in the table?

Donor Cross Markers

Recipient anth Allele in trp∙ Markers Recombinants

1

anth+— A34

anth−— A223

72 anth+: 332 anth−

18

2

anth+— A46

anth−— A223

196 anth+: 180 anth−

52

3

anth+— A223

anth−— A34

380 anth+: 379 anth−

50

4

anth+— A223

anth−— A46

60 anth+: 280 anth−

20

% anth∙

ANS: anth—A34—A223—A46.

8.19 Bacteriophage P1 mediates generalized transduction in E. coli. A P1 transducing lysate was prepared by growing P1 phage on pur+ pro- his- bacteria. Genes pur, pro, and his encode enzymes required for the synthesis of purines, proline, and histidine, respectively.

The phage and transducing particles in this lysate were then allowed to infect pur− pro+ his+ cells. After incubating the infected bacteria for a period of time sufficient to allow transduction to occur, they were plated on minimal medium supplemented with proline and histidine, but no purines to select for pur+ transductants. The pur+ colonies were then transferred to minimal medium with and without proline and with and without histidine to determine the frequencies of each of the outside markers. Given the following results, what is the order of the three genes on the E. coli chromosome?

crosses as described in Problem 8.18. The results of these crosses are summarized in the following table. What is the linear order of anth and the three mutant sites in the trp A gene?

Donor Cross Markers

Recipient Markers

anth Allele in trp∙ Recombinants

% anth∙

1

anth+— A487

anth−— A223

72 anth+: 332 anth−

82

2

anth+— A58

anth−— A223

196 anth+: 180 anth−

48

3

anth+— A223

anth−— A487

380 anth+: 379 anth−

50

4

anth+— A223

anth−— A58

60 anth+: 280 anth−

80

ANS: anth—A487—A223—A58.

8.21 You have identified a mutant E. coli strain that cannot synthesize histidine (His−). To determine the location of the his− mutation on the E. coli chromosome, you perform interrupted mating experiments with five different Hfr strains. The following chart shows the time of entry (minutes, in parentheses) of the wild-type alleles of the first five markers (mutant genes) into the His− strain.

Hfr A -------- his (1)

man (9) gal (28)

lac (37)

thr (45) arg (49)

Hfr B -------- man (15) his (23)

cys (38)

ser (42)

Hfr C -------- thr (3)

lac (11)

gal (20)

man (39) his (47)

Hfr D -------- cys (3)

his (18)

man (26) gal (45)

Hfr E -------- thr (6)

rha (18) arg (36)

ser (43)

lac (54) cys (47)

On the map below of the circular E. coli chromosome, indicate (1) the relative location of each gene relative to thr (located at 0/100 Min), (2) the position where the sex factor is integrated in each of the five Hfr’s, and (3) the direction of chromosome transfer for each Hfr (indicate direction with an arrow or arrowhead).

Genotype

Number Observed

pro+ his+

100

pro+ his+

22

pro+ his-

150

thr

pro- his-

1

100/0

lac

ANS: pro—pur—his.

8.20 Two additional mutations in the trp A gene of E. coli, trp A58 and trp A487, were ordered relative to trp A223 and the outside marker anth by three-factor transduction

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Answers to All Questions and Problems WC-31

Chapter 9

ANS: C rha

thr 0/100

88

E

9.1 (a) How did the transformation experiments of Griffith differ from those of Avery and his associates? (b) What was the significant contribution of each? (c) Why was Griffith’s work not evidence for DNA as the genetic material, whereas the experiments of Avery and coworkers provided direct proof that DNA carried the genetic information?

lac 8

gal 17 B

Min 70

arg

63 ser

36 59

44

D cys

man his

A

8.22 Mutations nrd 11 (gene nrd B, encoding the beta subunit of the enzyme ribonucleotide reductase), am M69 (gene 63, encoding a protein that aids tail-fiber attachment), and nd 28 (gene denA, encoding the enzyme endonuclease II) are known to be located between gene 31 and gene 32 on the bacteriophage T4 chromosome. Mutations am N54 and am A453 are located in genes 31 and 32, respectively. Given the three-factor cross data in the following table, what is the linear order of the five mutant sites?

Three-Factor Cross Data Cross

% Recombinationa

1. am A453—am M69 × nrd 11

2.6

2. am A453—am M69 × nrd 11

4.2

3. am A453—am M69 × nd 28

2.5

4. am A453—nd 28 × am M69

3.5

5. am A453—nrd 11 × nd 28

2.9

6. am A453—nd 28 × nrd 11

2.1

7. am N54—am M69 × nrd 11

3.5

8. am N54—nrd 11 × am M69

1.9

9. am N54—nd 28 × am M69

1.7

10. am N54—am M69 × nd 28

2.7

11. am N54—nd 28 × nrd 11

2.9

12. am N54—nrd 11 × nd 28

1.9

a

All recombination frequencies are given as 2

(wild type progeny) total progeny

× 100.

ANS: amA453—nrd11—nd28—amM69—amN54.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 31

ANS: (a) Griffith’s in vivo experiments demonstrated the occurrence of transformation in pneumococcus. They provided no indication as to the molecular basis of the transformation phenomenon. Avery and colleagues carried out in vitro experiments, employing biochemical analyses to demonstrate that transformation was mediated by DNA. (b) Griffith showed that a transforming substance existed; Avery et al. defined it as DNA. (c) Griffith’s experiments did not include any attempt to characterize the substance responsible for transformation. Avery et al. isolated DNA in “pure” form and demonstrated that it could mediate transformation.

9.2 A cell-free extract is prepared from Type IIIS pneumococcal cells. What effect will treatment of this extract with (a) protease, (b) RNase, and (c) DNase have on its subsequent capacity to transform recipient Type IIR cells to Type IIIS? Why? ANS: (a) No effect; (b) no effect; (c) DNase will destroy the capacity of the extract to transform type IIR cells to Type IIIS by degrading the DNA in the extract. Protease and RNase will degrade the proteins and RNA, respectively, in the extract. They will have no effect, since the proteins and RNA are not involved in transformation.

9.3 How could it be demonstrated that the mixing of heatkilled Type III pneumococcus with live Type II resulted in a transfer of genetic material from Type III to Type II rather than a restoration of viability to Type III by Type II? ANS: Purified DNA from Type III cells was shown to be sufficient to transform Type II cells. This occurred in the absence of any dead Type III cells.

9.4 What is the macromolecular composition of a bacterial virus or bacteriophage such as phage T2?

ANS: About 1/2 protein, 1/2 DNA. A single long molecule of DNA is enclosed within a complex “coat” composed of many proteins.

9.5 (a) What was the objective of the experiment carried out

by Hershey and Chase? (b) How was the objective accomplished? (c) What is the significance of this experiment?

ANS: (a) The objective was to determine whether the genetic material was DNA or protein. (b) By labeling phosphorus, a constituent of DNA, and sulfur, a constituent of protein, in a virus, it was possible to demonstrate that only the labeled phosphorus was introduced into the

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32-WC Answers to All Questions and Problems host cell during the viral reproductive cycle. The DNA was enough to produce new phages. (c) Therefore DNA, not protein, is the genetic material.

9.11 RNA was extracted from TMV (tobacco mosaic virus) particles and found to contain 20 percent cytosine (20 percent of the bases were cytosine). With this information, is it possible to predict what percentage of the bases in TMV are adenine? If so, what percentage? If not, why not?

9.6 How did the reconstitution experiment of Fraenkel– Conrat and colleagues show that the genetic information of tobacco mosaic virus (TMV) is stored in its RNA rather than its protein?

ANS: No. TMV RNA is single-stranded. Thus, the base-pair stoichiometry of DNA does not apply.

ANS: When tobacco leaves were infected with reconstituted virus particles containing RNA from type A viruses and protein from type B viruses, the progeny viruses were type A, showing that RNA, not protein, carries the genetic information in TMV.

9.12 DNA was extracted from cells of Staphylococcus afermentans and analyzed for base composition. It was found that 37 percent of the bases are cytosine. With this information, is it possible to predict what percentage of the bases are adenine? If so, what percentage? If not, why not?

9.7 (a) What background material did Watson and Crick have available for developing a model of DNA? (b) What was their contribution to building the model?

ANS: Yes. Because DNA in bacteria is double-stranded, the 1:1 base-pair stoichiometry applies. Therefore, if 37% of the bases are cytosine, then 37% are guanine. This means that the remaining 26% of the bases are adenine and thymine. Thus, 26%/2 = 13% of the bases are adenine.

ANS: (a) The ladder-like pattern was known from X-ray diffraction studies. Chemical analyses had shown that a 1:1 relationship existed between the organic bases adenine and thymine and between cytosine and guanine. Physical data concerning the length of each spiral and the stacking of bases were also available. (b) Watson and Crick developed the model of a double helix, with the rigid strands of sugar and phosphorus forming spirals around an axis, and hydrogen bonds connecting the complementary bases in nucleotide pairs.

9.8 (a) Why did Watson and Crick choose a double helix for their model of DNA structure? (b) Why were hydrogen bonds placed in the model to connect the bases?

ANS: (a) A multistranded, spiral structure was suggested by the X-ray diffraction patterns. A double-stranded helix with specific base-pairing nicely fits the 1:1 stoichiometry observed for A:T and G:C in DNA. (b) Use of the known hydrogen-bonding potential of the bases provided a means of holding the two complementary strands in a stable configuration in such a double helix.

9.9 (a) If a virus particle contained double-stranded DNA with 200,000 base pairs, how many nucleotides would be present? (b) How many complete spirals would occur on each strand? (c) How many atoms of phosphorus would be present? (d) What would be the length of the DNA configuration in the virus?

ANS: (a) 400,000; (b) 20,000; (c) 400,000; (d) 68,000 nm.

9.10 What are the differences between DNA and RNA? ANS: DNA has one atom less of oxygen than RNA in the sugar part of the molecule; the sugar in DNA is 2-deoxyribose, whereas the sugar in RNA is ribose. In DNA, thymine replaces the uracil that is present in RNA. (In certain bacteriophages, DNA-containing uracil is present.)

DNA is most frequently double-stranded, but bacteriophages such as ΦX174 contain single-stranded DNA. RNA is most frequently single-stranded. Some viruses, such as the Reoviruses, however, contain double-stranded RNA chromosomes.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 32

9.13 If one strand of DNA in the Watson–Crick double helix has a base sequence of 5-GTCATGAC-3, what is the base sequence of the complementary strand?

ANS: 3′-C A G T A C T G-5′ 9.14 Indicate whether each of the following statements about the structure of DNA is true or false. (Each letter is used to refer to the concentration of that base in DNA.) (a) A + T = G + C (b) A = G; C = T (c) A/T = C/G (d) T/A = C/G (e) A + G = C + T (f) G/C = 1 (g) A = T within each single strand. (h) Hydrogen bonding provides stability to the double helix in aqueous cytoplasms. (i) Hydrophobic bonding provides stability to the double helix in aqueous cytoplasms. (j) When separated, the two strands of a double helix are identical. (k) Once the base sequence of one strand of a DNA double helix is known, the base sequence of the second strand can be deduced. (l) The structure of a DNA double helix is invariant. (m) Each nucleotide pair contains two phosphate groups, two deoxyribose molecules, and two bases.

ANS: (a) False (f) True (k) Frue

(b) False (g) False (l) False

(c) True (h) True (m) True

(d) True (i) True

(e) True (j) False

9.15 The nucleic acids from various viruses were extracted and examined to determine their base composition.

Given the following results, what can you hypothesize about the physical nature of the nucleic acids from these viruses?

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Answers to All Questions and Problems WC-33

(a) 35% A, 35% T, 15% G, and 15% C. (b) 35% A, 15% T, 25% G, and 25% C. (c) 35% A, 30% U, 30% G, and 5% C. ANS: (a) Double-stranded DNA; (b) single-stranded DNA; (c) single-stranded RNA. 9.16 Compare and contrast the structures of the A, B, and Z forms of DNA. ANS: The B form of DNA helix is that proposed by Watson and Crick and is the conformation that DNA takes under physiological conditions. It is a right-handed double helical coil with 10 bases per turn of the helix and a diameter of 1.9 nm. It has a major and a minor groove. Z-DNA is left-handed, has 12 bases per turn, a single deep groove, and is 1.8 nm in diameter. Its sugar–phosphate backbone takes a zigzagged path, and it is G:C rich. A-DNA is a right-handed helix with 11 base pairs per turn. It is a shorter, thicker double helix with a diameter of 0.23 nm and has a narrow, deep major groove and a broad, shallow minor groove. A-DNA forms in vitro under high salt concentrations or in a partially dehydrated state.

9.17 The temperature at which one-half of a double-stranded DNA molecule has been denatured is called the melting temperature, Tm. Why does Tm depend directly on the GC content of the DNA? ANS: The value of Tm increases with the GC content because GC base pairs, connected by three hydrogen bonds, are stronger than AT base pairs connected by two hydrogen bonds.

9.18 A diploid rye plant, Secale cereale, has 2n = 14 chromosomes and approximately 1.6 × 1010 bp of DNA. How much DNA is in a nucleus of a rye cell at (a) mitotic metaphase, (b) meiotic metaphase I, (c) mitotic telophase, and (d) meiotic telophase II? ANS: (a) 3.2 × 1010 bp (b) 3.2 × 1010 bp (c) 1.6 × 1010 bp (d) 0.8 × 1010 bp 9.19 The available evidence indicates that each eukaryotic chromosome (excluding polytene chromosomes) contains a single giant molecule of DNA. What different levels of organization of this DNA molecule are apparent in chromosomes of eukaryotes at various times during the cell cycle?

ANS: DNA during interphase is not yet organized into individual chromosomes but consists of a series of ellipsoidal “beads on a string” that form an 11-nm fiber. Here, 146 bp of DNA is wrapped 1.65 turns around the nucleosome core of eight histones. However, during metaphase of meiosis and mitosis, DNA becomes organized into chromosomes. The 11-nm fiber is folded and supercoiled to produce a 30-nm chromatin fiber, the basic structural unit of the metaphase chromosome. A third and final level of packaging involves nonhistone chromosomal proteins that form a scaffold to condense the 30-nm fibers into tightly packaged metaphase chromosomes, the highest level of DNA condensation observed.

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9.20 A diploid nucleus of Drosophila melanogaster contains about 3.4 × 108 nucleotide pairs. Assume (1) that all nuclear DNA is packaged in nucleosomes and (2) that an average internucleosome linker size is 60 nucleotide pairs. How many nucleosomes would be present in a diploid nucleus of D. melanogaster? How many molecules of histone H2a, H2b, H3, and H4 would be required?

ANS: In the diploid nucleus of D. melanogaster, 1.65 × 106 nucleosomes would be present; these would contain 3.3 × 106 molecules of each histone, H2a, H2b, H3, and H4.

9.21 The relationship between the melting Tm and GC content can be expressed, in its much simplified form, by the formula Tm = 69 + 0.41 (% GC). (a) Calculate the melting temperature of E. coli DNA that has about 50% GC. (b) Estimate the %GC of DNA from a human kidney cell where Tm = 85°C. ANS: (a) 89.5°C. (b) About 39% 9.22 Experimental evidence indicates that most highly repetitive DNA sequences in the chromosomes of eukaryotes do not produce any RNA or protein products. What does this indicate about the function of highly repetitive DNA? ANS: It indicates that most highly repetitive DNA sequences do not contain structural genes specifying RNA and polypeptide gene products.

9.23 The satellite DNAs of Drosophila virilis can be isolated, essentially free of main-fraction DNA, by density-gradient centrifugation. If these satellite DNAs are sheared into approximately 40-nucleotide-pair-long fragments and are analyzed in denaturation–renaturation experiments, how would you expect their hybridization kinetics to compare with the renaturation kinetics observed using similarly sheared main-fraction DNA under the same conditions? Why? ANS: The satellite DNA fragments would renature much more rapidly than the main-fraction DNA fragments. In D. virilus satellite DNAs, all three have repeating heptanucleotide-pair sequences. Thus, essentially every 40 nucleotide-long (average) single-stranded fragment from one strand will have a sequence complementary (in part) with every single-stranded fragment from the complementary strand. Many of the nucleotide-pair sequences in main-fraction DNA will be unique sequences (present only once in the genome).

9.24 (a) What functions do (1) centromeres and (2) telomeres provide? (b) Do telomeres have any unique structural features? (c) When chromosomes are broken by exposure to high-energy radiation such as X rays, the resulting broken ends exhibit a pronounced tendency to stick to each other and fuse. Why might this occur?

ANS: (a) (1) Centromeres function as spindle-fiber attachment sites on chromosomes; they are required for the separation of homologous chromosomes to opposite poles of

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34-WC Answers to All Questions and Problems the spindle during anaphase I of meiosis and for the separation of sister chromatids during anaphase of mitosis and anaphase II of meiosis. (2) Telomeres provide at least three important functions: (i) prevention of exonucleolytic degradation of the ends of the linear DNA molecules in eukaryotic chromosomes, (ii) prevention of the fusion of ends of DNA molecules of different chromosomes, and (iii) provision of a mechanism for replication of the distal tips of linear DNA molecules in eukaryotic chromosomes. (b) Yes. Most telomeres studied to date contain DNA sequence repeat units (e.g., TTAGGG in human chromosomes), and at least in some species, telomeres terminate with single-stranded 3′ overhangs that form “hairpin” structures. The bases in these hairpins exhibit unique patterns of methylation that presumably contribute to the structure and stability of telomeres. (c) The broken ends resulting from irradiation will not contain telomeres; as a result, the free ends of the DNA molecules are apparently subject to the activities of enzymes such as exonucleases, ligases, and so on, which modify the ends. They can regain stability by fusing to broken ends of other DNA molecules that contain terminal telomere sequences.

9.25 Are eukaryotic chromosomes metabolically most active during prophase, metaphase, anaphase, telophase, or interphase?

ANS: Interphase. Chromosomes are for the most part metabolically inactive (exhibiting little transcription) during the various stages of condensation in mitosis and meiosis.

9.26 Are the scaffolds of eukaryotic chromosomes composed of histone or nonhistone chromosomal proteins? How has this been determined experimentally?

ANS: Nonhistone chromosomal proteins. The “scaffold” structures of metaphase chromosomes can be observed by light microscopy after removal of the histones by differential extraction procedures.

9.27 (a) Which class of chromosomal proteins, histones or nonhistones, is the more highly conserved in different eukaryotic species? Why might this difference be expected? (b) If one compares the histone and nonhistone chromosomal proteins of chromatin isolated from different tissues or cell types of a given eukaryotic organism, which class of proteins will exhibit the greater heterogeneity? Why are both classes of proteins not expected to be equally homogeneous in chromosomes from different tissues or cell types?

ANS: (a) Histones have been highly conserved throughout the evolution of eukaryotes. A major function of histones is to package DNA into nucleosomes and chromatin fibers.

Since DNA is composed of the same four nucleotides and has the same basic structure in all eukaryotes, one might expect that the proteins that play a structural role

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 34

in packaging this DNA would be similarly conserved. (b) The nonhistone chromosomal proteins exhibit the greater heterogeneity in chromatin from different tissues and cell types of an organism. The histone composition is largely the same in all cell types within a given species—consistent with the role of histones in packaging DNA into nucleosomes. The nonhistone chromosomal proteins include proteins that regulate gene expression. Because different sets of genes are transcribed in different cell types, one would expect heterogeneity in some of the nonhistone chromosomal proteins of different tissues.

9.28 (a) If the haploid human genome contains 3 × 109 nucleotide pairs and the average molecular weight of a nucleotide pair is 660, how many copies of the human genome are present, on average, in 1 mg of human DNA? (b) What is the weight of one copy of the human genome? (c) If the haploid genome of the small plant Arabidopsis thaliana contains 7.7 × 107 nucleotide pairs, how many copies of the A. thaliana genome are present, on average, in 1 mg of A. thaliana DNA? (d) What is the weight of one copy of the A. thaliana genome? (e) Of what importance are calculations of the above type to geneticists?

ANS: (a) One microgram of human DNA will contain, on average, 3.04 × 105 copies of the genome. Using an average molecular weight per nucleotide pair of 660, the molecular weight of the entire human genome is 1.98 × 1012 (3 × 109 × 660). Thus, 1.98 × 1012 g (1 “mole” = number of grams equivalent to the “molecular” weight) of human DNA will contain, on average, 6.02 × 1023 molecules [Avogadro’s number = number of molecules (here, copies of the genome) present in one “mole” of a substance]. One gram will contain on average (3.04 × 1011)(6.02 × 1023/1.98 × 1012) copies of the genome; thus, 1 mg will contain, on average, 3.04 × 105 copies of the human genome. (b) One copy of the human genome weighs approximately (3.3 × 10−12 g) (1.98 × 1012 g per “mole”/6.02 × 1023 molecules per “mole”) or 3.3 × 10−6 mg. (c) By analogous calculations, 1 g of A. thaliana DNA contains, on average, 1.18 × 107 copies of the genome. (d) Similarly, one copy of the A. thaliana genome weighs approximately 8.4 × 10−8 mg. (e) In carrying out molecular analyses of the structures of genomes, geneticists frequently need to know how many copies of a genome are present, on average, in a given quantity of DNA. Chapter 10 10.1 DNA polymerase I of E. coli is a single polypeptide of molecular weight 103,000. (a) What enzymatic activities other than polymerase activity does this polypeptide possess? (b) What are the in vivo functions of these activities? (c) Are these activities of major importance to an E. coli cell? Why?

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Answers to All Questions and Problems WC-35

ANS: (a) Both 3′ → 5′ and 5′ → 3′ exonuclease activities. (b) The 3′ → 5′ exonuclease “proofreads” the nascent DNA strand during its synthesis. If a mismatched base pair occurs at the 3′-OH end of the primer, the 3′ → 5′ exonuclease removes the incorrect terminal nucleotide before polymerization proceeds again. The 5′ → 3′ exonuclease is responsible for the removal of RNA primers during DNA replication and functions in pathways involved in the repair of damaged DNA (see Chapter 13). (c) Yes, both exonuclease activities appear to be very important. Without the 3′ → 5′ proofreading activity during replication, an intolerable mutation frequency would occur. The 5′ → 3′ exonuclease activity is essential to the survival of the cell. Conditional mutations that alter the 5′ → 3′ exonuclease activity of DNA polymerase I are lethal to the cell under conditions where the exonuclease is nonfunctional.

10.2 Escherichia coli cells are grown for many generations in a medium in which the only available nitrogen is the heavy isotope 15N. They are then transferred to a medium containing 14N as the only source of nitrogen. (a) What distribution of N and N would be expected in the DNA molecules of cells that had grown for one generation in the 14N-containing medium assuming that DNA replication was (i) conservative, (ii) semiconservative, or (iii) dispersive?

15

14

(b) What distribution would be expected after two generations of growth in the 14N-containing medium assuming (i) conservative, (ii) semiconservative, or (iii) dispersive replication?

ANS: (a) (i) One-half of the DNA molecules with 15N in both strands and the other half with 14N in both strands; (ii) all DNA molecules with one strand containing 15N and the complementary strand containing 14N; (iii) all DNA molecules with both strands containing roughly equal amounts of 15N and 14N. (b) (i) 1/4 of the DNA molecules with 15N in both strands and 3/4 with 14N in both strands; (ii) half of the DNA molecules with one strand containing 15N and the complementary strand containing 14N and the other half with 14N in both strands; (iii) all DNA molecules with both strands containing about 1/4 15N and 3/4 14N. 10.3 Why do DNA molecules containing 15N band at a different position than DNA molecules containing 14N when centrifuged to equilibrium in 6M CsCl? ANS:

Nitrogen contains eight neutrons instead of the seven neutrons in the normal isotope of nitrogen, 14N. Therefore, 15N has an atomic mass of about 15, whereas 14N has a mass of about 14. This difference means that purines and pyrimidines containing 15N have a greater density (weight per unit volume) than those containing 14N. Equilibrium density-gradient centrifugation in 6M CsCl separates DNAs or other macromolecules based on their

15

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 35

densities, and E. coli DNA, for example, that contains 15N has a density of 1.724 g/cm2, whereas E. coli DNA that contains 14N has a density of 1.710 g/cm2.

10.4 A DNA template plus primer with the structure 3´ P —TGCGAATTAGCGACAT— P 5´ ............. .. .. ..

5´ P —ATCGGTACGACGCTTAAC—OH 3´

(where P = a phosphate group) is placed in an in vitro DNA synthesis system (Mg2+, an excess of the four deoxyribonucleoside triphosphates, etc.) containing a mutant form of E. coli DNA polymerase I that lacks exonuclease activity. The polymerase and exonuclease activities of this aberrant enzyme are identical to those of normal E. coli DNA polymerase I. It simply has no exonuclease activity. (a) What will be the structure of the final product? (b) What will be the first step in the reaction sequence?

ANS: (a)

3´ P -TGCGAATTAGCGACAT- P 5´ ................. .. ....................

5´ P - ATCGGTACGACGCTTAATCGCTGTA -OH 3´;

Note that DNA synthesis will not occur on the left end since the 3′-terminus of the potential primer strand is blocked with a phosphate group—all DNA polymerases require a free 3′-OH terminus. (b) The first step will be the removal of the mismatched C (exiting as dCMP) from the 3′-OH primer terminus by the 3′ 5′ exonuclease (“proofreading”) activity.

10.5 How might continuous and discontinuous modes of DNA replication be distinguished experimentally?

ANS: If nascent DNA is labeled by exposure to 3H-thymidine for very short periods of time, continuous replication predicts that the label would be incorporated into chromosome-sized DNA molecules, whereas discontinuous replication predicts that the label would first appear in small pieces of nascent DNA (prior to covalent joining, catalyzed by polynucleotide ligase).

10.6 E. coli cells contain five different DNA polymerases—I, II, III, IV, and V. Which of these enzymes catalyzes the semiconservative replication of the bacterial chromosome during cell division? What are the functions of the other four DNA polymerases in E. coli?

ANS: DNA polymerase III is the true replicase. DNA polymerase I removes the RNA primers and replaces them with DNA. The other DNA polymerases play important roles in DNA repair pathways (see Chapter 13).

10.7 The Boston barberry is an imaginary plant with a diploid chromosome number of 4, and Boston barberry cells are easily grown in suspended cell cultures. 3H-thymidine was added to the culture medium in which a G1-stage cell of this plant was growing. After one cell

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36-WC Answers to All Questions and Problems generation of growth in 3H-thymidine-containing medium, colchicine was added to the culture medium.

The medium now contained both 3H-thymidine and colchicine. After two “generations” of growth in 3H-thymidine-containing medium (the second “generation” occurring in the presence of colchicine as well), the two progeny cells (each now containing eight chromosomes) were transferred to culture medium containing nonradioactive thymidine (3H-thymidine) plus colchicine.

Note that a “generation” in the presence of colchicine consists of a normal cell cycle’s chromosomal duplication but no cell division. The two progeny cells were allowed to continue to grow, proceeding through the “cell cycle,” until each cell contained a set of metaphase chromosomes that looked like the following.

If autoradiography were carried out on these metaphase chromosomes (four large plus four small), what pattern of radioactivity (as indicated by silver grains on the autoradiograph) would be expected? (Assume no recombination between DNA molecules.)

ANS: Two

Plus two

For both the large and small chromosomes

10.8 Suppose that the experiment described in Problem 10.7 was carried out again, except this time replacing the 3 H-thymidine with nonradioactive thymidine at the same time that the colchicine was added (after one cell generation of growth in 3H-thymidine-containing medium). The cells were then maintained in colchicine plus nonradioactive thymidine until the metaphase shown in Problem 10.7 occurred. What would the autoradiographs of these chromosomes look like?

ANS: Two

Plus two

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For both the large and small chromosomes

10.9 Suppose that the DNA of cells (growing in a cell culture) in a eukaryotic species was labeled for a short period of time by the addition of 3H-thymidine to the medium.

Next assume that the label was removed and the cells were resuspended in nonradioactive medium. After a short period of growth in nonradioactive medium, the DNA was extracted from these cells, diluted, gently layered on filters, and autoradiographed. If autoradiographs of the type ....

....

....

were observed, what would this indicate about the nature of DNA replication in these cells? Why? ANS: The DNA replication was unidirectional rather than bidirectional. As the intracellular pools of radioactive 3 H-thymidine are gradually diluted after transfer to nonradioactive medium, less and less 3H-thymidine will be incorporated into DNA at each replicating fork. This will produce autoradiograms with tails of decreasing grain density at each growing point. Since such tails appear at only one end of each track, replication must be unidirectional. Bidirectional replication would produce such tails at both ends of an autoradiographic track (see Figure 10.31).

10.10 Arrange the following enzymes in the order of their action during DNA replication in E. coli: (1) DNA polymerase I, (2) DNA polymerase III, (3) DNA primase, (4) DNA gyrase, and (5) DNA helicase.

ANS: The correct sequence of action is 4, 5, 3, 2, 1.

10.11 Fifteen distinct DNA polymerases—a, b, g, d, e, k, z, h, q, k, l, m, s, f, and Rev1—have been characterized in mammals. What are the intracellular locations and functions of these polymerases?

ANS: Current evidence suggests that polymerases a, d, and/or e are required for the replication of nuclear DNA. Polymerase e is thought to catalyze the continuous synthesis of the leading strand, and polymerases a and d are thought to catalyze the replication of the lagging strand.

Polymerase a forms a complex with primase and initiates the synthesis of Okazaki fragments during the discontinuous replication of the lagging strand. Polymerase a catalyzes the incorporation of approximately the first 30 nucleotides in each Okazaki fragment before being replaced by polymerase d, which then completes the synthesis of the fragments. Polymerase g catalyzes replication of organellar chromosomes. Polymerases b, k, z, h, q, k, l, m, s, f, and Rev1 function in various DNA repair pathways (see Chapter 13).

10.12 The E. coli chromosome contains approximately 4 × 106 nucleotide pairs and replicates as a single bidirectional replicon in approximately 40 minutes under a wide variety of growth conditions. The largest chromosome of D. melanogaster contains about 6 × 107 nucleotide pairs. (a) If this chromosome contains one giant molecule of

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Answers to All Questions and Problems WC-37

DNA that replicates bidirectionally from a single origin located precisely in the middle of the DNA molecule, how long would it take to replicate the entire chromosome if replication in Drosophila occurred at the same rate as replication in E. coli? (b) Actually, replication rates are slower in eukaryotes than in prokaryotes. If each replication bubble grows at a rate of 5000 nucleotide pairs per minute in Drosophila and 100,000 nucleotide pairs per minute in E. coli, how long will it take to replicate the largest Drosophila chromosome if it contains a single bidirectional replicon as described in (a) above? (c) In Drosophila embryos, the nuclei divide every 9 to 10 minutes. Based on your calculations in (a) and (b) earlier, what do these rapid nuclear divisions indicate about the number of replicons per chromosome in Drosophila?

ANS: (a) Given bidirectional replication of a single replicon, each replication fork must traverse 2 ×106 nucleotide pairs in E. coli and 3 × 107 nucleotide pairs in the largest Drosophila chromosome. If the rates were the same in both species, it would take 15 times (3 × 107/2 × 106) as long to replicate the Drosophila chromosome or 10 hours (40 minutes × 15 = 600 minutes).(b) If replication forks in E. coli move 20 times as fast as replication forks in ­Drosophila (100,000 nucleotide pairs per minute/5000 nucleotide pairs per minute), the largest Drosophila chromosome would require 8.3 days (10 hours × 20 = 200 hours) to complete one round of replication. (c) Each Drosophila chromosome must contain many replicons in order to complete replication in less than 10 minutes.

10.13 E. coli cells that have been growing in 14N for many generations are transferred to medium containing only 15N and allowed to grow in this medium for four generations.

Their DNA is then extracted and analyzed by equilibrium CsCl density-gradient centrifugation. What proportion of this DNA will band at the “light,” “hybrid,” and “heavy” positions in the gradient?

ANS: No DNA will band at the “light” position; 12.5 percent (2 of 16 DNA molecules) will band at the “hybrid” density; and 87.5 percent (14 of 16 DNA molecules) will band at the “heavy” position.

10.14 The bacteriophage lambda chromosome has several A:T-rich segments that denature when exposed to pH 11.05 for 10 minutes. After such partial denaturation, the linear packaged form of the lambda DNA molecule has the structure shown in Figure 10.9a. Following its injection into an E. coli cell, the lambda DNA molecule is converted into a covalently closed circular molecule by hydrogen bonding between its complementary singlestranded termini and the action of DNA ligase. It then replicates as a q-shaped structure. The entire lambda chromosome is 17.5 mm long. It has a unique origin of replication located 14.3 mm from the left end of the linear form shown in Figure 10.9a. Draw the structure that would be observed by electron microscopy after both (1) replication of an approximately 6-mm-long segment

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 37

of the lambda chromosomal DNA molecule (in vivo) and (2) exposure of this partially replicated DNA molecule to pH 11.05 for 10 minutes (in vitro), (a) if replication had proceeded bidirectionally from the origin, and (b) if replication had proceeded unidirectionally from the origin.

ANS: (a) and (b)

ef c

d

i

gh

j

a

b

i

b

ef gh

c d

j a

i

or

j

a

h g f e

d b c

10.15 What enzyme activity catalyzes each of the following steps in the semiconservative replication of DNA in prokaryotes? (a) The formation of negative supercoils in progeny DNA molecules. (b) The synthesis of RNA primers. (c) The removal of RNA primers. (d) The covalent extension of DNA chains at the 3′-OH termini of primer strands. (e) Proofreading of the nucleotides at the 3′-OH termini of DNA primer strands?

ANS: (a) DNA gyrase; (b) primase; (c) the 5′ → 3′ exonuclease activity of DNA polymerase I; (d) the 5′ → 3′ polymerase activity of DNA polymerase III; (e) the 3′ → 5′ exonuclease activity of DNA polymerase III. 10.16 One species of tree has a very large genome consisting of 2.0 × 1010 base pairs of DNA. (a) If this DNA was organized into a single linear molecule, how long (meters) would this molecule be? (b) If the DNA is evenly distributed among 10 chromosomes and each chromosome has one origin of DNA replication, how long would it take to complete the S phase of the cell cycle, assuming that DNA polymerase can synthesize 2 × 104 bp of DNA per minute? (c) An actively growing cell can complete the S phase of the cell cycle in approximately 300 minutes. Assuming that the origins of replication are evenly distributed, how many origins of replication are present on each chromosome? (d) What is the average number of base pairs between adjacent origins of replication?

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38-WC Answers to All Questions and Problems ANS: (a) (34 nm/100 bp)(2 × 1010 bp) = 6.8 × 109 = 6.8 meters. (b) 2 × 1010/10 chromosomes = 2 × 109 bp; 4 × 104 bp/ min (bidirectional); 2 × 109/4 × 104 = 5 × 104 min = 50,000 min. (c) (50,000 min)(1 ori) = (300 min) (X ori); X = 50,000/300 = 167 ori. (d) (2 × 109 bp/ chrom.)/(167 ori/chrom.) = 1.2 × 107 bp/ori.

10.17 Why must each of the giant DNA molecules in ­eukaryotic chromosomes contain multiple origins of replication?

ANS: In eukaryotes, the rate of DNA synthesis at each replication fork is about 2500–3000 nucleotide pairs per minute. Large eukaryotic chromosomes often contain 107–108 nucleotide pairs. A single replication fork could not replicate the giant DNA in one of these large chromosomes fast enough to permit the observed cell generation times.

10.18 In E. coli, viable polA mutants have been isolated that produce a defective gene product with little or no 5′→3′ polymerase activity, but normal 5′→3′ exonuclease activity. However, no polA mutant has been identified that is completely deficient in the 5′→3′ exonuclease activity, while retaining 5′→3′ polymerase activity, of DNA polymerase I. How can these results be explained?

ANS: The 5′ → 3′ exonuclease activity of DNA polymerase I is essential to the survival of the bacterium, whereas the 5′→3′ polymerase activity of the enzyme is not essential.

10.19 Other polA mutants of E. coli lack the 3′ → 5′ exonuclease activity of DNA polymerase I. Will the rate of DNA synthesis be altered in these mutants? What effect(s) will these polA mutations have on the phenotype of the organism?

ANS: No, the rate of DNA synthesis will not be altered. E. coli strains carrying polA mutations that eliminate the 3′ → 5′ exonuclease activity of DNA polymerase I will exhibit unusually high mutation rates.

10.20 Many of the origins of replication that have been characterized contain AT-rich core sequences. Are these AT-rich cores of any functional significance? If so, what?

ANS: Because AT base pairs are held together by only two hydrogen bonds instead of the three hydrogen bonds present in GC base pairs, the two strands of AT-rich regions of double helices are separated more easily, providing the single-stranded template regions required for DNA replication.

10.21 (a) Why isn’t DNA primase activity required to initiate rolling-circle replication?(b) DNA primase is required for the discontinuous synthesis of the lagging strand, which occurs on the single-stranded tail of the rolling circle. Why? ANS: Rolling-circle replication begins when an endonuclease cleaves one strand of a circular DNA double helix. This cleavage produces a free 3′-OH on one end of the cut

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 38

strand, allowing it to function as a primer. The discontinuous synthesis of the lagging strand requires the de novo initiation of each Okazaki fragment, which requires DNA primase activity.

10.22 DNA polymerase I is needed to remove RNA primers during chromosome replication in E. coli. However, DNA polymerase III is the true replicase in E. coli. Why does not DNA polymerase III remove the RNA primers?

ANS: DNA polymerase III does not have a 5′ → 3′ exonuclease activity that acts on double-stranded nucleic acids. Thus, it cannot excise RNA primer strands from replicating DNA molecules. DNA polymerase I is present in cells at much higher concentrations and functions as a monomer. Thus, DNA polymerase I is able to catalyze the removal of RNA primers from the vast number of Okazaki fragments formed during the discontinuous replication of the lagging strand.

10.23 In E. coli, three different proteins are required to unwind the parental double helix and keep the unwound strands in an extended template form. What are these proteins, and what are their respective functions?

ANS: DNA helicase unwinds the DNA double helix, and single-strand DNA-binding protein coats the unwound strands, keeping them in an extended state. DNA gyrase catalyzes the formation of negative supercoiling in E. coli DNA, and this negative supercoiling behind the replication forks is thought to drive the unwinding process because superhelical tension is reduced by unwinding the complementary strands.

10.24 How similar are the structures of DNA polymerase I and DNA polymerase III in E. coli? What is the structure of the DNA polymerase III holoenzyme? What is the function of the dnaN gene product in E. coli?

ANS: DNA polymerase I is a single polypeptide of molecular weight 109,000, whereas DNA polymerase III is a complex multimeric protein. The DNA polymerase III holoenzyme has a molecular mass of about 900,000 daltons and is composed of at least 20 different polypeptides. The dnaN gene product, the b subunit of DNA polymerase III, forms a dimeric clamp that encircles the DNA molecule and prevents the enzyme from dissociating from the template DNA during replication.

10.25 The dnaA gene product of E. coli is required for the initiation of DNA synthesis at oriC. What is its function?

How do we know that the DnaA protein is essential to the initiation process?

ANS: DnaA protein initiates the formation of the replication bubble by binding to the 9-bp repeats of OriC. DnaA protein is known to be required for the initiation process because bacteria with temperature-sensitive mutations in the dnaA gene cannot initiate DNA replication at restrictive temperatures.

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Answers to All Questions and Problems WC-39

10.26 What is a primosome, and what are its functions? What essential enzymes are present in the primosome? What are the major components of the E. coli replisome? How can geneticists determine whether these components are required for DNA replication?

ANS: The primosome is a protein complex that initiates the synthesis of Okazaki fragments during lagging strand synthesis. The major components of the E. coli DNA ­primosome are DNA primase and DNA helicase. Geneticists have been able to show that both DNA primase and DNA helicase are required for DNA replication by demonstrating that mutations in the genes encoding these enzymes result in the arrest of DNA synthesis in mutant cells under conditions where the altered proteins are inactive.

10.27 The chromosomal DNA of eukaryotes is packaged into nucleosomes during the S phase of the cell cycle. What obstacles do the size and complexity of both the replisome and the nucleosome present during the semiconservative replication of eukaryotic DNA? How might these obstacles be overcome?

ANS: Nucleosomes and replisomes are both large macromolecular structures, and the packaging of eukaryotic DNA into nucleosomes raises the question of how a replisome can move past a nucleosome and replicate the DNA in the nucleosome in the process. The most obvious solution to this problem would be to completely or partially disassemble the nucleosome to allow the replisome to pass. The nucleosome would then reassemble after the replisome had passed. One popular model has the nucleosome partially disassembling, allowing the replisome to move past it (see ­Figure 10.33b).

10.28 Two mutant strains of E. coli each have a temperaturesensitive mutation in a gene that encodes a product required for chromosome duplication. Both strains replicate their DNA and divide normally at 25°C but are unable to replicate their DNA or divide at 42°C. When cells of one strain are shifted from growth at 25°C to growth at 42°C, DNA synthesis stops immediately. When cells of the other strain are subjected to the same temperature shift, DNA synthesis continues, albeit at a decreasing rate, for about a half hour. What can you conclude about the functions of the products of these two genes?

ANS: The product of the first gene is required for DNA chain extension, whereas the product of the second gene is only required for the initiation of DNA synthesis.

10.29 In what ways does chromosomal DNA replication in eukaryotes differ from DNA replication in prokaryotes?

ANS: (1) DNA replication usually occurs continuously in rapidly growing prokaryotic cells but is restricted to the S phase of the cell cycle in eukaryotes. (2) Most eukaryotic chromosomes contain multiple origins of replication, whereas most prokaryotic chromosomes

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 39

contain a single origin of replication. (3) Prokaryotes utilize two catalytic complexes that contain the same DNA polymerase to replicate the leading and lagging strands, whereas eukaryotes utilize two or three distinct DNA polymerases for leading and lagging strand synthesis. (4) Replication of eukaryotic chromosomes requires the partial disassembly and reassembly of nucleosomes as replisomes move along parental DNA molecules. In prokaryotes, replication probably involves a similar partial disassembly/reassembly of nucleosome-like structures. (5) Most prokaryotic chromosomes are circular and thus have no ends. Most eukaryotic chromosomes are linear and have unique termini called telomeres that are added to replicating DNA molecules by a unique, RNA-containing enzyme called telomerase.

10.30 (a) The chromosome of the bacterium Salmonella typhimurium contains about 4 × 106 nucleotide pairs.

Approximately how many Okazaki fragments are produced during one complete replication of the S. typhimurium chromosome? (b) The largest chromosome of D. melanogaster contains approximately 6 × 107 nucleotide pairs. About how many Okazaki fragments are produced during the replication of this chromosome?

ANS: (a) 2000–4000 Okazaki fragments. (b) 300,000–600,000 Okazaki fragments.

10.31 In the yeast S. cerevisiae, haploid cells carrying a mutation called est1 (for ever-shorter telomeres) lose distal telomere sequences during each cell division. Predict the ultimate phenotypic effect of this mutation on the progeny of these cells.

ANS: The chromosomes of haploid yeast cells that carry the est1 mutation become shorter during each cell division.

Eventually, chromosome instability results from the complete loss of telomeres, and cell death occurs because of the deletion of essential genes near the ends of chromosomes.

10.32 Assume that the sequence of a double-stranded DNA shown below is present at one end of a large DNA molecule in a eukaryotic chromosome.

5′-(centromere sequence)-gattccccgggaagcttggggggcccatcttcgtacgtctttgca-3′ 3′-(centromere sequence)-ctaaggggcccttcgaaccccccgggtagaagcatgagaaacgt-5′

You have reconstituted a eukaryotic replisome that is active in vitro. However, it lacks telomerase activity. If you isolate the DNA molecule shown above and replicate it in your in vitro system, what products would you expect?

ANS: Without telomerase, the 5′ end of the newly replicated strand will be missing some bases. The exact number of missing bases does not matter: 5′-(centromere sequence)-gattccccgggaagcttggggggcccatcttcgtacgtctttgca-3′ 3′-(centromere sequence)-ctaaggggcccttcgaaccccccgcgggtagaagcatg-5′ 5′-(centromere sequence)-gattccccgggaagcttggggggcccatcttcgtacgtctttgca-3′ 3′-(centromere sequence)-ctaaggggcccttc-5′

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40-WC Answers to All Questions and Problems Chapter 11 11.1 Distinguish between DNA and RNA (a) chemically, (b) functionally, and (c) by location in the cell.

ANS: (a) RNA contains the sugar ribose, which has a hydroxyl (OH) group on the 2-carbon; DNA contains the sugar 2-deoxyribose, with only hydrogens on the 2-carbon.

RNA usually contains the base uracil at positions where thymine is present in DNA. However, some DNAs contain uracil, and some RNAs contain thymine. DNA exists most frequently as a double helix (double-stranded molecule); RNA exists more frequently as a single-stranded molecule. But, some DNAs are single-stranded and some RNAs are double-stranded. (b) The main function of DNA is to store genetic information and to transmit that information from cell to cell and from generation to generation. RNA stores and transmits genetic information in some viruses that contain no DNA. In cells with both DNA and RNA: (1) mRNA acts as an intermediary in protein synthesis, carrying the information from DNA in the chromosomes to the ribosomes (sites at which proteins are synthesized). (2) tRNAs carry amino acids to the ribosomes and function in codon recognition during the synthesis of polypeptides. (3) rRNA molecules are essential components of the ribosomes. (4) snRNAs are important components of spliceosomes. (5) miRNAs play key roles in regulating gene expression (see Chapter 18). (c) DNA is located primarily in the chromosomes, which are found in the nucleus of eukaryotic cells; however, some DNA is also found in cytoplasmic organelles, such as mitochondria and chloroplasts. RNA is located throughout cells.

11.2 What bases in the mRNA transcript would represent the following DNA template sequence: 5′-TGCAGACA-3′?

ANS: 3′-ACGUCUGU-5′ 11.3 What bases in the transcribed strand of DNA would give rise to the following mRNA base sequence: 5′-CUGAU-3′?

ANS: 3′-GACTA-5′ 11.4 On the basis of what evidence was the messenger RNA hypothesis established?

ANS: The genetic information of cells is stored in DNA, which is located predominantly in the chromosomes. The gene products (polypeptides) are synthesized primarily in the cytoplasm on ribosomes. Some intermediate must therefore carry the genetic information from the chromosomes to the ribosomes. RNA molecules (mRNAs) were shown to perform this function by means of RNA pulselabeling and pulse-chase experiments combined with autoradiography. The enzyme RNA polymerase was subsequently shown to catalyze the synthesis of mRNA using chromosomal DNA as a template. Finally, the mRNA molecules synthesized by RNA polymerase were shown to faithfully direct the synthesis of specific polypeptides when used in in vitro protein synthesis systems.

11.5 At what locations in a eukaryotic cell does protein synthesis occur?

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 40

ANS: Protein synthesis occurs on ribosomes. In eukaryotes, most of the ribosomes are located in the cytoplasm and are attached to the extensive membranous network of endoplasmic reticulum. Some protein synthesis also occurs in cytoplasmic organelles such as chloroplasts and mitochondria.

11.6 List three ways in which the mRNAs of eukaryotes differ from the mRNAs of prokaryotes.

ANS: (1) Eukaryotes have a 5′ cap on their mRNAs; prokaryotes do not. (2) Messenger RNAs of eukaryotes generally have a 3′ poly-A tail, prokaryotic mRNAs do not. (3) Messenger RNA formation in eukaryotes involves removal of introns (when present) and the splicing together of exons. Prokaryotic genes (with very rare exceptions) do not have introns.

11.7 What different types of RNA molecules are present in prokaryotic cells? in eukaryotic cells? What roles do these different classes of RNA molecules play in the cell?

ANS: Both prokaryotic and eukaryotic organisms contain messenger RNAs, transfer RNAs, and ribosomal RNAs. In addition, eukaryotes contain small nuclear RNAs and micro RNAs. Messenger RNA molecules carry genetic information from the chromosomes (where the information is stored) to the ribosomes in the cytoplasm (where the information is expressed during protein synthesis).

The linear sequence of triplet codons in an mRNA molecule specifies the linear sequence of amino acids in the polypeptides produced during translation of that mRNA. Transfer RNA molecules are small (about 80 nucleotides long) molecules that carry amino acids to the ribosomes and provide the codon-recognition specificity during translation. Ribosomal RNA molecules provide part of the structure and function of ribosomes; they represent an important part of the machinery required for the synthesis of polypeptides. Small nuclear RNAs are structural components of spliceosomes, which excise noncoding intron sequences from nuclear gene transcripts. Micro RNAs are involved in the regulation of gene expression.

11.8 Many eukaryotic genes contain noncoding introns that separate the coding sequences or exons of these genes. At what stage during the expression of these split genes are the noncoding intron sequences removed?

ANS: The entire nucleotide-pair sequences—including the introns—of the genes are transcribed by RNA polymerase to produce primary transcripts that still contain the intron sequences. The intron sequences are then spliced out of the primary transcripts to produce the mature, functional RNA molecules. In the case of ­protein-encoding nuclear genes of higher eukaryotes, the introns are spliced out by complex macromolecular structures called spliceosomes.

11.9 For several decades, the dogma in biology has been that molecular reactions in living cells are catalyzed by enzymes composed of polypeptides. We now know that

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Answers to All Questions and Problems WC-41

the introns of some precursor RNA molecules such as the rRNA precursors in Tetrahymena are removed autocatalytically (self-spliced) with no involvement of any catalytic protein. What does the demonstration of autocatalytic splicing indicate about the dogma that biological reactions are always catalyzed by proteinaceous enzymes?

(h) The RNA polymerase in the nucleus that catalyzes the synthesis of the transfer RNA molecules and small nuclear RNAs.

ANS: “Self-splicing” of RNA precursors demonstrates that RNA molecules can also contain catalytic sites; this property is not restricted to proteins.

(j) The RNA polymerase that transcribes nuclear genes that encode proteins.

11.10 What role(s) do spliceosomes play in pathways of gene expression? What is their macromolecular structure?

(i) A polyadenosine tract 20–200 nucleotides long that is added to the 3′ end of most eukaryotic messenger RNAs.

(k) A conserved sequence in the nontemplate strand of eukaryotic promoters that is located about 80 nucleotides upstream from the transcription start site.

ANS: Spliceosomes excise intron sequences from nuclear gene transcripts to produce the mature mRNA molecules that are translated on ribosomes in the cytoplasm. Spliceosomes are complex macromolecular structures composed of snRNA and protein molecules (see Figure 11.22).

(l) Segments of an eukaryotic gene that correspond to the sequences in the final processed RNA transcript of the gene.

11.11 What components of the introns of nuclear genes that encode proteins in higher eukaryotes are conserved and required for the correct excision of intron sequences from primary transcripts by spliceosomes?

ANS: (a) 7; (b) 5; (c) 10; (d) 3; (e) 11; (f) 1; (g) 12; (h) 6; (i) 2; (j) 8; (k) 13; (l) 4; (m) 9.

ANS: The introns of protein-encoding nuclear genes of higher eukaryotes almost invariably begin (5′) with GT and end (3′) with AG. In addition, the 3′ subterminal A in the “TACTAAC box” is completely conserved; this A is involved in bond formation during intron excision.

11.12 Match one of the following terms with each of the descriptions given below. Terms: (1) sigma (s) factor; (2) poly(A) tail; (3) TATAAT; (4) exons; (5) TATAAAA; (6) RNA polymerase III; (7) intron; (8) RNA polymerase II; (9) heterogeneous nuclear RNA (hnRNA); (10) snRNA; (11) RNA polymerase I; (12) TTGACA; (13) GGCCAATCT (CAAT box).

Descriptions: (a) Intervening sequence found in many eukaryotic genes. (b) A conserved nucleotide sequence (-30) in eukaryotic promoters involved in the initiation of transcription. (c) Small RNA molecules that are located in the nuclei of eukaryotic cells, most as components of the spliceosome, that participate in the excision of introns from nuclear gene transcripts. (d) A sequence (-10) in the nontemplate strand of the promoter of E. coli that facilitates the localized unwinding of DNA when complexed with RNA polymerase. (e) The RNA polymerase in the nucleus that catalyzes the synthesis of all rRNAs except for the small 5S rRNA. (f) The subunit of prokaryotic RNA polymerase that is responsible for the initiation of transcription at promoters. (g) An E. coli promoter sequence located 35 nucleotides upstream from the transcription-initiation site; it serves as a recognition site for the sigma factor.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 41

(m) The population of primary transcripts in the nucleus of a eukaryotic cell.

11.13 (a) Which of the following nuclear pre-mRNA nucleotide sequences potentially contains an intron?

(1) 5′-UGACCAUGGCGCUAACACUGCCAAU UGGCAAUACUGACCUGAUAGCAUCAGCCAA-3′ (2) 5′-UAGUCUCAUCUGUCCAUUGACUUCGAAA CUGAAUCGUAACUCCUACGUCUAUGGA-3′ (3) 5′-UAGCUGUUUGUCAUGACUGACUGGUCA CUAUCGUACUAACCUGUCAUGCAAUGUC-3′ (4) 5′-UAGCAGUUCUGUCGCCUCGUGGUGCUG CUGGCCCUUCGUCGCUCGGGCUUAGCUA-3′ (5) 5′-UAGGUUCGCAUUGACGUACUUCUGAAAC UACUAACUACUAACGCAUCGAGUCUCAA-3′ (b) One of the five pre-mRNAs shown in (a) may undergo RNA splicing to excise an intron sequence. What mRNA nucleotide sequence would be expected to result from this splicing event?

ANS: (a) Sequence 5. It contains the conserved intron sequences: a 5′ GU, a 3′ AG, and a UACUAAC internal sequence providing a potential bonding site for intron excision.

Sequence 4 has a 5′ GU and a 3′ AG but contains no internal A for the bonding site during intron excision. (b) 5′—UAGUCUCAA—3′; the putative intron from the 5′ GU through the 3′ AG has been removed.

11.14 What is the function of the introns in eukaryotic genes?

ANS: This is a wide-open question at present! There is much speculation, but little hard evidence. One popular hypothesis is that introns enhance exon shuffling by increasing recombination events between sequences encoding adjacent domains of a polypeptide. Also, in one yeast mitochondrial gene, the introns contain open reading frames that encode “maturases” that splice out these introns—a neat negative feedback control. Other introns may be merely relics of evolution.

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42-WC Answers to All Questions and Problems 11.15 A particular gene is inserted into the phage lambda chromosome and is shown to contain three introns. (a) The primary transcript of this gene is purified from isolated nuclei. When this primary transcript is hybridized under R-loop conditions with the recombinant lambda chromosome carrying the gene, what will the R-loop structure(s) look like? Label your diagram. (b) The mRNA produced from the primary transcript of this gene is then isolated from cytoplasmic polyribosomes and similarly examined by the R-loop hybridization ­procedure using the recombinant lambda chromosome carrying the gene. Diagram what the R-loop structure(s) will look like when the cytoplasmic mRNA is used. Again, label the components of your diagram.

When this segment of DNA is transcribed by RNA polymerase, what will be the sequence of nucleotides in the RNA transcript?

ANS: Assuming that there is a −35 sequence upstream from the consensus −10 sequence in this segment of the DNA molecule, the nucleotide sequence of the transcript will be 5′-ACCCGACAUAGCUACGAUGACGAUAAGC GACAUAGC-3′.

11.18 A segment of DNA in E. coli has the following sequence of nucleotide pairs: 3´-A A C T G T A C G T G C T A C C T T G C T G A T A T T A C T5´-T T G A C A T G C A C G A T G G A A C G A C T A T A A T G AG C A A T G G G C T G T A T C G A T G C T A C T G C T A T-5´

ANS: Displaced single-stranded DNA ("R-loop")

Ex

on

Primary transcript

λ DNA 1 Intron1 Exon2 Intron2 Exon3 Intron3 Exo λ DNA n4

(a) Displaced single-stranded exon DNA ("R-loops") mRNA Exon1

λ DNA (b)

Exon2

Exon3 Exon4

Intron1

Intron2 Intron3

λDNA

11.16 A segment of DNA in E. coli has the following sequence of nucleotide pairs: 3´-A T G C T A C T G C T A T T C G C T G T A T C G-5´ 5´-T A C G A T G A C G A T A A G C G A C A T A G C-3´

When this segment of DNA is transcribed by RNA polymerase, what will be the sequence of nucleotides in the RNA transcript if the promoter is located to the left of the sequence shown?

ANS: If there is a promoter located upstream from this DNA segment, the nucleotide sequence of this portion of the RNA transcript will be 5′-UACGAUGACGAUAAGCGACAUAGC-3′. If there is no upstream promoter, this segment of DNA will not be transcribed.

11.17 A segment of DNA in E. coli has the following sequence of nucleotide pairs: 3´-A T A T T A C T G C A A T G G G C T G T A T C G5´-T A T A A T G A C G T T A C C C G A C A T A G CA T G C T A C T G C T A T T C G C T G T A T C G-5´ T A C G A T G A C G A T A A G C G A C A T A G C-3´

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C G T T A C C C G A C A T A G C T A C G A T G A C G A T A-3´

When this segment of DNA is transcribed by RNA polymerase, what will be the sequence of nucleotides in the RNA transcript?

ANS: Given the consensus −35 and −10 sequences in this segment of DNA and the fact that transcripts almost always start with a purine, the predicted nucleotide sequence of the transcript is 5′-ACCCGACAUAGCUACGAUGA CGAUA-3′.

11.19 A segment of human DNA has the following sequence of nucleotide pairs: 3´-A T A T T T A C G T G C T A C C T T G C T G A T A G G A C T5´-T A T A A A T G C A C G A T G G A A C G A C T A T C C T G AG C A A T G G G C T G T A T C G A T G C T A C T G C T A T-5´ C G T T A C C C G A C A T A G C T A C G A T G A C G A T A-3´

When this segment of DNA is transcribed by RNA polymerase, what will be the sequence of nucleotides in the RNA transcript?

ANS: Assuming that there is a CAAT box located upstream from the TATA box shown in this segment of DNA, the nucleotide sequence of the transcript will be 5′-ACCCGACAUAGCUACGAUGACGAUA-3′.

11.20 The genome of a human must store a tremendous amount of information using the four nucleotide pairs present in DNA. What does the language of computers tell us about the feasibility of storing large amounts of information using an alphabet composed of just four letters?

ANS: Given the vast amount of information that can be stored on a small computer chip by using a binary code, it is clear that large quantities of genetic information can be stored in the genomes of organisms by using the fourletter alphabet of the genetic code.

11.21 What is the central dogma of molecular genetics? What impact did the discovery of RNA tumor viruses have on the central dogma?

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Answers to All Questions and Problems WC-43

ANS: According to the central dogma, genetic information is stored in DNA and is transferred from DNA to RNA to protein during gene expression. RNA tumor viruses store their genetic information in RNA, and that information is copied into DNA by the enzyme reverse transcriptase after a virus infects a host cell. Thus, the discovery of RNA tumor viruses or retroviruses—retro for backwards flow of genetic information—provided an exception to the central dogma.

11.22 The biosynthesis of metabolite X occurs via six steps catalyzed by six different enzymes. What is the minimal number of genes required for the genetic control of this metabolic pathway? Might more genes be involved?

Why? ANS: If six different enzymes are required for the pathway, then minimally six genes are necessary for genetic control of the pathway. However, because the expression of these enzymes may rely on the product of other genes, such as transcription factors, more than six genes could be involved in genetic control of this pathway.

11.23 What do processes of DNA synthesis, RNA synthesis, and polypeptide synthesis have in common?

ANS: DNA, RNA, and protein synthesis all involve the synthesis of long chains of repeating subunits. All three processes can be divided into three stages: chain initiation, chain elongation, and chain termination.

11.24 What are the two stages of gene expression? Where do they occur in a eukaryotic cell? a prokaryotic cell?

ANS: The two stages of gene expression are as follows: (1) Transcription—the transfer of genetic information from DNA to RNA. (2) Translation—the transfer of genetic information from RNA to protein. In eukaryotes, transcription occurs in the nucleus and translation occurs in the cytoplasm on complex macromolecular structures called ribosomes.

In prokaryotes, transcription and translation are often coupled with mRNA molecules often being translated by ribosomes while still being synthesized during transcription.

11.25 Compare the structures of primary transcripts with those of mRNAs in prokaryotes and eukaryotes. On average, in which group of organisms do they differ the most?

ANS: The primary transcripts of eukaryotes undergo more extensive posttranscriptional processing than those of prokaryotes. Thus, the largest differences between mRNAs and primary transcripts occur in eukaryotes.

Transcript processing is usually restricted to the excision of terminal sequences in prokaryotes. In contrast, eukaryotic transcripts are usually modified by (1) the excision of intron sequences; (2) the addition of 7-methyl guanosine caps to the 5′ termini; (3) the addition of poly(A) tails to the 3′ termini. In addition, the sequences of some eukaryotic transcripts are modified by RNA editing processes.

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11.26 What five types of RNA molecules participate in the process of gene expression? What are the functions of each type of RNA? Which types of RNA perform their function(s) in (a) the nucleus and (b) the cytoplasm?

ANS: The five types of RNA molecules that are involved in gene expression are mRNAs, rRNAs, tRNAs, micro RNAs, and snRNAs. mRNA molecules carry genetic information from genes to the sites of protein synthesis and specify the amino acid sequences of polypeptides. rRNAs are major structural components of the ribosomes and provide functions required for translation. tRNA molecules are the adapters that provide amino acid-codon specificity during translation; each tRNA is activated by a specific amino acid and contains an anticodon sequence that is complementary or partially complementary to one, two, or three codons in mRNAs. Micro-RNAs are involved in regulative gene expression (see Chapter 18). snRNAs are structural components of the spliceosomes that excise introns from gene t­ ranscripts in eukaryotes. snRNAs perform their splicing functions in the nucleus. mRNAs carry information from the nucleus to the cytoplasm, so they function in both compartments of the cell. However, their most ­prominent function is to direct the synthesis of polypeptides during translation, which occurs in the cytoplasm. rRNAs and tRNAs perform their functions during translation in the cytoplasm. Micro-RNAs become incorporated into ribonucleoprotein complexes in the cytoplasm.

11.27 Why was the need for an RNA intermediary in protein synthesis most obvious in eukaryotes? How did researchers first demonstrate that RNA synthesis occurred in the nucleus and that protein synthesis occurred in the cytoplasm?

ANS: In eukaryotes, the genetic information is stored in DNA in the nucleus, whereas proteins are synthesized on ribosomes in the cytoplasm. How could the genes, which are separated from the sites of protein synthesis by a doublemembrane—the nuclear envelope, direct the synthesis of polypeptides without some kind of intermediary to carry the specifications for the polypeptides from the nucleus to the cytoplasm? Researchers first used labeled RNA and protein precursors and autoradiography to demonstrate that RNA synthesis and protein synthesis occurred in the nucleus and the cytoplasm, respectively.

11.28 Two eukaryotic genes encode two different polypeptides, each of which is 335 amino acids long. One gene contains a single exon; the other gene contains an intron of 41,324 nucleotide pairs long. Which gene would you expect to be transcribed in the least amount of time?

Why? When the mRNAs specified by these genes are translated, which mRNA would you expect to be translated in the least time? Why? ANS: Because transcription results in a primary transcript from the DNA template, the DNA sequence for the singleexon gene is the shortest, and so it will be transcribed in

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44-WC Answers to All Questions and Problems the least amount of time. However, because each polypeptide is the same length, the mature mRNAs for both genes will be the same length and will be translated in the same amount of time.

11.29 Design an experiment to demonstrate that RNA transcripts are synthesized in the nucleus of eukaryotes and are subsequently transported to the cytoplasm.

ANS: A simple pulse- and pulse/chase-labeling experiment will demonstrate that RNA is synthesized in the nucleus and is subsequently transported to the cytoplasm. This experiment has two parts: (1) pulse label eukaryotic culture cells by growing them in 3H-uridine for a few minutes and localize the incorporated radioactivity by autoradiography. (2) Repeat the experiment, but this time add a large excess of nonradioactive uridine to the medium in which the cells are growing after the labeling period and allow the cells to grow in the nonradioactive medium for about an hour. Then localize the incorporated radioactivity by autoradiography.

11.30 Total RNA was isolated from human cells growing in culture. This RNA was mixed with nontemplate strands (single strands) of the human gene encoding the enzyme thymidine kinase, and the RNA–DNA mixture was incubated for 12 hours under renaturation conditions. Would you expect any RNA–DNA duplexes to be formed during the incubation? If so, why? If not, why not? The same experiment was then performed using the template strand of the thymidine kinase gene. Would you expect any RNA–DNA duplexes to be formed in this second ­experiment? If so, why? If not, why not?

ANS: RNA–DNA duplexes will be formed when the template strand is used, but not when the nontemplate strand is used. (However, if some hybridization is observed with the nontemplate strand, this is because total RNA was used and is nonspecific to the thymidine kinase gene) Only one strand—the template strand—of most genes is transcribed. Thus, RNA will contain nucleotide sequences complementary to the template strand but not to the nontemplate strand.

11.31 Two preparations of RNA polymerase from E. coli are used in separate experiments to catalyze RNA synthesis in vitro using a purified fragment of DNA carrying the argH gene as template DNA. One preparation catalyzes the synthesis of RNA chains that are highly heterogeneous in size. The other preparation catalyzes the synthesis of RNA chains that are all the same length. What is the most likely difference in the composition of the RNA polymerases in the two preparations?

ANS: The first preparation of RNA polymerase is probably lacking the sigma subunit and, as a result, initiates the synthesis of RNA chains at random sites along both strands of the argH DNA. The second preparation probably contains the sigma subunit and initiates RNA chains only at the site used in vivo, which is governed by the position of the −10 and −35 sequences of the promoter.

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11.32 Transcription and translation are coupled in prokaryotes.

Why is this not the case in eukaryotes?

ANS: In eukaryotes, transcription occurs in the nucleus and translation occurs in the cytoplasm. Because these processes occur in different compartments of the cell, they cannot be coupled as they are in prokaryotes.

11.33 What two elements are almost always present in the promoters of eukaryotic genes that are transcribed by RNA polymerase II? Where are these elements located ­relative to the transcription start site? What are their functions?

ANS: TATA and CAAT boxes. The TATA and CAAT boxes are usually centered at positions −30 and −80, respectively, relative to the startpoint (+1) of transcription. The TATA box is responsible for positioning the transcription startpoint; it is the binding site for the first basal transcription factor that interacts with the promoter.

The CAAT box enhances the efficiency of transcriptional initiation.

11.34 In what ways are most eukaryotic gene transcripts modified? What are the functions of these posttranscriptional modifications?

ANS: (1) Intron sequences are spliced out of gene transcripts to provide contiguous coding sequences for translation. (2) The 7-methyl guanosine caps added to the 5′ termini of most eukaryotic mRNAs help protect them from degradation by nucleases and are recognized by proteins involved in the initiation of translation. (3) The poly(A) tails at the 3′ termini of mRNAs play an important role in their transport from the nucleus to the cytoplasm and enhance their stability.

11.35 How does RNA editing contribute to protein diversity in eukaryotes?

ANS: RNA editing sometimes leads to the synthesis of two or more distinct polypeptides from a single mRNA. 11.36 How do the mechanisms by which the introns of tRNA precursors, Tetrahymena rRNA precursors, and nuclear pre-mRNAs are excised differ? In which process are snRNAs involved? What role(s) do these snRNAs play?

ANS: The introns of tRNA precursors, Tetrahymena rRNA precursors, and nuclear pre-mRNAs are excised by completely different mechanisms. (1) Introns in tRNA are excised by cleavage and joining events catalyzed by splicing nucleases and ligases, respectively. (2) Introns in Tetrahymena rRNA precursors are excised autocatalytically.

(3) Introns of nuclear pre-mRNAs are excised by spliceosomes. snRNAs are involved in nuclear pre-mRNA splicing as structural components of spliceosomes. In addition, snRNA U1 is required for the cleavage events at the 5′ termini of introns; U1 is thought to base-pair with a partially complementary consensus sequence at this position in pre-mRNAs.

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Answers to All Questions and Problems WC-45

11.37 A mutation in an essential human gene changes the 5′ splice site of a large intron from GT to CC. Predict the phenotype of an individual homozygous for this mutation.

12.3 Is the number of potential alleles of a gene directly related to the number of nucleotide pairs in the gene? Is such a relationship more likely to occur in prokaryotes or in eukaryotes? Why?

ANS: This zygote will probably be nonviable because the gene product is essential, and the elimination of the 5′ splice site will almost certainly result in the production of a nonfunctional gene product.

ANS: It depends on how you define alleles. If every variation in nucleotide sequence is considered to be a different allele, even if the gene product and the phenotype of the organism carrying the mutation are unchanged, then the ­number of alleles will be directly related to gene size.

However, if the nucleotide sequence change must produce an altered gene product or phenotype before it is considered a distinct allele, then there will be a positive correlation, but not a direct relationship, between the number of alleles of a gene and its size in nucleotide pairs. The relationship is more likely to occur in prokaryotes where most genes lack introns. In eukaryotic genes, nucleotide sequence changes within introns are usually neutral; that is, they do not affect the activity of the gene product or the phenotype of the organism.

Thus, in the case of eukaryotic genes with introns, there may be no correlation between gene size and number of alleles producing altered phenotypes.

11.38 Total RNA was isolated from nuclei of human cells growing in culture. This RNA was mixed with a purified, denatured DNA fragment that carried a large intron of a housekeeping gene (a gene expressed in essentially all cells), and the RNA–DNA mixture was incubated for 12 hours under renaturation conditions. Would you expect any RNA–DNA duplexes to be formed during the incubation? If so, why? If not, why not? The same experiment was then performed using total cytoplasmic RNA from these cells. Would you expect any RNA–DNA duplexes to be formed in this second experiment? If so, why? If not, why not?

ANS: In the first experiment, yes, some hybridization would be expected because not all of the RNA in the preparation has been completely processed and will still contain the intron sequence. However, if only cytoplasmic RNA is used in the hybridization experiment, all of the mRNA in the preparation will have been processed (because processing occurs in the nucleus) and no hybridization will be observed.

Chapter 12 12.1 In a general way, describe the molecular organization of proteins and distinguish proteins from DNA, chemically and functionally. Why is the synthesis of proteins of particular interest to geneticists?

ANS: Proteins are long chainlike molecules made up of amino acids linked together by peptide bonds. Proteins are composed of carbon, hydrogen, nitrogen, oxygen, and usually sulfur. They provide the enzymatic capacity and much of the structure of living organisms. DNA is composed of phosphate, the pentose sugar 2-deoxyribose, and four nitrogen-containing organic bases (adenine, cytosine, guanine, and thymine). DNA stores and transmits the genetic information in most living organisms. Protein synthesis is of particular interest to geneticists because proteins are the primary gene products—the key intermediates through which genes control the phenotypes of living organisms.

12.2 At what locations in the cell does protein synthesis occur?

ANS: Protein synthesis occurs on ribosomes. In eukaryotes, most of the ribosomes are located in the cytoplasm and are attached to the extensive membranous network of endoplasmic reticulum. Some protein synthesis also occurs in cytoplasmic organelles such as chloroplasts and mitochondria.

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12.4 Why was it necessary to modify Beadle and Tatum’s one gene–one enzyme concept of the gene to one gene–one polypeptide?

ANS: Several enzymes were shown to contain two or more different polypeptides, and these polypeptides were sometimes controlled by genes that mapped to different chromosomes. Thus, the mutations clearly were not in the same gene.

12.5 (a) Why is the genetic code a triplet code instead of a singlet or doublet code? (b) How many different amino acids are specified by the genetic code? (c) How many different amino acid sequences are possible in a polypeptide 146 amino acids long?

ANS: (a) Singlet and doublet codes provide a maximum of 4 and (4)2 or 16 codons, respectively. Thus, neither code would be able to specify all 20 amino acids. (b) 20. (c) (20)146.

12.6 What types of experimental evidence were used to decipher the genetic code?

ANS: Synthetic RNA molecules (polyuridylic acid molecules) containing only the base uracil were prepared. When these synthetic molecules were used to activate in vitro protein synthesis systems, small polypeptide containing only the amino acid phenylalanine (polyphenylalanine molecules) was synthesized. Codons composed only of uracil were thus shown to specify phenylalanine. Similar experiments were carried out using synthetic RNA molecules with different base compositions. Later, in vitro systems activated with synthetic RNA molecules with known repeating base sequences were developed. Ultimately, in vitro systems in which specific aminoacyltRNAs where shown to bind to ribosomes activated with

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46-WC Answers to All Questions and Problems specific mini-mRNAs, which were trinucleotides of known base sequence, were developed and used in codon identification.

12.7 In what sense and to what extent is the genetic code (a) degenerate, (b) ordered, and (c) universal?

ANS: (a) The genetic code is degenerate in that all but 2 of the 20 amino acids are specified by two or more codons.

Some amino acids are specified by six different codons.

The degeneracy occurs largely at the third or 3′ base of the codons. “Partial degeneracy” occurs where the third base of the codon may be either of the two purines or either of the two pyrimidines and the codon still specifies the same amino acid. “Complete degeneracy” occurs where the third base of the codon may be any one of the four bases and the codon still specifies the same amino acid. (b) The code is ordered in the sense that related codons (codons that differ by a single base change) specify chemically similar amino acids. For example, the codons CUU, AUU, and GUU specify the structurally related amino acids leucine, isoleucine, and valine, respectively. (c) The code appears to be almost completely universal. Known exceptions to universality include strains carrying suppressor mutations that alter the reading of certain codons (with low efficiencies in most cases) and the use of UGA as a tryptophan codon in yeast and human mitochondria.

12.8 The thymine analog 5-bromouracil is a chemical mutagen that induces single base-pair substitutions in DNA called transitions (substitutions of one purine for another purine and one pyrimidine for another pyrimidine).

Using the known nature of the genetic code (Table 12.1), which of the following amino acid substitutions should you expect to be induced by 5-bromouracil with the highest frequency: (a) Met → Val; (b) Met → Leu; (c) Lys → Thr; (d) Lys → Gln; (e) Pro → Arg; or (f) Pro → Gln? Why? ANS: (a) Met → Val. This substitution occurs as a result of a transition. All other amino acid substitutions listed would require transversions.

12.9 Using the information given in Problem 12.8, would you expect 5-bromouracil to induce a higher frequency of His → Arg or His → Pro substitutions? Why? ANS: Expect a higher frequency of His → Arg substitutions.

His → Arg results from a transition; His → Pro would require a transversion (not induced by 5-bromouracil).

12.10 What is the minimum number of tRNAs required to recognize the six codons specifying the amino acid leucine?

ANS: Because of wobble (Table 12.2), one tRNA can recognize both UUA and UUG codons for leucine. However, it takes two more tRNAs to recognize CUU, CUC, CUA, and CUG. Therefore, a minimum of three tRNAs are required to recognize the six codons for leucine.

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12.11 Characterize ribosomes in general as to size, location, function, and macromolecular composition.

ANS: Ribosomes are from 10 to 20 nm in diameter. They are located primarily in the cytoplasm of cells. In bacteria, they are largely free in the cytoplasm. In eukaryotes, many of the ribosomes are attached to the endoplasmic reticulum in the cytoplasm. Ribosomes are complex structures composed of over 50 different polypeptides and three to five different RNA molecules. In both prokaryotes and eukaryotes, ribosomes are the site of translation.

12.12 (a) Where in the cells of higher organisms do ribosomes originate? (b) Where in the cells are ribosomes most active in protein synthesis?

ANS: (a) The nucleus, specifically the nucleoli. (b) The cytoplasm.

12.13 Identify three different types of RNA that are involved in translation and list the characteristics and functions of each.

ANS: Messenger RNA (mRNA) molecules carry genetic ­information from the chromosomes (where the information is stored) to the ribosomes in the cytoplasm (where the information is expressed during protein synthesis).

The linear sequence of triplet codons in an mRNA molecule specifies the linear sequence of amino acids in the polypeptide(s) produced during translation of that mRNA. Transfer RNA (tRNA) molecules are small (about 80 nucleotides long) molecules that carry amino acids to the ribosomes and provide the codonrecognition specificity during translation. Ribosomal RNA (rRNA) molecules provide part of the structure and function of ribosomes; they represent an important part of the machinery required for the synthesis of polypeptides.

12.14. (a) How is messenger RNA related to polysome formation? (b) How does rRNA differ from mRNA and tRNA in specificity? (c) How does the tRNA molecule differ from that of DNA and mRNA in size and helical arrangement?

ANS: (a) Polysomes are formed when two or more ribosomes are simultaneously translating the same mRNA molecule. Ribosomes are usually spaced about 90 nucleotides apart on an mRNA molecule. Thus, polysome size is determined by mRNA size. (b) A ribosome, which contains rRNA molecules, can participate in the synthesis of any polypeptide specified by the ribosome-associated mRNA. In that sense, rRNA is nonspecific. Messenger RNAs and tRNAs, in contrast, are specific, in directing the synthesis of a particular polypeptide or set of polypeptides (mRNA) or in attaching to a particular amino acid (tRNA). (c) Transfer RNA molecules are much smaller (about 80 nucleotides) than DNA or mRNA molecules.

They are single-stranded molecules but have complex secondary structures because of the base pairing between different segments of the molecules.

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Answers to All Questions and Problems WC-47

12.15 Outline the process of aminoacyl-tRNA formation.

ANS: A specific aminoacyl-tRNA synthetase catalyzes the formation of an amino acid-AMP complex from the appropriate amino acid and ATP (with the release of pyro­pho­spha­te).

The same enzyme then catalyzes the formation of the ­aminoacyl-tRNA complex, with the release of AMP. Both the amino acid-AMP and aminoacyl-tRNA linkages are high-energy phosphate bonds.

12.16 How is translation (a) initiated and (b) terminated?

ANS: (a) Translation is initiated by a complex reaction involving mRNA, ribosomes, initiation factors (IF-1, IF-2, and IF-3), GTP, the initiator codon AUG, and a special ­initiator tRNA (rRNAfMet). It also appears to involve a base-pairing interaction between a base sequence near the 3′-end of the 16S rRNA and a base sequence in the “leader sequence” of the mRNA. (b) Translation is ­terminated by recognition of one or more of the chaintermination codons (UAG, UAA, and UGA) by the appropriate protein release factor (RF-1 or RF-2).

12.17 Of what significance is the wobble hypothesis?

ANS: Crick’s wobble hypothesis explains how the anticodon of a given tRNA can base-pair with two or three different mRNA codons. Crick proposed that the base-pairing between the 5′ base of the anticodon in tRNA and the 3′ base of the codon in mRNA was less stringent than ­normal and thus allowed some “wobble” at this site. As a result, a single tRNA often recognizes two or three of the related codons specifying a given amino acid (see Table 12.2).

12.18 If the average molecular mass of an amino acid in a particular polypeptide is 100 daltons, about how many nucleotides will be present in an mRNA coding sequence specifying this polypeptide, which has a molecular mass of 27,000 daltons?

ANS: At least 813 nucleotides [= (270 aa × 3) + 3 nucleotides for termination codon].

12.19 The bases A, G, U, C, I (inosine) all occur at the 5′ positions of anticodons in tRNAs. (a) Which base can pair with three different bases at the 3′ positions of codons in mRNA? (b) What is the minimum number of tRNAs required to recognize all codons of amino acids specified by codons with complete degeneracy?

ANS: (a) Inosine. (b) Two. 12.20 Assume that in the year 2025, the first expedition of humans to Mars discovers several Martian life forms thriving in hydrothermal vents that exist below the planet’s surface. Several teams of molecular biologists extract proteins and nucleic acids from these organisms and make some momentous discoveries. Their first discovery is that the proteins in Martian life forms contain only 14 different amino acids instead of the 20 present in life

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forms on Earth. Their second discovery is that the DNA and RNA in these organisms have only two different nucleotides instead of the four nucleotides present in living organisms on Earth. (a) Assuming that transcription and translation work similarly in Martians and Earthlings, what is the minimum number of nucleotides that must be present in the Martian codon to specify all the amino acids in Martians? (b) Assuming that the Martian code proposed above has translational start-and-stop signals, would you expect the Martian genetic code to be degenerate like the genetic code used on Earth?

ANS: (a) Two nucleotides in all combinations of four (24) would produce 16 codons. Therefore, the minimum number of nucleotides comprising the Martian genetic code must be four. (b) Sixteen codons would allow code words for 14 amino acids, one initiation codon, and a translational termination codon. The Martian genetic code could not be degenerate.

12.21 What are the basic differences between translation in prokaryotes and in eukaryotes?

ANS: Translation occurs by very similar mechanisms in prokaryotes and eukaryotes; however, there are some differences. (1) In prokaryotes, the initiation of translation involves base-pairing between a conserved sequence (AGGAGG)—the Shine–Dalgarno box—in mRNA and a complementary sequence near the 3′ end of the 16S rRNA. In eukaryotes, the initiation complex forms at the 5′ end of the transcript when a cap-binding protein interacts with the 7-methyl guanosine on the mRNA. The complex then scans the mRNA processively and initiates translation (with a few exceptions) at the AUG closest to the 5′ terminus. (2) In prokaryotes, the amino group of the initiator methionyl-tRNAfMet is formylated; in eukaryotes, the amino group of methionyl-tRNAiMet is not formylated. (3) In prokaryotes, two soluble protein release factors (RFs) are required for chain termination.

RF-1 terminates polypeptides in response to UAA and UAG codons; RF-2 terminates chains in response to UAA and UGA codons. In eukaryotes, one release factor responds to all three termination codons.

12.22 What is the function of each of the following components of the protein-synthesizing apparatus: (a) Aminoacyl-tRNA synthetase. (b) Release factor 1. (c) Peptidyl transferase. (d) Initiation factors. (e) Elongation factor G ANS: (a) Attachment of an amino acid to the correct tRNA. (b) Recognition of termination codons UAA and UAG and release of the nascent polypeptide from the tRNA in the P site of the ribosome. (c) Formation of a peptide bond between the amino group of the aminoacyl-tRNA in the A site and the carboxyl group of the growing polypeptide on the tRNA in the P site. (d) Formation of the initiation complex required for translation; all steps leading up to peptide bond formation. (e) Translocation of the peptidyl-tRNA from the A site on the ribosome to the P site.

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48-WC Answers to All Questions and Problems 12.23 An E. coli gene has been isolated and shown to be 68 nm long. What is the maximum number of amino acids that this gene could encode?

ANS: Assuming 0.34 nm per nucleotide pair in B-DNA, a gene 68 nm long would contain 200 nucleotide pairs. Given the triplet code, this gene would contain 200/3 = 66.7 triplets, one of which must specify chain termination.

Disregarding the partial triplet, this gene could encode a maximum of 65 amino acids.

12.24 (a) What is the difference between a nonsense mutation and a missense mutation? (b) Are nonsense or missense mutations more frequent in living organisms? (c) Why? ANS: (a) A nonsense mutation changes a codon specifying an amino acid to a chain-termination codon, whereas a missense mutation changes a codon specifying one amino acid to a codon specifying a different amino acid. (b) Missense mutations are more frequent. (c) Of the 64 codons, only three specify chain termination. Thus, the number of possible missense mutations is much larger than the number of possible nonsense mutations. Moreover, nonsense mutations almost always produce nonfunctional gene products. As a result, nonsense mutations in essential genes are usually lethal in the homozygous state.

12.25 The human a-globin chain is 141 amino acids long. How many nucleotides in mRNA are required to encode human a-globin?

ANS: 426 nucleotides—3 × 141 = 423 specifying amino acids plus three (one codon) specifying chain termination.

12.26 What are the functions of the A, P, and E aminoacyltRNA binding sites on the ribosome?

ANS: The incoming aminoacyl-tRNA enters the A site of the ribosome, the nascent polypeptide-tRNA occupies the P site, and the uncharged exiting tRNA occupies the E site.

12.27 (a) In what ways does the order in the genetic code minimize mutational lethality? (b) Why do base-pair changes that cause the substitution of a leucine for a valine in the polypeptide gene product seldom produce a mutant phenotype?

ANS: (a) Related codons often specify the same or very similar amino acids. As a result, single base-pair substitutions frequently result in the synthesis of identical proteins (degeneracy) or proteins with amino acid substitutions involving very similar amino acids. (b) Leucine and valine have very similar structures and chemical properties; both have nonpolar side groups and fold into essentially the same three-dimensional structures when present in polypeptides. Thus, substitutions of leucine for valine or valine for leucine seldom alter the function of a protein.

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12.28 (a) What is the function of the Shine–Dalgarno sequence in prokaryotic mRNAs? (b) What effect does the deletion of the Shine–Dalgarno sequence from an mRNA have on its translation?

ANS: (a) The Shine–Dalgarno sequence is a conserved ­polypurine tract, consensus AGGAGG, that is located about seven nucleotides upstream from the AUG ­initiation codon in mRNAs of prokaryotes. It is complementary to, and is believed to base-pair with, a sequence near the 5′ terminus of the 16S ribosomal RNA. (b) Prokaryotic mRNAs with the Shine–­Dalgarno sequence deleted are either not translated or are translated inefficiently.

12.29 (a) In what ways are ribosomes and spliceosomes similar? (b) In what ways are they different?

ANS: (a) Both ribosomes and spliceosomes play essential roles in gene expression, and both are complex macromolecular structures composed of RNA and protein molecules. (b) Ribosomes are located in the cytoplasm; spliceosomes in the nucleus. Ribosomes are larger and more complex than spliceosomes.

12.30 The 5′ terminus of a human mRNA has the following sequence: 5 ′ - G A AG AG AC A AG G T C AU G G C C AUAU G C UUGUUCCAAUCGUUAGCUGCGCAGGAUCGCCCUGGG . . . . . . 3′ When this mRNA is translated, what amino acid sequence will be specified by this mRNA sequence?

ANS: NH 2-Met-Ala-Ile-Cys-Leu-Phe-Gln-Ser-Leu-AlaAla-Gln-Asp-Arg-Pro-Gly-COOH. 12.31 A partial (5′ subterminal) nucleotide sequence of a prokaryotic mRNA is as follows: 5′-.....AGGAGGCUCGAACAUGUCAAUAUG­CUU GUUCCAAUCGUUAGCUGCGCAGGACCGUCC CG­GA. . . . . . 3′ When this mRNA is translated, what amino acid sequence will be specified by this portion of the mRNA? ANS: Met-Ser-lle-Cys-Leu-Phe-Gln-Ser-Leu-Ala-AlaGln-Asp-Arg-Pro-Gly 12.32 The following DNA sequence occurs in the nontem­plate strand of a gene in a bacterium (the promoter sequence is located to the left but is not shown): ↓ 5′-GAATGTCAGAACTGCCATGCTTCATATGAATAGACCTCTAG-3′ (a) What is the ribonucleotide sequence of the mRNA molecule that is transcribed from this piece of DNA? (b) What is the amino acid sequence of the polypeptide encoded by this mRNA? (c) If the nucleotide indicated

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Answers to All Questions and Problems WC-49

by the arrow undergoes a mutation that changes T to A, what will be the resulting amino acid sequence following transcription and translation?

ANS: (a) 5′-GAAUGUCAGAACUGCCAUGCUUCAUAUG AAUAGACCUCUAG-3′ (b) NH2-fMet-Ser-Glu-LeuPro-Cys-Phe-Ile-COOH (c) NH2-fMet-Ser-Glu-LeuPro-Cys-Phe-Ile-Arg-Ile-Asp-Leu-COOH. 12.33 Alan Garen extensively studied a particular nonsense (chain-termination) mutation in the alkaline phosphatase gene of E. coli. This mutation resulted in the termination of the alkaline phosphatase polypeptide chain at a position where the amino acid tryptophan occurred in the wild-type polypeptide. Garen induced revertants (in this case, mutations altering the same codon) of this mutant with chemical mutagens that induced single base-pair substitutions and sequenced the polypeptides in the revertants. Seven different types of revertants were found, each with a different amino acid at the tryptophan position of the wild-type polypeptide (termination position of the mutant polypeptide fragment). The amino acids present at this position in the various revertants included tryptophan, serine, tyrosine, leucine, glutamic acid, glutamine, and lysine. Did the nonsense mutation studied by Garen contain a UAG, a UAA, or a UGA nonsense mutation? Explain the basis of your deduction.

ANS: (UAG). This is the only nonsense codon that is related to tryptophan, serine, tyrosine, leucine, glutamic acid, glutamine, and lysine codons by a single base-pair substitution in each case.

12.34 The following DNA sequence occurs in a bacterium (the promoter sequence is located to the left but is not shown). ↓

Chapter 13 13.1 Identify the following point mutations represented in DNA and in RNA as (1) transitions, (2) transversions, or (3) reading frameshifts. (a) A to G; (b) C to T; (c) C to G; (d) T to A; (e) UAU ACC UAU to UAU AAC CUA; (f) UUG CUA AUA to UUG CUG AUA. ANS: (a) Transition, (b) transition, (c) transversion, (d) transversion, (e) frameshift, (f) transition.

13.2 Of all possible missense mutations that can occur in a segment of DNA encoding the amino acid tryptophan, what is the ratio of transversions to transitions if all single base-pair substitutions occur at the same frequency?

ANS: 6:1. UGG transitions: UGA (nonsense), UAG (nonsense), CGG (Arg). UGG transversions: UGC (Cys), UGU (Cys), UCG (Ser), UUG (Leu), AGG (Arg), GGG (Gly).

13.3 Both lethal and visible mutations are expected to occur in fruit flies that are subjected to irradiation. Outline a method for detecting (a) X-linked lethals and (b) X-linked visible mutations in irradiated Drosophila.

ANS: (a) ClB method, (b) attached X method (see Chapter 6).

13.4 H. J. Muller used the ClB technique to identify many radiation-induced recessive lethal mutations on Drosophila’s X chromosome, which is now known to contain more than a thousand genes. These mutations could be propagated in stock cultures by keeping them in heterozygous condition with the ClB chromosome.

Would you expect all these lethal mutations to be alleles of one essential X-linked gene, or to be alleles of different essential X-linked genes? Why couldn’t H. J. Muller determine the answer to this question experimentally?

(c) If the nucleotide indicated by the arrow undergoes a mutation that causes this C:G base pair to be deleted, what will be the polypeptide encoded by the mutant gene?

ANS: The radiation-induced recessive lethal mutations that Muller recovered in his experiments were likely scattered across the entire X chromosome. Thus, they likely affected different genes. However, Muller could not study this issue because he could not carry out a complementation test to determine if any of the lethal mutations were alleles of the same gene. The reason is that to perform the complementation test, Muller would have had to cross females that carried a particular lethal mutation balanced with the ClB chromosome to males that carried a different lethal mutation. Because such males are not viable, the required cross cannot be performed.

ANS: (a) 5′-CAAUCAUGGACUGCCAUGCUUCAUAUG AAUAGUUGACAU-3′ (b) NH2-fMet-Asp-Cys-His-AlaSer-Tyr-Glu-COOH (c) NH2-fMet-Asp-Cys­-Met-LeuHis­-Met-Asn-Ser- COOH.

13.5 Published spontaneous mutation rates for humans are generally higher than those for bacteria. Does this indicate that individual genes of humans mutate more frequently than those of bacteria? Explain.

5′-CAATCATGGACTGCCATGCTTCATATGAATAGTTGACAT-3′ 3′-GTTAGTACCTGACGGTACGAAGTATACTTATCAACTGTA-5′

(a) What is the ribonucleotide sequence of the mRNA molecule that is transcribed from the template strand of this piece of DNA? Assume that both translational start and termination codons are present. (b) What is the amino acid sequence of the polypeptide encoded by this mRNA?

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50-WC Answers to All Questions and Problems ANS: Probably not. A human is larger than a bacterium, with more cells and a longer life span. If mutation frequencies are calculated in terms of cell generations, the rates for human cells and bacterial cells are similar.

13.6 A precancerous condition (intestinal polyposis) in a particular human family group is caused by a single dominant gene. Among the descendants of one woman who died with cancer of the colon, 10 people have died with the same type of cancer and 6 now have intestinal polyposis. All other branches of the large kindred have been carefully examined, and no cases have been found. Suggest an explanation for the origin of the defective gene.

ANS: A dominant mutation presumably occurred in the woman in whom the condition was first known.

13.7 Juvenile muscular dystrophy in humans depends on an X-linked recessive gene. In an intensive study, 33 cases were found in a population of some 800,000 people. The investigators were confident that they had found all cases that were well enough advanced to be detected at the time the study was made. The symptoms of the disease were expressed only in males. Most of those with the disease died at an early age, and none lived beyond 21 years of age. Usually, only one case was detected in a family, but sometimes two or three cases occurred in the same family. Suggest an explanation for the sporadic occurrence of the disease and the tendency for the gene to persist in the population.

ANS: The X-linked gene is carried by mothers, and the disease is expressed in half of their sons. Such a disease is difficult to follow in pedigree studies because of the recessive nature of the gene, the tendency for the expression to skip generations in a family line, and the loss of the males who carry the gene. One explanation for the sporadic occurrence and tendency for the gene to persist is that, by mutation, new defective genes are constantly being added to the load already present in the population.

13.8 Products resulting from somatic mutations, such as the navel orange and the Delicious apple, have become widespread in citrus groves and apple orchards. ­However, traits resulting from somatic mutations are seldom maintained in animals. Why? ANS: Plants can be propagated vegetatively, but no such methods are available for widespread use in animals.

13.9 If a single short-legged sheep should occur in a flock, suggest experiments to determine whether the short legs are the result of a mutation or an environmental effect. If due to a mutation, how can one determine whether the mutation is dominant or recessive?

ANS: The sheep with short legs could be mated to unrelated animals with long legs. If the trait is expressed in the first generation, it could be presumed to be inherited and to depend on a dominant gene. On the other hand, if it does not appear in the first generation, F1 sheep could be

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crossed back to the short-legged parent. If the trait is expressed in one-half of the backcross progeny, it is probably inherited as a simple recessive. If two short-legged sheep of different sex could be obtained, they could be mated repeatedly to test the hypothesis of dominance. In the event that the trait is not transmitted to the progeny that result from these matings, it might be considered to be environmental or dependent on some complex genetic mechanism that could not be identified by the simple test used in the experiments.

13.10 How might enzymes such as DNA polymerase be in­volved in the mode of action of both mutator and antimutator genes (mutant genes that increase and decrease, respectively, mutation rates)?

ANS: Enzymes may discriminate among the different nucleotides that are being incorporated. Mutator enzymes may utilize a higher proportion of incorrect nucleotides, whereas antimutator enzymes may select fewer incorrect bases in DNA replication. In the case of the phage T4 DNA polymerase, the relative efficiencies of polymerization and proofreading by the polymerase’s 3′ → 5′ exonuclease activity play key roles in determining the mutation rate.

13.11 How could spontaneous mutation rates be optimized by natural selection?

ANS: If both mutators and antimutators operate in the same living system, an optimum mutation rate for a particular organism in a given environment may result from natural selection.

13.12 A mutator gene Dt in maize increases the rate at which the gene for colorless aleurone (a) mutates to the dominant allele (A), which yields colored aleurone. When reciprocal crosses were made (i.e., seed parent dt/dt, a/a × Dt/Dt, a/a and seed parent Dt/Dt, a/a × dt/dt, a/a), the cross with Dt/Dt seed parents produced three times as many dots per kernel as the reciprocal cross. Explain these results.

ANS: Dt is a mutator gene that induces somatic mutations in developing kernels.

13.13 The deficiency Df(1)wrJ1 removes 16 contiguous bands from a region near the left end of the Drosophila X chromosome. Females homozygous for this deficiency die.

However, females heterozygous for it and a ClB chromosome are viable and fertile. If such females are mated to males that carry wild-type X and Y chromosomes, what kinds of progeny will appear and in what proportions?

ANS: The cross is Df(1)wrJ1/ClB females × +/Y (wild-type) males. The genotypes of the daughters are Df(1)wrJ1/+ (phenotypically wild-type) and ClB/+ (bar-eyed). These two classes of daughters will occur in equal proportions.

The sons inherit a Y chromosome and either the Df(1) wrJ1 or ClB X chromosomes, both of which act as recessive lethals. Thus, no sons will appear in the progeny.

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Answers to All Questions and Problems WC-51

13.14 In Drosophila, the Y chromosome Y·w+ has a small piece of the X chromosome translocated to it; this piece contains the wild-type alleles of all the genes missing in Df(1)wrJ1 mentioned in Problem 13.13. If males carrying Y·w+ and a wild-type X chromosome are crossed to Df(1) wrJ1/ClB females, what kinds of progeny will appear, and in what proportions? How would your answer change if the wild-type X chromosome in the males carried a radiation-induced recessive lethal mutation located within the region that is missing in Df(1)wrJ1? How could these unusual chromosomes be used to devise a scheme that would allow you to carry out complementation tests between two independently induced recessive lethal mutations that map within this region?

ANS: In the cross Df(1)wrJ1/ClB females × +/Y·w+ (wild-type) males, the daughters will be either Df(1)wrJ1/+ (phenotypically wild-type) or ClB/+ (bar-eyed); these two types of daughters will appear in equal proportions. The sons from the cross will be either Df(1)wrJ1/ Y·w+ (viable and phenotypically wild-type because the small piece of the X chromosome translocated to the Y chromosome contains the genes that are missing in the deficiency Df(1) wrJ1) or ClB/ Y·w+. This latter genotype will be viable if the lethal mutation in the ClB chromosome resides in the region defined by the deficiency Df(1)wrJ1. In that case, bar-eyed males would appear and be as frequent as wildtype males. If the lethal mutation in the ClB chromosome resides outside the region defined by Df(1)wrJ1, then no bar-eyed males will appear among the progeny.

If the males in the cross carry a radiation-induced lethal (ril) mutation in the region defined by Df(1)wrJ1—that is, if the cross is Df(1)wrJ1/ClB females × ril/Y·w+ males (viable because the small piece of the X chromosome carried by the Y chromosome includes the wild-type allele of ril)—there will not be any wild-type daughters (genotype Df(1)wrJ1/ril) among the progeny. However, bar-eyed daughters (genotypically ClB/ril should appear unless the lethal mutation in the ClB chromosome is allelic to ril.

An induced lethal mutation that lies within the region defined by Df(1)wrJ1 could be tested for complementation with another such lethal mutation. Let’s call the first mutation lethal-1 and the second mutation lethal-2. The required cross for the complementation test is ClB/lethal-1 females × lethal-2/Y·w+ males. This cross is possible because the small piece of the X chromosome carried by the Y·w+ chromosome contains the wild-type allele of lethal-2; thus, the lethal-2/Y·w+ males are viable. Among the progeny, we would look for daughters with the genotype lethal-1/lethal-2, which, because they lack the ClB chromosome, will have normal (non-bar) eyes. If these females appear, we know that lethal-1 and lethal-2 complement one another; that is, they are mutations in different genes. If these females do not appear, we can conclude that lethal-1 and lethal-2 are alleles of the same gene.

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13.15 If CTT is a DNA triplet (transcribed strand of DNA) specifying glutamic acid, what DNA and mRNA base triplet alterations could account for valine and lysine in position 6 of the b-globin chain?

ANS: Amino Acid

mRNA

DNA

Glumatic acid

GAA

Valine

GAA ..... ..

Transcribed strand CTT Mutation

GUA

Lysine

AAA

GTA ..... ..

CAT Mutation AAA ... .. ..

TTT

13.16 The bacteriophage T4 genome contains about 50 percent A:T base pairs and 50 percent G:C base pairs. The base analog 2-aminopurine induces A:T → G:C and G:C → A:T base-pair substitutions by undergoing tautomeric shifts. Hydroxylamine is a mutagenic chemical that reacts specifically with cytosine and induces only G:C → A:T substitutions. If a large number of independent mutations were produced in bacteriophage T4 by treatment with 2-aminopurine, what percentage of these mutations should you expect to be induced to mutate back to the wild-type genotype by treatment with hydroxylamine?

ANS: About half of the induced mutations would be expected to mutate back to the wild-type genotype.

13.17 Assuming that the b-globin chain and the a-globin chain shared a common ancestor, what mechanisms might explain the differences that now exist in these two chains?

What changes in DNA and mRNA codons would account for the differences that have resulted in unlike amino acids at corresponding positions?

ANS: Mutations: transitions, transversions, and frameshifts.

13.18 In a given strain of bacteria, all of the cells are usually killed when a specific concentration of streptomycin is present in the medium. Mutations that confer resistance to streptomycin occur. The streptomycin-resistant mutants are of two types: some can live with or without streptomycin; others cannot survive unless this drug is present in the medium. Given a streptomycin-sensitive strain of this species, outline an experimental procedure by which streptomycin-resistant strains of the two types could be established.

ANS: Irradiate the nonresistant strain and plate the irradiated organisms on a medium containing streptomycin. Those that survive and produce colonies are resistant. They could then be replicated to a medium without streptomycin. Those that survive would be of the first type; those that can live with streptomycin but not without it would be the second type.

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52-WC Answers to All Questions and Problems 13.19 One stock of fruit flies was treated with 1000 roentgens (r) of X-rays. The X-ray treatment increased the mutation rate of a particular gene by 2 percent. What percentage increases in the mutation rate of this gene would be expected if this stock of flies was treated with X-ray doses of 1500 r, 2000 r, and 3000 r?

ANS: 3%; 4%; 6%.

13.20 Why does the frequency of chromosome breaks induced

by X-rays vary with the total dosage and not with the rate at which it is delivered?

ANS: Each quantum of energy from the X-rays that is absorbed in a cell has a certain probability of hitting and breaking a chromosome. Hence, the greater the number of quanta of energy or dosage, the more likely the breaks are to occur. The rate at which this dosage is delivered does not change the probability of each quantum inducing a break.

13.21 A reactor overheats and produces radioactive tritium (H3), radioactive iodine (I131), and radioactive xenon (Xn133). Why should we be more concerned about radioactive iodine than the other two radioactive isotopes?

13.24 How does nitrous acid induce mutations? What specific end results might be expected in DNA and mRNA from the treatment of viruses with nitrous acid?

ANS: Nitrous acid brings about a substitution of an OH group for an NH2 group in those bases (A, C, and G) having NH2 side groups. In so doing, adenine is converted into hypoxanthine, which base-pairs with cytosine, and cytosine is converted into uracil, which base-pairs with adenine. The net effects are GC ↔ AT base-pair substitutions (see Figure 13.16).

13.25 Are mutational changes induced by nitrous acid more likely to be transitions or transversions?

ANS: Transitions.

13.26 You are screening three new pesticides for potential mutagenicity by using the Ames test. Two his− strains resulting from either a frameshift or a transition mutation were used and produced the following results (number of revertant colonies): Transition Mutant Control (no chemical)

Transion Mutant ∙ Chemical

Transion Mutant ∙ Chemical ∙ Rat liver Enzymes

ANS: Radioactive iodine is concentrated by living organisms and food chains.

Strain 1

13.22 One person was in an accident and received 50 roentgens (r) of X-rays at one time. Another person received 5 r in each of 20 treatments. Assuming no intensity effect, what proportionate number of mutations would be expected in each person?

Pesticide #1

21

180

19

Pesticide #2

18

19

17

Pesticide #3

25

265

270

ANS: The person receiving a total of 100 r would be expected to have twice as many mutations as the one receiving 50 r.

Strain 2

Frameshift Mutant Control (no chemical)

Frameshift Mutant ∙ Chemical

Frameshift Mutant ∙ Chemical ∙ Rat liver Enzymes

Pesticide #1

5

4

5

Pesticide #2

7

5

93

Pesticide #3

6

9

7

13.23 A cross was performed in Neurospora crassa between a strain of mating type A and genotype x+ m+ z and a strain of mating type a and genotype x m z+. Genes x, m, and z are closely linked and are present in the order x–m–z on the chromosome. An ascus produced from this cross contained two copies (“identical twins”) of each of the four products of meiosis. If the genotypes of the four products of meiosis showed that gene conversion had occurred at the m locus and that reciprocal recombination had occurred at the x and z loci, what might the genotypes of the four products look like? In the parentheses below, write the genotypes of the four haploid products of meiosis in an ascus showing gene conversion at the m locus and reciprocal recombination of the flanking markers (at the x and z loci).

Ascus Spore Pairs 1–2 ( )

3–4 (

5–6 )

(

7–8 )

( )

ANS: ( x+ m+ z) (x+ m+ z+) (x m+ z) (x m z+) or equivalent.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 52

What type of mutations, if any, do the three pesticides induce?

ANS: P #1—Causes transition mutation. Liver enzymes convert it into nonmutagen. Does not cause frameshift mutations.

P #2—Does not cause transition mutations. Liver enzymes convert it into a frameshift mutagen. P #3—Causes ­transition mutations. Liver enzymes have no effect on mutagenicity. Does not cause frameshift mutations.

13.27 How does the action and mutagenic effect of 5-bromouracil differ from that of nitrous acid?

ANS: Nitrous acid acts as a mutagen on either replicating or nonreplicating DNA and produces transitions from A to G or C to T, whereas 5-bromouracil does not affect nonreplicating DNA but acts during the replication process causing GC ↔ AT transitions. 5-Bromouracil must be incorporated into DNA during the replication process in order to induce mispairing of bases and thus mutations.

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13.28 Sydney Brenner and A. O. W. Stretton found that nonsense mutations did not terminate polypeptide synthesis in the rII gene of the bacteriophage T4 when these mutations were located within a DNA sequence interval in which a single nucleotide insertion had been made on one end and a single nucleotide deletion had been made on the other. How can this finding be explained?

ANS: The reading frame will be shifted between the two frameshift mutations. This shift in reading frame does not read the normal nonsense codon and termination does not occur.

13.29 Seymour Benzer and Ernst Freese compared spontaneous and 5-bromouracil-induced mutants in the rII gene of the bacteriophage T4; the mutagen increased the mutation rate (rII+ → rII) several hundred times above the spontaneous mutation rate. Almost all (98 percent) of the 5-bromouracil-induced mutants could be induced to revert to wild-type (rII → rII+) by 5-bromouracil treatment, but only 14 percent of the spontaneous mutants could be induced to revert to wild-type by this treatment.

Discuss the reason for this result.

ANS: 5-BU causes GC ↔ AT transitions. 5-BU can, therefore, revert almost all of the mutations that it induces by enhancing the transition event that is the reverse of the one that produced the mutation. In contrast, the spontaneous mutations will include transversions, frameshifts, deletions, and other types of mutations, including ­transitions. Only the spontaneous transitions will show enhanced reversion after treatment with 5-BU. 13.30 How do acridine-induced changes in DNA result in inactive proteins?

Mutant 5-ACGTTATGCCCGTACTGCCAGCTAA CTGCTAAAGAACAATTA….-3 (a) What type of mutation is present in the mutant hemoglobin gene? (b) What are the codons in the translated portion of the mRNA transcribed from the normal and mutant genes? (c) What are the amino acid sequences of the normal and mutant polypeptides?

ANS: (a) Frameshift due to the insertion of C at the 9th, 10th, or 11th nucleotide from the 5′ end. (b) Normal: 5′-AUGCCGUACUGCCAGCUAACUGCUAAAGAACAAUUA-3′.

Mutant: 5′-AUGCCCGUACUGCCAGCUAACUGC­ UAAAGAACAAUUA-3′. (c) Normal: NH2-Met-Pro-TyrCys-Gln-Leu-Thr-Ala-Lys-Glu-Gln-Leu. Mutant: NH2Met-Pro-Val-Leu-Pro-Ala-Asn-Cys. 13.32 Bacteriophage MS2 carries its genetic information in RNA. Its chromosome is analogous to a polygenic molecule of mRNA in organisms that store their genetic information in DNA. The MS2 minichromosome encodes four polypeptides (i.e., it has four genes). One of these four genes encodes the MS2 coat protein, a polypeptide of 129 amino acids long. The entire nucleotide sequence in the RNA of MS2 is known. Codon 112 of the coat protein gene is CUA, which specifies the amino acid leucine. If you were to treat a replicating population of bacteriophage MS2 with the mutagen 5-bromouracil, what amino acid substitutions would you expect to be induced at position 112 of the MS2 coat protein (i.e., Leu → other amino acid)? (Note: Bacteriophage MS2 RNA replicates using a complementary strand of RNA and base-pairing as DNA.) ANS: Proline and serine.

ANS: Mutations induced by acridine dyes are primarily insertions or deletions of single base-pairs. Such mutations alter the reading frame (the in-phase triplets specifying mRNA codons) for that portion of the gene distal (relative to the direction of transcription and translation) to the mutation (see Figure 13.7b). This would be expected to totally change the amino acid sequences of polypeptides distal to the mutation site and produce inactive polypeptides. In addition, such frameshift mutations frequently produce in-frame termination codons that result in truncated proteins.

13.33 Would the different amino acid substitutions induced by 5-bromouracil at position 112 of the coat polypeptide that you indicated in Problem 13.32 be expected to occur with equal frequency? If so, why? If not, why not? Which one(s), if any, would occur more frequently?

Use the known codon-amino acid assignments (Table 12.1) to work the following problems.

13.34 Would such mutations occur if a nonreplicating suspension of MS2 phage was treated with 5-bromouracil?

13.31 Mutations in the genes encoding the α- and β-subunits of hemoglobin lead to blood diseases such as thalassemia and sickle-cell anemia. You have found a family in China in which some members suffer from a new genetic form of anemia. The DNA sequences at the 5′ end of the nontemplate strand of the normal and mutant DNA encoding the α subunit of hemoglobin are as follows:

ANS: No. 5-Bromouracil is mutagenic only to replicating nucleic acids.

Normal 5-ACGTTATGCCGTACTGCCAGCTAAC TGCTAAAGAACAATTA……-3

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 53

ANS: No. Leucine → proline would occur more frequently.

Leu (CUA) — 5-BU → Pro (CCA) occurs by a single base-pair transition, whereas Leu (CUA) — 5-BU → Ser (UCA) requires two base-pair transitions. Recall that 5-bromouracil (5-BU) induces only transitions (see Figure 13.15).

13.35 Recall that nitrous acid deaminates adenine, cytosine, and guanine (adenine → hypoxanthine, which base-pairs with cytosine; cytosine → uracil, which base-pairs with adenine; and guanine → xanthine, which base-pairs with cytosine).

Would you expect nitrous acid to induce any mutations that result in the substitution of another amino acid for a glycine residue in a wild-type polypeptide (i.e., glycine →

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54-WC Answers to All Questions and Problems another amino acid) if the mutagenesis were carried out on a suspension of mature (nonreplicating) T4 bacteriophage? (Note: After the mutagenic treatment of the phage suspension, the nitrous acid is removed. The treated phage is then allowed to infect E. coli cells to express any induced mutations.) If so, by what mechanism? If not, why not?

ANS: Yes: DNA:

GGX

GGX HNO2

CCX'

UCX'

mRNA:

GGX

AGX

Polypeptide:

Gly

Ser or Arg (depending on X) or

DNA:

GGX

GGX HNO2

CCX'

CUX'

mRNA:

GGX

GAX

Polypeptide:

Gly

Asp or Glu (depending on X) or

DNA:

GGX CCX'

GGX HNO2

UUX'

mRNA:

GGX

AAX

Polypeptide:

Gly

Asn or Lys (depending on X)

Note: The X at the third position in each codon in mRNA and in each triplet of base pairs in DNA refers to the fact that there is complete degeneracy at the third base in the glycine codon. Any base may be present in the codon, and it will still specify glycine.

13.36 Keeping in mind the known nature of the genetic code, the information given about phage MS2 in Problem 13.32, and the information you have learned about nitrous acid in Problem 13.35, would you expect nitrous acid to induce any mutations that would result in amino acid substitutions of the type glycine → another amino acid if the mutagenesis were carried out on a suspension of mature (nonreplicating) MS2 bacteriophage? If so, by what mechanism? If not, why not?

ANS: No. The glycine codon is GGX, where X can be any one of the four bases. Because of this complete degeneracy at the third position of the glycine codon, changing X to any other base will have no effect (i.e., the codon will still specify glycine). Nitrous acid deaminates guanine (G) to xanthine, but xanthine still base-pairs with cytosine. Thus, guanine is not a target for mutagenesis by nitrous acid.

13.37 Would you expect nitrous acid to induce a higher frequency of Tyr → Ser or Tyr → Cys substitutions? Why?

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ANS: Tyr → Cys substitutions; Tyr to Cys requires a transition, which is induced by nitrous acid. Tyr to Ser would require a transversion, and nitrous acid is not expected to induce transversions.

13.38 Which of the following amino acid substitutions should you expect to be induced by 5-bromouracil with the highest frequency? (a) Met → Leu; (b) Met → Thr; (c) Lys → Thr; (d) Lys → Gln; (e) Pro → Arg; or (f) Pro → Gln? Why? ANS: (b) Met → Thr. 5-Bromouracil induces transitions, not transversions. All other changes listed require transversions.

13.39 The wild-type sequence of part of a protein is NH2-Trp-Trp-Trp-Met-Arg-Glu-Trp-Thr-Met Each mutant in the following table differs from wildtype by a single point mutation. Using this information, determine the mRNA sequence coding for the wild-type polypeptide. If there is more than one possible nucleotide, list all possibilities.

Mutant

Amino Acid Sequence of Polypeptide

1

Trp-Trp-Trp Met

2

Trp-Trp-Trp-Met-Arg-Asp-Trp-Thr-Met

3

Trp-Trp-Trp-Met-Arg-Lys-Trp-Thr-Met

4

Trp-Trp-Trp-Met-Arg-Glu-Trp-Met-Met

ANS: 5′-UGG-UGG-UGG-AUG-CGA(or AGA)-GAA(or GAG)-UGG-AUG-3′ 13.40 Acridine dyes such as proflavin are known to induce primarily single base-pair additions and deletions. Suppose that the wild-type nucleotide sequence in the mRNA produced from a gene is

5′-AUGCCCUUUGGGAAAGGGUUUCCCUAA-3′

Also, assume that a mutation is induced within this gene by proflavin, and, subsequently, a revertant of this mutation is similarly induced with proflavin and shown to result from a second-site suppressor mutation within the same gene. If the amino acid sequence of the polypeptide encoded by this gene in the revertant (double mutant) strain is

NH2-Met-Pro-Phe-Gly-Glu-Arg-Phe-Pro-COOH what would be the most likely nucleotide sequence in the mRNA of this gene in the revertant (double mutant)?

ANS: 5′-AUGCCCUUUGGGGAAAGGUUUCCCUAA-3′ 13.41 Eight independently isolated mutants of E. coli, all of which are unable to grow in the absence of histidine (his−), were examined in all possible cis and trans heterozygotes (partial diploids). All of the cis heterozygotes were able to grow in the absence of histidine. The trans heterozygotes yielded two different responses: some of them grew in the absence of histidine; others did not. The experimental results, using “+” to indicate growth and “0” to indicate no growth, are given in the accompanying table. How many genes are defined by these eight mutations? Which mutant strains carry mutations in the same gene(s)?

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Growth of Trans Heterozygotes (without Histidine) Mutant

1

2

3

4

5

6

7

8

8

0

0

0

0

0

0

1

0

7

+

+

+

+

+

+

0

6

0

0

0

0

0

0

5

0

0

0

0

0

4

0

0

0

0

3

0

0

0

2

0

0

1

0

ANS: Two genes; mutations 1, 2, 3, 4, 5, 6, and 8 are in one gene; mutation 7 is in a second gene.

13.42 Assume that the mutants described in Problem 13.41 yielded the following results. How many genes would they have defined? Which mutations would have been in the same gene(s)?

Mutant

1

2

3

4

5

6

7

8

8

+

+

+

+

+

+

0

0

7

+

+

+

+

+

+

0

6

+

+

+

+

0

0

5

+

+

+

+

0

4

+

+

0

0

3

+

+

0

2

0

0

1

0

ANS: Four genes; mutations 1 and 2 in one gene; mutations 3 and 4 in a second gene; mutations 5 and 6 in a third gene; and mutations 7 and 8 in a fourth gene.

13.43 In Drosophila, white, white cherry, and vermilion are all sexlinked mutations affecting eye color. All three mutations are recessive to their wild-type allele(s) for red eyes. A white-eyed female crossed with a vermilion-eyed male produces white-eyed male offspring and red-eyed (wildtype) female offspring. A white-eyed female crossed with a white cherry-eyed male produces white-eyed sons and light cherry-eyed daughters. Do these results indicate whether or not any of the three mutations affecting eye color are located in the same gene? If so, which mutations?

ANS: The complementation test for allelism involves placing mutations pairwise in a common protoplasm in the trans configuration and determining whether the resulting trans heterozygotes have wild-type or mutant phenotypes. If the two mutations are in different genes, the two mutations will complement each other, because the wild-type copies of each gene will produce functional

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 55

gene products (see Figure 13.21a). However, if the two mutations are in the same gene, both copies of the gene in the trans heterozygote will produce defective gene products, resulting in a mutant phenotype (see Figure 13.21b).

When complementation occurs, the trans heterozygote will have the wild-type phenotype. Thus, the complementation test allows one to determine whether any two recessive mutations are located in the same gene or in different genes. Because the mutations of interest are sex-linked, all the male progeny will have the same phenotype as the female parent. They are hemizygous, with one X chromosome obtained from their mother. In contrast, the female progeny are trans heterozygotes. In the cross between the white-eyed female and the vermilion-eyed male, the female progeny have red eyes, the wild-type phenotype. Thus, the white and vermilion mutations are in different genes, as illustrated in the ­following diagram: trans heterozygote X chromosome from parent X chromosome from parent

v+

w

v+ gene product

v

w+

w+ gene product

Complementation yields wild-type phenotype; both v+ and w+ gene products are produced in the trans heterozygote.

In the cross between a white-eyed female and a white cherry-eyed male, the female progeny have light cherrycolored eyes (a mutant phenotype), not wild-type red eyes as in the first cross. Since the trans heterozygote has a mutant phenotype, the two mutations, white and white cherry, are in the same gene: trans heterozygote X chromosome from parent

X chromosome from parent

w No active (w+) gene product wch

No w+ gene product; therefore, mutant phenotype.

13.44 The loz (lethal on Z) mutants of bacteriophage X are conditional lethal mutants that can grow on E. coli strain Y but cannot grow on E. coli strain Z. The results shown in the following table were obtained when seven loz mutants were analyzed for complementation by infecting E. coli strain Z with each possible pair of mutants.

A “+” indicates that progeny phage were produced in the infected cells, and a “0” indicates that no progeny phage were produced. All possible cis tests were also done, and all cis heterozygotes produced wild-type yields of progeny phage.

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1

2

3

4

5

6

7

Chapter 14

7

+

+

0

+

0

0

0

6

+

+

+

+

+

0

14.1 (a) In what ways is the introduction of recombinant DNA molecules into host cells similar to mutation? (b) In what ways is it different?

5

+

+

0

+

0

4

0

0

+

0

3

+

+

0

2

0

0

1

0 Propose three plausible explanations for the apparently anomalous complementation behavior of loz mutant number 7. (b) What simple genetic experiments can be used to distinguish between the three possible explanations? (c) Explain why specific outcomes of the proposed experiments will distinguish between the three possible explanations.

ANS: Mutations 1, 2, and 4 do not complement one another and thus appear to be located in the same gene, as do mutations 3 and 5. The anomaly is that mutation 7 does not complement mutations 3, 5, and 6, even though mutation 6 does complement mutations 3 and 5. (a) There are three simple explanations of the seemingly anomalous complementation behavior of mutation 7. (1) It is a deletion spanning all or parts of two genes. (2) It is a double mutation with defects in two genes. (3) It is a polar mutation—a nonsense mutation that interferes with the expression of downstream genes—in the promoter-proximal gene of a multigenic transcription unit. (b) Three simple genetic operations will distinguish between these three possibilities.

(1) Reversion. Plate a large number of mutant 7 phage on E. coli strain Z and look for wild-type revertants.

(2) Backcross mutant 7 to wild-type phage and test the mutant progeny for the ability to complement mutations 3 and 6. (3) Introduce F’s carrying tRNA nonsense suppressor genes into E. coli strain Z and determine whether any of them suppress the loz7 mutation. (c) If mutation 7 is a deletion, it will not revert, and, if it is a double mutation, the reversion rate will probably be below the level of detection in your experiment. On the other hand, if it is a polar nonsense mutation, loz+ revertants will be obtained. If mutation 7 is a deletion, no new genotypes will be produced in the backcross to wild-type. However, if it is a double mutation, some recombinant single-mutant progeny will be produced in the backcross to wildtype, and these single mutations will complement either mutation 3 or mutation 6. If mutation 7 is a polar nonsense mutation, it should be suppressed by one or more of the tRNA suppressor genes introduced into E. coli strain Z.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 56

ANS: (a) Both introduce new genetic variability into the cell. In both cases, only one gene or a small segment of DNA representing a small fraction of the total genome is changed or added to the genome. The vast majority of the genes of the organism remain the same. (b) The introduction of recombinant DNA molecules, if they come from a very different species, is more likely to result in a novel, functional gene product in the cell, if the introduced gene (or genes) is capable of being expressed in the foreign protoplasm. The introduction of recombinant DNA molecules is more analogous to duplication mutations (see Chapter 6) than to other types of mutations.

14.2 Listed in this question are four different single strands of DNA. Which of these, in their double-stranded form, would you expect to be cleaved by a restriction endonuclease? (a) ACTCCAGAATTCACTCCG (b) GCCTCATTCGAAGCCTGA (c) CTCGCCAATTGACTCGTC (d) ACTCCACTCCCGACTCCA ANS: Restriction endonucleases cleave at palindromes in double-stranded DNA. A palindrome (indicated in boldfaced letters) can be found in all the double-stranded sequences except d. Therefore, sequence d would not be cleaved by a restriction endonuclease, whereas a, b, and c could be cleaved by the appropriate enzyme. (a) A CTCC AGA ATTC A CTCCG TGCGGTCTTA AGTGAGGC (b) G CCTC ATTCGA AGCCTGA CGGAGTA AGCTTCGGA CT (c) C TCGCCA ATTGA CTCGTC GAGCGGTTA ACTGAGC AG (d) A CTCC A CTCCCGA CTCC A TGAGGAGAGGGCTGAGGT 14.3 If the sequence of base pairs along a DNA molecule occurs strictly at random, what is the expected frequency of a specific restriction enzyme recognition sequence of length (a) four and (b) six base pairs?

ANS: (a) (1/4)4 = 1/256; (b) (1/4)6 = 1/4096 14.4 In what ways do restriction endonucleases differ from other endonucleases?

ANS: Restriction endonucleases recognize and cut specific nucleotide sequences in DNA. Most other endonucleases are not sequence-specific; many cut DNA sequences at random.

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14.5 Of what value are recombinant DNA and gene-cloning technologies to geneticists?

microorganisms do not have a sophisticated immune system like that of higher animals (see Chapter 22).

ANS: Recombinant DNA and gene-cloning techniques allow geneticists to isolate essentially any gene or DNA sequence of interest and to characterize it structurally and functionally. Large quantities of a given gene can be obtained in pure form, which permits one to determine its nucleotide-pair sequence (to “sequence it” in common lab jargon). From the nucleotide sequence and our knowledge of the genetic code, geneticists can predict the amino acid sequence of any polypeptide encoded by the gene. By using an appropriate subclone of the gene as a hybridization probe in northern blot analyses, geneticists can identify the tissues in which the gene is expressed. Based on the predicted amino acid sequence of a polypeptide encoded by a gene, geneticists can synthesize oligopeptides and use these to raise antibodies that, in turn, can be used to identify the actual product of the gene and localize it within cells or tissues of the organism. Thus, recombinant DNA and gene-cloning technologies provide very powerful tools with which to study the genetic control of essentially all biological processes. These tools have played major roles in the explosive progress in the field of biology during the last three decades.

14.8 Why is the DNA of a microorganism not degraded by a restriction endonuclease that it produces, even though its DNA contains recognition sequences normally cleaved by the endonuclease?

14.6 What determines the sites at which DNA molecules will be cleaved by a restriction endonuclease?

ANS: The nucleotide-pair sequence. Restriction endonucleases recognize a specific nucleotide-pair sequence in DNA regardless of the source of the DNA. In most cases, this is a 4 or 6 nucleotide-pair sequence; in a few cases, the recognition sequence is longer (e.g., 8 nucleotide pairs). Most restriction enzymes cleave the two strands of the DNA at a specific position (between the same two adjacent nucleotides in each strand) within the recognition sequence. A few restriction enzymes bind at a specific recognition sequence but cut the DNA at a nearby site outside of the recognition sequences. Some restriction endonucleases cut both strands between the same two nucleotide pairs (“blunt end” cutters), whereas others cut the two strands at different positions and yield complementary single-stranded ends (“sticky or staggered end” cutters). See Table 14.1 for examples.

14.7 Restriction endonucleases are invaluable tools for biologists. However, genes encoding restriction enzymes obviously did not evolve to provide tools for scientists.

Of what possible value are restriction endonucleases to the microorganisms that produce them?

ANS: Restriction endonucleases are believed to provide a kind of primitive immune system to the microorganisms that produce them—protecting their genetic material from “invasion” by foreign DNAs from viruses or other pathogens or just DNA in the environment that might be taken up by the microorganism. Obviously, these

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 57

ANS: Microorganisms that produce restriction endonucleases also produce enzymes that modify one or more bases in the recognition sequence for that endonuclease so that it can no longer cleave the DNA at that site. In most cases, the modifying enzyme is a methylase that attaches a methyl group to one or more of the bases in the recognition sequence. For example, E. coli strains that produce the restriction endonuclease EcoRI also produce EcoRI methylase, an enzyme that transfers a methyl group from S-adenosylmethionine to the second (most 3′) adenine residue in each strand of the recognition sequence (5-GAATCC-3) producing N6-methyladenines at these positions. EcoRI cannot cleave DNA that contains N6-methyladenine at these positions even if the EcoRI recognition sequence is present in this DNA (see Figure 14.1). Thus, if one wishes to digest DNA with EcoRI, that DNA must not be isolated from an E. coli strain that is producing EcoRI methylase.

14.9 One of the procedures for cloning foreign DNA segments takes advantage of restriction endonucleases such as HindIII (see Table 14.1) that produce complementary singlestranded ends. These enzymes produce identical complementary ends on cleaved foreign DNAs and on the vector DNAs into which the foreign DNAs are inserted.

Assume that you have inserted your favorite gene into the HindIII site in the polycloning region of the Bluescript cloning vector with DNA ligase, have amplified the plasmid containing your gene in E. coli, and have isolated a large quantity of gene/Bluescript DNA. How could you excise your favorite gene from the Bluescript vector?

ANS: A foreign DNA cloned using an enzyme that produces single-stranded complementary ends can always be excised from the cloning vector by cleavage with the same restriction enzyme that was originally used to clone it. For example, the HindIII fragment carrying your favorite can be excised from the Bluescript DNA by cleavage with restriction endonuclease HindIII. The human HindIII fragment will be flanked in the recombinant plasmid DNA clone by HindIII cleavage sites.

14.10 You are working as part of a research team studying the structure and function of a particular gene. Your job is to clone the gene. A restriction map is available for the region of the chromosome in which the gene is located; the map is as follows: Xba I

Pst I EcoRI

HindIII

Sal I

EcoRI

HindIII

GENE

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58-WC Answers to All Questions and Problems Your first task is to prepare a genomic DNA library that contains clones carrying the entire gene. Describe how you would prepare such a library in plasmid vector pBluescript II (see Figure 14.3), indicating which restriction enzymes, media, and host cells you would use.

14.13 Genomic clones of the chloroplastic glutamine synthetase gene (gln2) of maize are cleaved into two fragments by digestion with restriction endonuclease HindIII, whereas full-length maize gln2 cDNA clones are not cut

by HindIII. Explain these results.

ANS: Step 1: Digest genomic DNA isolated from your research organism with EcoRI. Step 2: Treat pBluescript II DNA with EcoRI. Step 3: Mix the EcoRI-digested genomic and vector DNAs under annealing conditions and incubate with DNA ligase. Step 4: Transform amp5 E. coli cells carrying the lacZM15 gene with the resulting ligation products. Step 5: Plate the transformed cells on nutrient agar medium containing Xgal and ampicillin. Only transformed cells will produce colonies in the presence of ampicillin. Step 6: Prepare your genomic DNA library by using bacteria from white colonies; these bacteria will contain pBluescript II DNA with genomic DNA inserts.

Bacteria harboring Bluescript II plasmids with no insert will produce blue colonies (see Figure 14.4).

ANS: The maize gln2 gene contains many introns, and one of the introns contains a HindIII cleavage site. The intron sequences (and thus the HindIII cleavage site) are not present in mRNA sequences and thus are also not present in full-length gln2 cDNA clones.

14.14 In the following illustration, the upper line shows a gene composed of segments A–D. The lower circle shows a mutant version of this gene, consisting of two fused pieces (A-B, C-D), carried on a plasmid. You attempt a targeted mutagenesis of a diploid cell by transforming cells with the cloned mutant gene. The following diagram shows the desired pairing of the plasmid and chromosome just prior to recombination.

14.11 Compare the nucleotide-pair sequences of genomic DNA clones and cDNA clones of specific genes of higher plants and animals. What is the most frequent difference that you would observe?

ANS: Most genes of higher plants and animals contain noncoding intron sequences. These intron sequences will be present in genomic clones, but not in cDNA clones, because cDNAs are synthesized using mRNA templates and intron sequences are removed during the processing of the primary transcripts to produce mature mRNAs. 14.12 Most of the genes of plants and animals that were cloned soon after the development of recombinant DNA technologies were genes encoding products that are synthesized in large quantities in specialized cells. For example, about 90 percent of the protein synthesized in mature red blood cells of mammals consists of α- and b-globin chains, and the globin genes were among the first mammalian genes cloned. Why were genes of this type so prevalent among the first eukaryotic genes that were cloned?

ANS: Higher eukaryotes have very large genomes; for example, the genomes of mammals contain approximately 3 × 109 nucleotide pairs. Thus, trying to identify a particular ­single-copy gene from a clone library is like looking for the proverbial “needle-in-a-haystack.” To accomplish this, one needs a nucleic acid hybridization probe specific for the gene or an antibody probe specific for the gene product. Given a specific cell or tissue type producing the mRNA and/or the protein gene product in large amounts, it was relatively easy to obtain pure mRNA or pure protein to use in making a hybridization or antibody probe, respectively, with which to screen a library for the gene or cDNA of interest. These approaches are much more difficult for the majority of the genes that encode products that represent only a small proportion of the total gene products in any given cell type.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 58

x

A A'

x

B

C

B'

C'

Probe

Plasmid

x x

D D'

You prepare DNA from the cells, digest it with an enzyme that cuts at x, and hybridize the cleaved DNA with the probe shown above. The following diagram shows a Southern blot of possible results.

1

____

2

3

____

____

____

4

5

____ ____ ____

(a) Which lane shows fragments produced from DNA in the cell before transformation? (b) Which lane shows fragments produced from DNA in the cell in which the anticipated targeted mutagenesis occurred? (c) Which of these blot patterns might be expected if two crossovers occurred, one between A and B, and the other between C and D? ANS: (a) 1 (b) 2 (c) 2 14.15 (a) What experimental procedure is carried out in Southern, northern, and western blot analyses? (b) What is the major difference between Southern, northern, and western blot analyses?

ANS: (a) Southern, northern, and western blot procedures all share one common step, namely, the transfer of macromolecules (DNAs, RNAs, and proteins, respectively) that have been separated by gel electrophoresis to a solid support—usually a nitrocellulose or nylon

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membrane—for further analysis. (b) The major d ­ ifference between these techniques is the class of macromolecules that are separated during the electrophoresis step: DNA for Southern blots, RNA for northern blots, and protein for western blots.

14.16 What major advantage does the polymerase chain reaction (PCR) have over other methods for analyzing nucleic acid structure and function?

ANS: The PCR technique has much greater sensitivity than any other method available for analyzing nucleic acids.

Thus, PCR procedures permit analysis of nucleic acid structure given extremely minute amounts of starting material. DNA sequences can be amplified and structurally analyzed from very small amounts of tissue like blood or sperm in assault and rape cases. In addition, PCR methods permit investigators to detect the presence of rare gene transcripts (e.g., in specific types of cells) that could not be detected by less sensitive procedures such as northern blot analyses or in situ hybridization studies.

14.17 The cloning vectors in use today contain an origin of replication, a selectable marker gene (usually an antibiotic-resistance gene), and one additional component.

What is this component, and what is its function?

ANS: All modern cloning vectors contain a “polycloning site” or “multiple cloning site” (MCS)—a cluster of unique cleavage sites for a number of different restriction endonucleases in a nonessential region of the vector into which the foreign DNA can be inserted. In general, the greater the complexity of the MCS—that is, the more restriction endonuclease cleavage sites that are present— the greater the utility of the vector for cloning a wide variety of different restriction fragments. For example, see the MCS present in plasmid Bluescript II shown in Figure 14.3.

14.18 The drawing in this problem shows a restriction map of a segment of a DNA molecule. Eco refers to locations where the restriction endonuclease EcoRI cuts the DNA, and Pst refers to locations where the restriction enzyme PstI cuts the DNA. Potential restriction sites are numbered 1–6. Distances between restriction sites are shown on the bottom scale in base pairs (bp). The thick line represents the part of the molecule that has homology with a probe.

Eco 1

Pst

5000 bp

2

Eco

3000 bp

3

4000 bp

Pst

Eco

4

5

2000 bp

Pst

5000 bp

6

(a) Assume that individual 1 has restriction sites 1 through 6. If this individual’s DNA is digested with PstI, what are the expected sizes of the DNA fragments that will hybridize with the probe? (b) Assume that individual 2 has a mutation that eliminates site 4. If this individual’s

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 59

DNA is digested with PstI, what are the expected sizes of the DNA fragments that will hybridize with the probe? (c) Assume that individual 3 has a mutation that eliminates site 5. If the DNA of this individual is digested with PstI, what are the expected sizes of the DNA fragments that will hybridize with the probe? (d) If the DNA of individual 1 is digested with both PstI and EcoRI, what are the expected sizes of the DNA fragments that will hybridize with the probe? (e) If the DNA of individual 3 is digested with both PstI and EcoRI, what are the expected sizes of the DNA fragments that will hybridize with the probe?

ANS: (a) 7000 bp + 7000 bp; (b) 14,000 bp; (c) 7000 bp + 7000 bp; (d) 4000 bp + 2000 bp + 5000 bp; (e) 4000 bp + 7000 bp.

14.19 The cystic fibrosis (CF) gene (location: chromosome 7, region q31) has been cloned and sequenced, and studies of CF patients have shown that about 70 percent of them are homozygous for a mutant CF allele that has a specific three-nucleotide-pair deletion (equivalent to one codon).

This deletion results in the loss of a phenylalanine residue at position 508 in the predicted CF gene product. Assume that you are a genetic counselor responsible for advising families with CF in their pedigrees regarding the risk of CF among their offspring. How might you screen putative CF patients and their parents and relatives for the presence of the CFDF508 mutant gene? What would the detection of this mutant gene in a family allow you to say about the chances that CF will occur again in the family?

ANS: Because the nucleotide-pair sequences of both the normal CF gene and the CF∆508 mutant gene are known, labeled oligonucleotides can be synthesized and used as hybridization probes to detect the presence of each allele (normal and ∆508). Under high-stringency hybridization conditions, each probe will hybridize only with the CF allele that exhibits perfect complementarity to itself. Since the sequences of the CF gene flanking the ∆508 site are known, oligonucleotide PCR primers can be synthesized and used to amplify this segment of the DNA obtained from small tissue explants of putative CF patients and their relatives

by PCR. The amplified DNAs can then be separated by agarose gel electrophoresis, transferred to nylon membranes, and hybridized to the respective labeled ­ ­oligonucleotide probes, and the presence of each CF allele can be detected by autoradiography. For a demonstration of the utility of this procedure, see Focus on Detection of a Mutant Gene Causing Cystic Fibrosis. In the procedure described there, two synthetic oligonucleotide probes— oligo-N = 3′-CTTTTATAGTAGAAACCAC-5′ and oligo-DF = 3′-TTCTTTTATAGTA—ACCACAA-5′ (the dash indicates the deleted nucleotides in the CFD508 mutant allele) were used to analyze the DNA of CF patients and their parents. For confirmed CF families, the results of these Southern blot hybridizations with the oligo-N (normal) and oligo-DF (CFD508) labeled probes were often as follows:

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60-WC Answers to All Questions and Problems

Oligo-N probe: Oligo-∆F probe:

Both parents were heterozygous for the normal CF allele and the mutant CFD508 allele as would be expected for a rare recessive trait, and the CF patient was homozygous for the CFD508 allele. In such families, one-fourth of the children would be expected to be homozygous for the D508 mutant allele and exhibit the symptoms of CF, whereas three-fourths would be normal (not have CF). However, two-thirds of these normal children would be expected to be heterozygous and transmit the allele to their children.

Only one-fourth of the children of this family would be homozygous for the normal CF allele and have no chance of transmitting the mutant CF gene to their offspring.

Note that the screening procedure described here can be used to determine which of the normal children are carriers of the CFD508 allele: that is, the mutant gene can be detected in heterozygotes as well as in homozygotes.

14.20 Cereal grains are major food sources for humans and other animals in many regions of the world. However, most cereal grains contain inadequate supplies of certain of the amino acids that are essential for monogastric animals such as humans. For example, corn contains insufficient amounts of lysine, tryptophan, and threonine.

Thus, a major goal of plant geneticists is to produce corn varieties with increased kernel lysine content. As a prerequisite to the engineering of high-lysine corn, molecular biologists need more basic information about the regulation of the biosynthesis and the activity of the enzymes involved in the synthesis of lysine. The first step in the anabolic pathway unique to the biosynthesis of lysine is catalyzed by the enzyme dihydrodipicolinate synthase. Assume that you have recently been hired by a major U.S. plant research institute and that you have been asked to isolate a clone of the nucleic acid sequence encoding dihydrodipicolinate synthase in maize. Briefly describe four different approaches you might take in attempting to isolate such a clone and include at least one genetic approach.

ANS: You could attempt to isolate either a dihydropicolinate synthase (DHPS) cDNA clone or a DHPS genomic clone. Once you have isolated either DHPS clone, it can be used as a hybridization probe to isolate the other (genomic or cDNA) by screening an appropriate library by in situ colony hybridization. Four approaches that have proven effective in isolating other eukaryotic coding sequences of interest are the following: (1) You could obtain a clone of the DHPS gene of a lower eukaryote (a clone of the DHPS gene of Saccharomyces cerevisae is available) or even a prokaryote and use it as a heterologous hybridization probe to screen a maize cDNA

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 60

library using low stringency conditions. Sometimes this approach is successful; sometimes it is not successful.

Whether or not this approach works depends on how similar the coding sequences of the specific gene of interest are in the two species. (2) You could purify the DHPS enzyme from corn and use the purified protein to produce an antibody to DHPS. This DHPS-specific antibody could then be used to screen a maize cDNA expression library by a protocol analogous to the western blot procedure. (An expression library contains the cDNA coding sequences fused to appropriate transcription and translation signals so that they are expressed in E. coli or other host cells in which the cDNA library is prepared.) (3) You could purify the DHPS enzyme from maize and determine the amino acid sequence of its NH2-terminus by microsequencing techniques. From the amino acid sequence and the known genetic code, you could predict the possible nucleotide sequences encoding this segment of the protein. Because of the degeneracy in the code, there would be a set of nucleotide sequences that would all specify the same amino acid sequence, and you would not know which one was present in the maize DHPS gene. However, the synthesis of oligonucleotides is now routine and quite inexpensive.

Thus, you could synthesize a mixture of oligonucleotides containing all possible coding sequences and use this mixture as a set of hybridization probes to screen an appropriate library by in situ colony hybridization.

(4) Finally, you might try a simple and very quick genetic approach based on the ability of cDNAs in an expression library to rescue DHPS mutants of E. coli or other species that can be transformed at high frequencies. You would obtain a DHPS-deficient mutant of E. coli (available from the E. coli Genetics Stock Center at Yale University), transform it with your cDNA expression library, and plate the transformed cells on medium lacking diaminopimelic acid (the product of DHPS). DHPS-deficient E. coli mutants cannot grow in the absence of diaminopimelic acid; thus, any colonies that grow on your selection plates should be the result of rescue of the DHPS mutant bacteria by corn DHPS encoded by cDNAs in the library.

This entire screening procedure can be carried out in 3 or 4 days; thus, it is much simpler than the preceding approaches. In fact, David A. Frisch first isolated a maize DHPS cDNA by this simple, but powerful genetic approach. However, that this approach would only be expected to work in the case of enzymes that are active as monomers or homomultimers; it is not applicable when the active form of the enzyme is a heteromultimer.

14.21 You have just isolated a mutant of the bacterium Shigella dysenteriae that is resistant to the antibiotic kanamycin, and you want to characterize the gene responsible for this resistance. Design a protocol using genetic selection to identify the gene of interest.

ANS: Genetic selection is the most efficient approach to cloning genes of this type. Prepare a genomic library in an expression vector such as Bluescript (see Figure 14.3)

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using DNA from the kanamycin-resistant strain of Shigella dysenteriae. Then, screen the library for the kanamycin-resistance gene by transforming kanamycin-sensitive E. coli cells with the clones in the library and plating the transformed cells on medium containing kanamycin.

Only cells that are transformed with the kanamycinresistance gene will produce colonies in the presence of kanamycin.

14.22 You have isolated a cDNA clone encoding a protein of interest in a higher eukaryote. This cDNA clone is not cleaved by restriction endonuclease EcoRI. When this cDNA is used as a radioactive probe for blot hybridization analysis of EcoRI-digested genomic DNA, three radioactive bands are seen on the resulting Southern blot. Does this result indicate that the genome of the eukaryote in question contains three copies of the gene encoding the protein of interest?

ANS: No. The genome of the species in question may contain one, two, or three copies of the gene (or family of closely related genes) encoding this protein. The possibilities are as follows: (1) One copy of the gene with two EcoRI cleavage sites located within intron sequences. (2) Two copies of the gene with one EcoRI cleavage site located within an intron sequence of one of the copies. (3) Three copies of the gene with no EcoRI cleavage site in any of the copies, that is, each copy present on a single EcoRI restriction fragment.

14.23 A linear DNA molecule is subjected to single and double digestions with restriction endonucleases, and the following results are obtained: Enzymes EcoRI

2.9, 4.5, 7.4, 8.0

HindIII

3.9, 6.0, 12.9

EcoRI and HindIII

1.0, 2.0, 2.9, 3.5, 6.0, 7.4

6.0

H 2.0

E

HindIII

8

5

12

4

4 3

E 4

E B

14.25 You are studying a circular plasmid DNA molecule of size 10.5 kilobase pairs (kb). When you digest this plasmid with restriction endonucleases BamHI, EcoRI, and HindIII, singly and in all possible combinations, you obtain linear restriction fragments of the following sizes: Enzymes

Fragment Sizes (in kb)

BamHI

7.3, 3.2

EcoRI

10.5

HindIII

5.1, 3.4, 2.0

BamHI + EcoRI

6.7, 3.2, 0.6

BamHI + HindIII

4.6, 2.7, 2.0, 0.7, 0.5

EcoRI + HindIII

4.0, 3.4, 2.0, 1.1

BamHI + EcoRI + HindIII

4.0, 2.7, 2.0, 0.7, 0.6, 0.5

Draw a restriction map for the plasmid that fits your data.

ANS: There are two possible restriction maps for these data as shown below:

4.0

H

B 0.5

E H

4.0

2.7 2.0

0.7 H

0.6

B 0.5

H

2.0 H

0.7 B

H

2.7

B

Restriction enzyme cleavage sites for BamHI, EcoRI, and HindIII are denoted by B, E, and H, respectively. The numbers give distances in kilobase pairs.

E

BamHI

EcoRI and BamHI

HindIII and BamHI

6

6

6

6

4

5

2

1

Draw the restriction map of this DNA molecule.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 61

2

5

0.6

14.24 A DNA molecule is subjected to single and double digestions with restriction enzymes, and the products are separated by gel electrophoresis. The results are as follows (fragment sizes are in kb): EcoRI and HindIII

B

1

E

H 3.5 1.02.9

7.4

E

EcoRI

H

Fragment Sizes (in kb)

Draw the restriction map defined by these data.

ANS:

ANS:

14.26 The automated DNA sequencing machines utilize fluorescent dyes to detect the nascent DNA chains synthesized in the presence of the four dideoxy (ddX) chain terminators, each labeled with a different fluorescent dye. The dyes fluoresce at different wavelengths, which are recorded by a photocell as the products of the reactions are separated based on length by capillary gel electrophoresis (see Figure 14.17). In the standard sequencing reaction, the chains terminating with ddG fluoresce dark blue (peaks appear black in computer printout), those terminating with ddC fluoresce light blue, those terminating with ddA fluoresce green, and those terminating with ddT fluoresce red. The computer printout for the

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62-WC Answers to All Questions and Problems sequence of a short segment of DNA is shown as follows.

First nucleotide

c­ hain-terminators—ddGTP, ddCTP, ddATP, and ddTTP— each labeled with a different fluorescent dye. The dyes ­fluoresce at different wavelengths, which are recorded by a photocell as the products of the reactions are separated by capillary gel electrophoresis (see Figure 14.17). In the standard sequencing reaction, the chains terminating with ddG fluoresce dark blue (peaks appear black in the computer printouts), those terminating with ddC fluoresce light blue, those terminating with ddA fluoresce green, and those terminating with ddT fluoresce red. The computer printout for sequencing reaction 1, which contained the Watson strand as template, is shown as follows.

Last nucleotide

What is the nucleotide sequence of the nascent strand of DNA? What is the nucleotide sequence of the DNA template strand?

ANS: Nascent strand: 5′-AGTTCTAGAGCGGCCGCCACCGCGTGGAGCTCCAGCTTTTGTTCCCTTT-3′

Template strand:

Nucleotide: 1

3′-TCAAGATCTCGCCGGCGGTGGCGCACCTCGAGGTCGAAAACAAGGGAAA-5′

10

Draw the predicted computer printout for reaction 2, which contained the Crick strand as template, in the following box. Remember that all DNA synthesis occurs in the 5′ → 3′ direction and that the sequence of the nascent strand reads 5′ to 3′ from left to right in the printout.

14.27 Ten micrograms of a decanucleotide-pair HpaI restriction fragment were isolated from the double-stranded DNA chromosome of a small virus. Octanucleotide poly(A) tails were then added to the 3′ ends of both strands using terminal transferase and dATP; that is, 5′-X X X X X X X X X X-3′ 3′-X′ X′ X′ X′ X′ X′ X′ X′ X′ X′-5′ ↓ terminal transferase, dATP

Nucleotide: 1

5′-X X X X X X X X X X A A A A A A A A-3′ 3′-A A A A A A A A X′ X′ X′ X′ X′ X′ X′ X′ X′ X′-5′

10

ANS:

where X and X′ can be any of the four standard nucleotides, but X′ is always complementary to X. The two complementary strands (“Watson” strand and “Crick” strand) were then separated and sequenced by the 2′,3′-dideoxyribonucleoside triphosphate chain-termination method. The reactions were primed using a synthetic poly(T) octamer; that is, Watson strand 3′-A A A A A A A A X′ X′ X′ X′ X′ X′ X′ X′ X′ X′-5′ 5′-T T T T T T T T-OH Crick strand 5′-X X X X X X X X X X A A A A A A A A-3′

HO-T T T T T T T T-5′

Two DNA sequencing reactions were carried out.

Reaction 1 contained the Watson strand template/primer shown above; reaction 2 contained the Crick strand template/primer. Both sequencing reactions contained DNA polymerase and all other substrates and components required for DNA synthesis in vitro plus the standard four 2′,3′-dideoxyribonucleoside triphosphate

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 62

Nucleotide: 1

10

Chapter 15 15.1 Distinguish between a genetic map, a cytogenetic map, and a physical map. How can each of these types of maps be used to identify a gene by positional cloning?

ANS: Genetic map distances are determined by crossover frequencies. Cytogenetic maps are based on chromosome morphology or physical features of chromosomes. Physical maps are based on actual physical distances—the number of nucleotide pairs (0.34 nm per base pair)— separating genetic markers. If a gene or other DNA sequence of interest is shown to be located near a mutant gene, a specific band on a chromosome, or a particular DNA restriction fragment, that genetic or physical marker (mutation, band, or restriction fragment) can be used to initiate a chromosome walk to the gene of interest.

15.2 In the technique of positional cloning, a researcher begins with a DNA library and selects a clone that is tightly linked to the gene of interest. That clone, or a

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Answers to All Questions and Problems WC-63

piece of it, is then used as a probe to isolate an overlapping clone from a different DNA library. The second clone is used to isolate a third overlapping clone from the first library, and so on, until the researcher has “walked” along the chromosome to the desired locus. (a) How can the researcher walk consistently in the same direction along the chromosome during the cloning process? (b) What could happen if a long repetitive DNA sequence such as a transposon was situated between the starting clone and the gene of interest?

I A

ANS: A contig (contiguous clones) is a physical map of a chromosome or part of a chromosome prepared from a set of overlapping genomic DNA clones. An RFLP (restriction fragment length polymorphism) is a variation in the length of a specific restriction fragment excised from a chromosome by digestion with one or more restriction endonucleases. A VNTR (variable number tandem repeat) is a short DNA sequence that is present in the genome as tandem repeats and in highly variable copy number. An STS (sequence tagged site) is a unique DNA sequence that has been mapped to a specific site on a chromosome. An EST (expressed sequence tag) is a cDNA sequence—a genomic sequence that is transcribed. Contig maps permit researchers to obtain clones harboring genes of interest directly from DNA Stock Centers—to “clone by phone.” RFLPs are used to construct the highdensity genetic maps that are needed for positional cloning. VNTRs are especially valuable RFLPs that are used to identify multiple sites in genomes. STSs and ESTs provide molecular probes that can be used to initiate chromosome walks to nearby genes of interest.

15.4 The following is a Southern blot of EcoRI-digested DNA of rye plants from two different inbred lines, A and B. Developed autoradiogram I shows the bands resulting from probing the blot with 32P-labeled cDNA1. Autoradiogram II shows the same Southern blot after it was stripped of probe and reprobed with 32P-labeled cDNA2.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 63

B

A

B

b1

a1

b2 a2 a3

b3

a4

ANS: (a) To ensure that the researcher “walks” in the same direction along the chromosome to the desired locus, it is necessary to check at each step in the process, usually by blot hybridization with subclones, which end of the clone used as a probe overlaps with the next recovered clone. (b) A long repetitive sequence could present a problem in chromosome walking because if that sequence is used as a probe to isolate a new clone, it will hybridize with restriction fragments from all over the genome, and thereby thwart the effort to walk systematically in one direction along a single chromosome. To overcome this problem, a researcher would have to “jump over” the repetitive element and use a piece of unique DNA beyond the element as a probe in the next step of the walking procedure.

15.3 What is a contig? What is an RFLP? What is a VNTR? What is an STS? What is an EST? How is each of these used in the construction of chromosome maps?

II

(a) Which bands would you expect to see in the autoradiogram of a similarly probed Southern blot prepared using EcoRI-digested DNA from F1 hybrid plants produced by crossing the two inbred lines? (b) What can you conclude about the gene(s) represented by band a1 on blot I in the two inbreds? (c) The F1 plants were crossed to plants possessing only bands a1, a4, and b3. DNA was isolated from several individual progeny and digested with EcoRI. The resulting DNA fragments were separated by gel electrophoresis, transferred to a nylon membrane, and hybridized with radioactive cDNA1 and cDNA2 probes. The following table summarizes the bands present in autoradiograms obtained using DNA from individual progeny.

Plant No.

Bands Present a1

a2

a3

a4

b1

b2

b3

1

+

+

+

+

+

2

+

+

+

+

+

3

+

+

+

+

+

4

+

+

+

+

+

5

+

+

+

+

+

+

+

6

+

+

+

+

+

7

+

+

+

+

+

8

+

+

+

+

+

9

+

+

+

+

+

10

+

+

+

Interpret these data. Do the data provide evidence for RFLPs? At how many loci? Are any of the RFLPs linked? If so, what are the linkage distances defined by the data?

ANS: (a) a1, a2, a3, a4, b1, b2, and b3. (b) Band a1 represents a locus whose DNA is homologous to cDNA1. Since the marker is not polymorphic in the parents used, it cannot be mapped in this cross. (c) The cDNA1 probe detects one RFLP locus with alleles that are visualized as band a4 and bands a2/a3. The cDNA2 detects a second RFLP

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64-WC Answers to All Questions and Problems locus with alleles that are visualized as band b3 and bands b1/b2. The two loci are linked with 20% recombination observed in this cross.

15.5 As part of the Human Genome Mapping Project, you are trying to clone a gene involved in colon cancer. Your first step is to localize the gene using RFLP markers. In the following table, RFLP loci are defined by STS number (e.g., STS1), and the gene for colon cancer is designated C. % Recombi­ nation

Loci

% Recombi­ nation

C, STS1

50

STS1, STS5

10

C, STS2

15

STS2, STS3

30

C, STS3

15

STS2, STS4

14

C, STS4

1

STS2, STS5

50

C, STS5

40

STS3, STS4

16

STS1, STS2

50

STS3, STS5

25

STS1, STS3

35

STS4, STS5

41

STS1, STS4

50

Loci

(a) Given the percentage recombination between different RFLP loci and the gene for colon cancer shown in the table, draw a genetic map showing the order and genetic distances between adjacent RFLP markers and the gene for colon cancer. (b) Given that the human genome contains approximately 3.3 × 109 base pairs of DNA and that the human genetic map contains approximately 3300 cM, approximately how many base pairs of DNA are located along the stretch of chromosome defined by this RFLP map? (Hint: First figure how many base pairs of DNA are present per cM in the human genome.) (c) How many base pairs of DNA are present in the region between the colon cancer gene and the nearest STS? ANS: (a)

10 cM STS1

25 cM STS5

15 cM STS3

1 cM C

14 cM STS4

STS2

(b) 3.3 × 109 bp/3.3 × 103 cM = 1 × 106 bp/cM. The total map length is 65 cM, which equates to about 65 × 106 or 65 million bp. (c) The cancer gene (C) and STS4 are separated by 1 cM or about 1 million base pairs.

15.6 What are STRs? Why are they sometimes called microsatellites?

ANS: STRs are polymorphic tandem repeats of sequences of only two to four nucleotide pairs. They are called microsatellites because they are short in length and are components of the highly repetitive satellite DNAs of eukaryotes (see Chapter 9).

15.7 You have cloned a previously unknown human gene.

What procedure will allow you to position this gene on the cytological map of the human genome without performing any pedigree analyses? Describe how you would carry out this procedure.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 64

ANS: With a clone of the gene available, fluorescent in situ hybridization (FISH) can be used to determine which human chromosome carries the gene and to localize the gene on the chromosome. Single-stranded copies of the clone are coupled to a fluorescent probe and hybridized to denatured DNA in chromosomes spread on a slide.

After hybridization, free probe is removed by washing, and the location of the fluorescent probe is determined by photography using a fluorescence microscope (see Focus on: In Situ Hybridization).

15.8 You have identified a previously unknown human EST. What must be done before this new EST can be called an STS? ANS: An EST must be placed on the physical map of a chromosome before it can be called an STS. If it is also positioned on the genetic map of the chromosome, then it is called an anchor marker.

15.9 VNTRs and STRs are specific classes of polymorphisms.

What is the difference between a VNTR and an STR? ANS: Variable number tandem repeats (VNTRs) are composed of repeated sequences of 10–80 nucleotide pairs, and short tandem repeats (STRs) are composed of repeated sequences of 2–10 nucleotide pairs.

15.10 An RFLP and a mutant allele that causes albinism in humans cannot be shown to be separated by recombination based on pedigree analysis or by radiation hybrid mapping. Do these observations mean that the RFLP occurs within or overlaps the gene harboring the mutation that causes albinism? If so, why? If not, why not?

ANS: The resolution of genetic mapping in humans is quite low—in the range of 1–10 million base pairs. Radiation hybrid mapping provides higher resolution—to about 50 kb. However, even with 50-kb resolution, there could be several genes separating the RFLP from the mutation responsible for albinism.

15.11 A cloned 6-kb fragment of DNA from human chromosome 9 contains a single site recognized by the restriction enzyme EcoRI. This cloned fragment is demarcated by sites for the restriction enzyme BamHI. There are no other BamHI recognition sites within the clone.

A researcher has collected DNA samples from 10 people.

He digests each sample with a combination of EcoRI and BamHI enzymes. The doubly digested DNA is then fractionated by gel electrophoresis and blotted to a membrane. After fixing the DNA to the membrane, the researcher hybridizes it with a radioactive probe made from the entire cloned BamHI fragment. The autoradiogram obtained by exposing an X-ray film to this membrane yielded the following results. Three of the DNA samples contained a 4-kb fragment and a 2-kb fragment that hybridize with the probe, three of the DNA samples contained a 6-kb DNA fragment that hybridizes with the probe, and four of the DNA samples contain 6-, 4-, and 2-kb DNA fragments that hybridize with the probe.

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Answers to All Questions and Problems WC-65

ANS: The results of this analysis reveal that the EcoRI cleavage site within the original 6-kb BamHI clone is polymorphic— that is, it is present in some chromosomes 9 but absent in others. We can represent these two types of chromosomes 9 as BEB (with the EcoRI site between flanking BamHI sites) and B-B (without the EcoRI site). The first three samples came from people who were homozygous for the BEB version of chromosome 9, the next three samples came from people who were homozygous for the B-B version, and the last four samples came from people who where heterozygous for the two versions— that is, they had the genotype BEB/B-B. In the human population that was sampled, the EcoRI site is therefore the basis for a restriction fragment length polymorphism (RFLP).

15.12 Both an RFLP and a mutation that causes deafness in humans map to the same location on the same chromosome. How can you determine whether or not the RFLP overlaps with the gene containing the deafness mutation?

ANS: You can start a chromosome walk using the hybridization probe that detects the RFLP. However, if a physical map of this region of the chromosome already exists (see Figure 15.5), a chromosome walk might not be necessary. cDNAs can be used to locate candidate genes in the region covered by the chromosome walk, and the sequences of genes in individuals with this form of inherited deafness can be compared with the sequences of homologous genes of individuals with normal hearing, looking for changes that would be expected to cause a loss of gene function. The overall process is illustrated in Figure 15.6.

15.13 What were the goals of the Human Genome Project?

What impact has achieving these goals had on the practice of medicine to date? What are some of the predicted future impacts? What are some of the possible misuses of human genome data?

ANS: The goals of the Human Genome Project were to prepare genetic and physical maps showing the locations of all the genes in the human genome and to determine the nucleotide sequences of all 24 chromosomes in the human genome. These maps and nucleotide sequences of the human chromosomes helped scientists identify mutant genes that result in inherited diseases. Hopefully, the identification of these mutant disease genes will lead to successful treatments, including gene therapies, for at least some of these diseases in the future. Potential misuses of these data include invasions of privacy by governments and businesses—especially employment agencies and insurance companies. Individuals must not be denied educational opportunities, employment, or insurance because of inherited diseases or mutant genes that result in a predisposition to mental or physical abnormalities.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 65

15.14 What difficulty does repetitive DNA pose for the assembly of whole genome shotgun sequences by computer analysis?

ANS: Repetitive DNA can make it difficult to assemble long stretches of DNA sequence from the sequences of DNA fragments that contain these sequences. The reason is that in the computer analysis, a repetitive sequence in one DNA fragment will match with the same repetitive sequence in another DNA fragment even if the two DNA fragments are not contiguous in the genome. The effort to determine which short DNA fragments are authentic neighbors (contiguous to each other) can therefore be thrown off by these spurious matches.

15.15 Which type of molecular marker, RFLP or EST, is most likely to mark a disease-causing mutant gene in humans?

Why? ANS: An EST is more likely than an RFLP to occur in a disease-causing human gene. All ESTs correspond to expressed sequences in a genome. RFLPs occur throughout a genome, in both expressed and unexpressed sequences. Because less than 2 percent of the human genome encodes proteins, most RFLPs occur in noncoding DNA. 15.16 Bacteriophage FX174 contains 11 genes in a genome of 5386 bp; E. coli has a predicted 4288 genes in a genome of about 4.639 kb; S. cerevisiae has about 6000 genes in a genome of size 12.1 mb; C. elegans has about 19,000 genes present in a genome of about 100 mb; and H. sapiens has an estimated 22,000 genes in its 3000-mb genome.

Which genome has the highest gene density? Which genome has the lowest gene density? Does there appear to be any correlation between gene density and developmental complexity? If so, describe the correlation.

ANS: The bacteriophage FX174 genome has the highest gene density—one gene per 490 nucleotide pairs. The human genome has the lowest gene density—about one gene per 100 kb. In the species mentioned in this question, there is a striking correlation between genome size and developmental complexity. With some exceptions, including species with polyploid genomes, there does appear to be a rough correlation between genome size and developmental complexity.

15.17 A contig map of one segment of chromosome 3 of Arabidopsis is as follows.

Genomic clones in YAC vectors

What has this analysis revealed? What are the genotypes of the three different types of DNA samples?

1 A

B

2

C

Chromosome segment 3 4 5 6 7 8

D

E

9

F

10

G

H

(a) If an EST hybridizes with genomic clones C, D, and E, but not with the other clones, in which segment of chromosome 3 is the EST located? (b) If a clone of gene ARA hybridizes only with genomic clones C and D, in

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66-WC Answers to All Questions and Problems which chromosome segment is the gene located? (c) If a restriction fragment hybridizes with only one of the genomic clones shown above, in which chromosome segment(s) could the fragment be located?

ANS: (a) Segment 5; (b) segment 4; (c) segment 1, 6, or 10.

Marker

15.18 Eight human–Chinese hamster radiation hybrids were tested for the presence of six human ESTs designated A through F. The results are shown in the following table, where a plus indicates that a marker was present and a minus indicates that it was absent.

Radiation hybrid 3 4 5 6

7

8

+

–

–

+

+

–

+

–

+

–

–

+

–

+

1

2

A

–

B

+

C

–

+

+

+

–

–

–

+

D

+

–

+

+

–

–

+

–

E

+

–

+

+

–

–

+

–

F

–

+

–

+

+

+

+

–

Based on these data, do any of the ESTs appear to be closely linked? Which ones? What would be needed for you to be more certain of your answer?

ANS: EST markers D and E appear to be closely linked. The eight human–Chinese hamster radiation hybrids contain either both D and E or neither marker. More radiation hybrids would need to be tested for the presence of these ESTs to obtain convincing evidence of this linkage.

ANS: The DNA sequences in human chromosome-specific cDNA libraries can be coupled to fluorescent dyes and hybridized in situ to the chromosomes of other ­primates.

The hybridization patterns can be used to detect changes in genome structure that have occurred during the evolution of the various species of primates from common ancestors (see Figure 6.4). Such comparisons are especially effective in detecting new linkage relationships resulting from translocations and centric fusions.

15.22 Of the cereal grass species, only maize contains two copies of each block of linked genes. What does this duplication of sets of maize genes indicate about the origin of this agronomically important species?

ANS: The presence of two copies of each block of genes in the corn genome indicates that maize has evolved from a tetraploid ancestor. The presence of one set of genes primarily in the large chromosomes and the second set largely in the small chromosomes suggests that maize has evolved from an allotetraploid (see Chapter 6) produced by combining the diploid genomes of two ancestral cereal grass species.

15.23 Five human genomic DNA clones present in PAC vectors were tested by hybridization for the presence of six sequence-tagged sites designated STS1 through STS6.

The results are given in the following table: a plus indicates the presence of the STS, and a minus indicates the absence of the STS.

ANS: The major advantage of gene chips as a microarray hybridization tool is that a single gene chip can be used to quantify thousands of distinct nucleotide sequences simultaneously. The gene-chip technology allows researchers to investigate the levels of expression of a large number of genes more efficiently than was possible using earlier microarray procedures.

15.20 What major advantage does the green fluorescent protein of the jellyfish have over other methods for studying protein synthesis and localization?

ANS: The green fluorescent protein (GFP) can be used to study protein localization and movement over time in living cells. Most other procedures for studying protein localization require that cells be permeabilized and/or fixed and exposed to antibodies coupled to radioactive or fluorescent compounds prior to visualization. As a result, these procedures only provide information about the location of a protein at a single time point. In contrast, GFP-tagged proteins can be used to study the synthesis and movement of proteins in living cells over time (hours to days).

15.21 You are given chromosome-specific cDNA libraries for all 24 human chromosomes. How might these libraries be used to study chromosome evolution in primates?

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 66

YAC clone

15.19 What is the major advantage of gene chips as a microarray hybridization tool?

1

2

STS 3 4

5

6

A

+

–

+

+

–

–

B

+

–

–

–

+

–

C

–

–

+

+

–

+

D

–

+

–

–

+

–

E

–

–

+

–

–

+

(a) What is the order of the STS sites on the chromosome? (b) Draw the contig map defined by these data.

ANS: (a) Order of STS sites: 2-5-1-4-3-6. (b) STS markers: 2 YAC clones

B D

5 1 A

4

3

6 C E

15.24 The complete sequences of six mitochondrial genomes of H. neanderthalensis have been available for some time; the first H. neanderthalensis mtDNA sequence was published in 2008. How similar are the sequences of the mtDNAs of H. neanderthalensis and H. sapiens? Are the genomes similar in size? Is the amount of diversity observed in the mtDNAs of Neanderthals and humans the same? If not, what might this tell us about the sizes of Neanderthal and human populations? How many genes are present in the H. neanderthalensis mitochondrial

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Answers to All Questions and Problems WC-67

genome? How many of these genes encode proteins?

How many specify structural RNA molecules? Are there any pseudogenes in H. neanderthalensis mtDNA? All of these questions can be answered by visiting the http:// www.ncbi.nlm.nih.gov/ web site.

ANS: The mtDNAs of Homo neanderthalensis and H. sapiens are very similar, both in size and in nucleotide sequence. The H. neanderthalensis mtDNA contains 16,565 nucleotide pairs, whereas the H. sapiens mtDNA contains 16,569 nucleotide pairs, and the two DNAs exhibit 99 percent sequence identity. There is less sequence diversity in the mtDNAs of Neanderthals than in the mtDNAs of humans, which is consistent with Neanderthal populations being smaller than human populations. There are 37 genes in the Neanderthal mtDNA: 13 encode proteins and 24 specify structural RNAs (tRNAs and rRNAs). The Neanderthal mitochondrial genome does not contain any pseudogenes.

15.25 Assume that you have just sequenced a small fragment of DNA that you had cloned. The nucleotide sequence of this segment of DNA is as follows. aagtagtcgaaaccgaattccgtagaaacaactcgcacgctccggtttc gtgttgcaacaaaataggcattcccatcgcggcagttagaatcaccga gtgcccagagtcacgttcgtaagcaggcgcagtttacaggcagca gaaaaatcgattgaacagaaatggctggcggtaaagcaggcaagga ttcgggcaaggccaaggcgaaggcggtatcgcgttccgcgcgcgcggg In an attempt to learn something about the identity or possible function of this DNA sequence, you decide to perform a BLAST (nucleotide blast) search on the NCBI web site (http://www.ncbi.nlm.nih.gov). Paste or type this sequence into the query sequence box. Run the search and examine the sequences most closely related to your query sequence. Are they coding sequences? What proteins do they encode? Repeat the BLAST search with only half of your sequence as the query sequence. Do you still identify the same sequences in the databases? If you use one-fourth of your sequence as a query, do you still retrieve the same sequences? What is the shortest DNA sequence that you can use as a query and still identify the same sequences in the databanks?

ANS: All of the sequences identified by the megablast search encode histone H2a proteins. The query sequence is identical to the coding sequence of the Drosophila ­melanogaster histone H2aV gene (a member of the gene family encoding histone H2a proteins). The query sequence encodes a Drosophila histone H2a polypeptide designated variant V. The same databank sequences are identified when one-half or one-fourth of the given nucleotide sequence is used as the query in the megablast search. Query sequences as short as 15–20 nucleotides can be used to identify the Drosophila gene encoding the histone H2a variant. However, the results will vary depending on the specific nucleotide sequence used as the query sequence.

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 67

15.26 The NCBI web site (http://www.ncbi.nlm.nih.gov) can also be used to search for protein sequences. Instead of performing a BLAST search with a nucleic acid query, one performs a protein blast with a polypeptide (amino acid sequence) query. Assume that you have the following partial sequence of a polypeptide: GYDVEKNNSRIKLGLKSLVSKGILVQTKGT GASGSFKLNKKAASGEAKPQAKKAGAAKA Go to the NCBI web site and access the BLAST tool.

Then click on protein blast and enter the query sequence in the box at the top. Then click BLAST. What is the identity of your query sequence?

ANS: Your query sequence is a portion of histone H1.2 of the mouse (Mus musculus). This portion of the mouse histone H1.2 polypeptide differs by one amino acid from the corresponding part of the human H1 histone.

15.27 The sequence of a gene in Drosophila melanogaster that encodes a histone H2A polypeptide is as follows: aagtagtcgaaaccgaattccgtagaaacaactcgcacgctccggtttc gtgttgcaacaaaataggcattcccatcgcggcagttagaatcacc gagtgcccagagtcacgttcgtaagcaggcgcagtttacaggcagcag aaaaatcgattgaacagaaatggctggcggtaaagcaggcaagg attcgggcaaggccaaggcgaaggcggtatcgcgttccgcgcg cgcgggtcttcagttccccgtgggtcgcatccatcgtcatctcaag agccgcactacgtcacatggacgcgtcggagccactgcagccgtg tactccgctgccatattggaatacctgaccgccgaggtcctggagtt ggcaggcaacgcatcgaaggacttgaaagtgaaacgtatcactcc tcgccacttacagctcgccattcgcggagacgaggagctggacag cctgatcaaggcaaccatcgctggtggcggtgtcattccgcacata cacaagtcgctgatcggcaaaaaggaggaaacggtgcaggatccgc agcggaagggcaacgtcattctgtcgcaggcctactaagccagtcgg caatcggacgccttcgaaacatgcaacactaatgtttaattcagattt cagcagagacaagctaaaacaccgacgagttgtaatcatttctgtgcg ccagcatatatttcttatatacaacgtaatacataattatgtaattctagca tctccccaacactcacatacatacaaacaaaaaatacaaacacacaaaac gtatttacccgcacgcatccttggcgaggttgagtatgaaacaaa aacaaaacttaatttagagcaaagtaattacacgaataaatttaataa aaaaaactataataaaaacgcc.

Let’s use the translation software available on the Internet at http://www.expasy.org/tools/dna.html to translate this gene in all six possible reading frames and see which reading frame specifies histone H2A. Just type or paste the DNA sequence in the “ExPASy Translate” tool box and click TRANSLATE SEQUENCE. The results will show the products of translation in all six reading frames with Met’s and Stop’s boldfaced to highlight potential open-reading frames. Which reading frame specifies histone H2A? ANS: Reading frame 5′ → 3′ number 1 has a large open reading frame with a methionine codon near the 5′ end. You can verify that this is the correct reading frame by using the predicted translation product as a query to search one of the protein databases (see Problem 15.26).

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68-WC Answers to All Questions and Problems Chapter 16 16.1 What are CpG islands? Of what value are CpG islands in positional cloning of human genes?

ANS: CpG islands are clusters of cytosines and guanines that are often located just upstream (5′) from the coding regions of human genes. Their presence in nucleotide sequences can provide hints as to the location of genes in human chromosomes.

16.2 Why is the mutant gene that causes Huntington’s disease called huntingtin? Why might this gene be renamed in the future?

ANS: The gene was named huntingtin after the disease that it causes when defective. The gene will probably be renamed after the function of its gene product has been determined.

16.3 How was the nucleotide sequence of the CF gene used to obtain information about the structure and function of its gene product?

ANS: The CF gene was identified by map position-based cloning, and the nucleotide sequences of CF cDNAs were used to predict the amino acid sequence of the CF gene product. A computer search of the protein data banks revealed that the CF gene product was similar to several ion channel proteins. This result focused the attention of scientists studying cystic fibrosis on proteins involved in the transport of salts between cells and led to the discovery that the CF gene product was a transmembrane conductance regulator—now called the CFTR protein.

16.4 How might the characterization of the CF gene and its product lead to the treatment of cystic fibrosis by somatic-cell gene therapy? What obstacles must be overcome before cystic fibrosis can be treated successfully by gene therapy?

ANS: Once the function of the CF gene product has been established, scientists should be able to develop procedures for introducing wild-type copies of the CF gene into the appropriate cells of cystic fibrosis patients to alleviate the devastating effects of the mutant gene. A major obstacle to somatic-cell gene-therapy treatment of cystic fibrosis is the size of the CF gene—about 250 kb, which is too large to fit in the standard gene transfer vectors. Perhaps a shortened version of the gene constructed from the CF cDNA—about 6.5 kb—can be used in place of the wild-type gene. A second major obstacle is getting the transgene into enough of the target cells of the cystic fibrosis patient to alleviate the symptoms of the disease.

A third challenge is to develop an expression vector containing the gene that will result in long-term expression of the introduced gene in transgenic cells. Another concern is how to avoid possible side effects caused by overexpression or inappropriate expression of the transgene in cystic fibrosis patients. Despite these obstacles, many scientists are optimistic that cystic fibrosis will be effectively treated by somatic-cell gene therapy in the future.

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16.5 Myotonic dystrophy (MD), occurring in about 1 of 8000 individuals, is the most common form of muscular ­dystrophy in adults. The disease, which is characterized by progressive muscle degeneration, is caused by a dominant mutant gene that contains an expanded CAG repeat region. Wild-type alleles of the MD gene contain 5–30 copies of the trinucleotide. Mutant MD alleles contain 50 to over 2000 copies of the CAG repeat. The complete nucleotide sequence of the MD gene is available. Design a diagnostic test for the mutant gene responsible for myotonic dystrophy that can be carried out using genomic DNA from newborns, fetal cells obtained by amniocentesis, and single cells from eight-cell preembryos produced by in vitro fertilization.

ANS: Oligonucleotide primers complementary to DNA sequences on both sides (upstream and downstream) of the CAG repeat region in the MD gene can be synthesized and used to amplify the repeat region by PCR. One primer must be complementary to an upstream region of the template strand, and the other primer must be ­complementary to a downstream region of the nontemplate strand. After amplification, the size(s) of the CAG repeat regions can be determined by gel electrophoresis (see Figure 16.2). Trinucleotide repeat lengths can be ­measured by including repeat regions of known length on the gel. If fewer than 30 copies of the trinucleotide repeat are present on each chromosome, the newborn, fetus, or pre-embryo is homozygous for a wild-type MD allele or heterozygous for two different wild-type MD alleles. If more than 50 copies of the repeat are present on each of the homologous chromosomes, the individual, fetus, or cell is homozygous for a dominant mutant MD allele or heterozygous for two different mutant alleles. If one chromosome contains less than 30 copies of the CAG repeat and the homologous chromosome contains more than 50 copies, the newborn, fetus, or pre-embryo is h ­eterozygous, carrying one ­wild-type MD allele and one mutant MD allele.

16.6 In humans, the absence of an enzyme called purine nucleoside phosphorylase (PNP) results in a severe T-cell immunodeficiency similar to that of severe combined immunodeficiency disease (SCID). PNP deficiency exhibits an autosomal recessive pattern of inheritance, and the gene encoding human PNP has been cloned and sequenced. Would PNP deficiency be a good candidate for treatment by gene therapy? Design a procedure for the treatment of PNP deficiency by somatic-cell gene therapy.

ANS: Yes. A somatic-cell gene therapy procedure similar to that used for X-linked SCID (see Figure 16.7) might be effective in treating purine nucleoside phosphorylase (PNP) deficiency. White blood cells could be isolated from the patient, transfected with a vector carrying a wild-type PNP gene, grown in culture and assayed for the expression of the PNP transgene, and then infused back into the patient after the expression of the transgene had been verified.

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Answers to All Questions and Problems WC-69

16.7 Human proteins can now be produced in bacteria such as E. coli. However, one cannot simply introduce a human gene into E. coli and expect it to be expressed. What steps must be taken to construct an E. coli strain that will produce a mammalian protein such as human growth hormone?

ANS: The transcription initiation and termination and translation initiation signals or eukaryotes differ from those of prokaryotes such as E. coli. Therefore, to produce a human protein in E. coli, the coding sequence of the human gene must be joined to appropriate E. coli regulatory signals—promoter, transcription terminator, and translation initiator sequences. Moreover, if the gene contains introns, they must be removed or the coding sequence of a cDNA must be used, because E. coli does not possess the spliceosomes required for the excision of introns from nuclear gene transcripts. In addition, many eukaryotic proteins undergo posttranslational processing events that are not carried out in prokaryotic cells.

Such proteins are more easily produced in transgenic eukaryotic cells growing in culture.

16.8 You have constructed a synthetic gene that encodes an enzyme that degrades the herbicide glyphosate. You wish to introduce your synthetic gene into Arabidopsis plants and test the transgenic plants for resistance to glyphosate. How could you produce a transgenic Arabidopsis plant harboring your synthetic gene by A. tumefaciensmediated transformation?

ANS: You would first construct a chimeric gene containing your synthetic gene fused to a plant promoter such as the 35S promoter of cauliflower mosaic virus and a plant transcription termination and polyadenylation signal such as the one from the nos gene of the Ti plasmid. This chimeric gene would then be inserted into the T-DNA of a Ti plasmid carrying a dominant selectable marker gene (e.g., 35S/NPTII/nos, which confers resistance to kanamycin to host cells) and introduced into Agrobacterium tumefaciens cells by transformation. Tissue explants from Arabidopsis plants would be co-cultivated with A. tumefaciens cells harboring the recombinant Ti plasmid, and plant cells that carry T-DNAs inserted into their chromosomes would be selected by growth on medium containing the appropriate selective agent (e.g., kanamycin). Transgenic plants would then be regenerated from the transformed cells and tested for resistance to glyphosate.

16.9 A human STR locus contains a tandem repeat (TAGA)n, where n may be any number between 5 and 15. How many alleles of this locus would you expect to find in the human population?

ANS: Eleven, ranging in multiples of 3, from 15 to 45 nucleotides long.

16.10 A group of bodies are found buried in a forest. The police suspect that they may include the missing Jones family (two parents and two children). They extract DNA from

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bones and examine (using PCR) genes A and B, which are known to contain tandem triplet repeats of variable length. They also analyze DNA from two other men.

The results are shown below where the numbers indicate the number of copies of a tandem repeat in a particular allele; for example, male 1 has one allele with 8 and another allele with 9 copies of a tandem repeat in gene A. Gene A

Gene B

Male 1

8/9

5/7

Male 2

6/8

5/5

Male 3

7/10

7/7

Woman

8/8

3/5

Child 1

7/8

5/7

Child 2

8/8

3/7

Could the woman have been the mother of both children? Why or why not? Which man, if any, could have been the father of child 1?

ANS: The woman could be the mother of both children, having passed alleles A 8 and B 5 to child 1 and A 8 and B 3 to child 2. Male 3 could have been the father of child 1.

If so, he contributed alleles A 7 and B 7.

16.11 DNA profiles have played central roles in many rape and murder trials. What is a DNA profile? What roles do DNA profiles play in these forensic cases? In some cases, geneticists have been concerned that DNA profile data were being used improperly. What were some of their concerns, and how can these concerns be properly addressed?

ANS: DNA profiles are the specific patterns (1) of peaks present in electropherograms of chromosomal STRs or VNTRs amplified by PCR using primers tagged with fluorescent dyes and separated by capillary gel electrophoresis (see Figures 16.11 and 16.12) or (2) of bands on Southern blots of genomic DNAs that have been digested with specific restriction enzymes and hybridized to appropriate STR or VNTR sequences (see Figure 16.10).

DNA profiles, “like” epidermal fingerprints, are used as evidence for identity or nonidentity in forensic cases.

Geneticists have expressed concerns about the statistical uses of DNA profile data. In particular, they have questioned some of the methods used to calculate the probability that DNA from someone other than the suspect could have produced an observed profile. These concerns have been based in part on the lack of adequate DNA profile databases for various human subpopulations and the lack of precise information about the amount of variability in DNA profiles for individuals of different ethnic backgrounds. These concerns have been addressed by the acquisition of data on profile frequencies in different populations and ethnic groups from throughout the world.

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70-WC Answers to All Questions and Problems 16.12 The DNA profiles shown in this problem were prepared using genomic DNA from blood cells obtained from a woman, her daughter, and three men who all claim to be the girl’s father.

M

C

F1

F2

F3

16.15 The generation of transgenic plants using A. tumefaciensmediated transformation often results in multiple sites of insertion. These sites frequently vary in the level of transgene expression. What approaches could you use to determine whether or not transgenic plants carry more than one transgene and, if so, where the transgenes are inserted into chromosomes?

ANS: Probing Southern blots of restriction enzyme-digested DNA of the transgenic plants with 32P-labeled transgene may provide evidence of multiple insertions but would not reveal the genomic location of the inserts. Fluorescence in situ hybridization (FISH) is a powerful procedure for determining the genomic location of gene inserts. FISH is used to visualize the location of transgenes in chromosomes.

16.16 Disarmed retroviral vectors can be used to introduce genes into higher animals including humans. What advantages do retroviral vectors have over other kinds of gene-transfer vectors? What disadvantages?

Based on the DNA profiles, what can be determined about paternity in this case?

ANS: Neither F1 nor F2 could be the girl’s biological father; only individual F3 could be the child’s father.

16.13 Most forensic experts agree that profiles of DNA from blood samples obtained at crime scenes and on personal items can provide convincing evidence for murder convictions. However, the defense attorneys sometimes argue successfully that sloppiness in handling blood samples results in contamination of the samples. What problems would contamination of blood samples present in the interpretation of DNA profiles? Would you expect such errors to lead to the conviction of an innocent person or the acquittal of a guilty person?

ANS: Contamination of blood samples would introduce more variability into DNA profiles. This would lead to a lack of allelic matching of profiles obtained from the blood samples and from the defendant. Mixing errors would be expected to lead to the acquittal of a guilty person and not to the conviction of an innocent person. Only the mislabeling of samples could implicate someone who is innocent.

16.14 The Ti plasmid contains a region referred to as T-DNA. Why is this region called T-DNA, and what is its significance?

ANS: The T in T-DNA is an abbreviation for “transferred.”

The T-DNA region of the Ti plasmid is the segment that is transferred from the Ti plasmid of the bacterium to the chromosomes of the plant cells during Agrobacterium tumefaciens-mediated transformation.

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ANS: Disarmed retroviruses are lacking genes essential for reproduction in host cells but can still integrate into the DNA of the host cell in the proviral state. The retroviral genomes are small enough to allow them to be manipulated easily in vitro and yet will accept foreign DNA inserts of average gene size. The retroviruses contain strong promoters in their long terminal repeats that can be used to drive high levels of transcription of the foreign gene insert.

16.17 Transgenic mice are now routinely produced and studied in research laboratories throughout the world. How are transgenic mice produced? What kinds of information can be obtained from studies performed on transgenic mice? Does this information have any importance to the practice of medicine? If so, what?

ANS: Transgenic mice are usually produced by microinjecting the genes of interest into pronuclei of fertilized eggs or by infecting pre-implantation embryos with retroviral vectors containing the genes of interest. Transgenic mice provide invaluable tools for studies of gene expression, mammalian development, and the immune system of mammals. Transgenic mice are of major importance in medicine; they provide the model system most closely related to humans. They have been, and undoubtedly will continue to be, of great value in developing the tools and technology that will be used for human gene therapy in the future.

16.18 Two men claim to be the father of baby Joyce Doe. Joyce’s mother had her CODIS STR DNA profile analyzed and was homozygous for allele 8 at the TPOX locus (allele 8 contains 8 repeats of the GAAT sequence at this polymorphic locus). Baby Joyce is heterozygous for alleles 8 and 11 at this locus. In an attempt to resolve the disputed paternity, the two men were tested for their STR DNA profiles at the TPOX locus on chromosome 2. Putative father 1 was heterozygous for alleles 8 and 11 at the

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Answers to All Questions and Problems WC-71

TPOX locus, and putative father 2 was homozygous for allele 11 at this locus. Can these results resolve this case of disputed paternity. If so, who is the biological father? If not, why not?

ANS: The results cannot resolve the case of disputed paternity.

Baby Joyce received allele 8 at the TPOX locus from her mother. Given that baby Joyce is heterozygous for alleles 8 and 11 at the TPOX locus, she must have received allele 11 from her father. However, both of the putative fathers carry allele 11 and could have passed it on to baby Joyce.

16.19 Many valuable human proteins contain carbohydrate or lipid components that are added posttranslationally. Bacteria do not contain the enzymes needed to add these components to primary translation products. How might these proteins be produced using transgenic animals?

ANS: Posttranslationally modified proteins can be produced in transgenic eukaryotic cells growing in culture or in transgenic plants and animals. Indeed, transgenic sheep have been produced that secrete human blood-clotting factor IX and a1-antitrypsin in their milk. These sheep were produced by fusing the coding sequences of the respective genes to a DNA sequence that encodes the signal peptide required for secretion and introducing this chimeric gene into fertilized eggs that were then implanted and allowed to develop into transgenic animals. In principle, this approach could be used to produce any protein of interest.

16.20 Richard Meagher and coworkers have cloned a family of 10 genes that encode actins (a major component of the cytoskeleton) in Arabidopsis thaliana. The 10 actin gene products are similar, often differing by just a few amino acids. Thus, the coding sequences of the 10 genes are also very similar, so that the coding region of one gene will cross-hybridize with the coding regions of the other nine genes. In contrast, the noncoding regions of the 10 genes are quite divergent. Meagher has hypothesized that the 10 actin genes exhibit quite different temporal and spatial patterns of expression. You have been hired

by Meagher to test this hypothesis. Design experiments that will allow you to determine the temporal and spatial pattern of expression of each of the 10 actin genes in Arabidopsis.

ANS: You can subclone the 3′ noncoding regions (sequences between the translation-termination codons and the 3′ termini of the transcripts) of the actin genes and use these sequences as gene-specific hybridization probes, after verifying their specificity (no cross-hybridization to the other genes in the gene family). This procedure has worked elegantly for the 15 tubulin genes and the 10 actin genes of Arabidopsis, because the transcript sequences are very divergent in the 5′ and 3′ noncoding regions. You cannot, of course, use intron sequences because they are excised during pre-mRNA processing.

These 3′ noncoding gene-specific hybridization probes

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are then used to measure individual gene transcript levels in various organs and tissues of developing plants by either northern blot or in situ hybridization experiments.

Alternatively, the 5′ and 3′ noncoding regions can be used to design gene-specific PCR primers, and reverse transcription PCR (RT-PCR) can be used to measure individual gene transcript levels in organs and tissues.

Indeed, Meagher and colleagues have already used this approach to document the striking temporal and tissuespecific patterns of actin gene expression in Arabidopsis.

16.21 The first transgenic mice resulted from microinjecting fertilized eggs with vector DNA similar to that diagrammed in Figure 16.15 except that it contained a promoter for the mammalian metallothionein gene linked to the HGH gene. The resulting transgenic mice showed elevated levels of HGH in tissues of organs other than the pituitary gland—for example, in heart, lung, and liver— and the pituitary gland underwent atrophy. How might the production of HGH in transgenic animals be better regulated, with expression restricted to the pituitary gland?

ANS: The vector described contains the HGH gene; however, it does not contain a mammalian HGH promoter that will regulate the expression of the transgene in the appropriate tissues. Construction of vectors containing a properly positioned mammalian HGH-promoter sequence should result in transgenic mice in which HGH synthesis is restricted to the pituitary gland.

16.22 How do the reverse genetic approaches used to dissect biological processes differ from classical genetic approaches?

ANS: Classical genetic approaches use mutational dissection to probe the functions of genes. Mutant alleles that produce altered phenotypes are identified and used to investigate the functions of the wild-type alleles. Comparative molecular analyses of mutant and wild-type organisms sometimes allow researchers to determine the precise function of a gene. Reverse genetic approaches use the known nucleotide sequences of genes to design procedures to either produce null mutations in them or inhibit their expression. RNA interference protocols are used to block the expression of specific genes. T-DNA and transposon insertion protocols are used to produce null mutations (to “knock out” the function) of specific genes.

Basically, in classical genetics, you start with mutant alleles and hope to uncover the wild-type gene function; in reverse genetics, you start with the wild-type gene and generate mutant alleles.

16.23 How can RNAi gene silencing be used to determine the function of genes?

ANS: RNAi involves the use of double-stranded RNAs, where one strand is complementary to the mRNA and the other strand is equivalent to the mRNA, to silence the expression of target genes. RNAi makes use of the ­RNA-induced silencing complex (RISC) to block gene expression (see Figure 16.23).

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72-WC Answers to All Questions and Problems 16.24 How do insertional mutagenesis approaches differ from other reverse genetic approaches?

ANS: Insertional mutagenesis approaches produce mutant alleles—usually null alleles, or “knockouts,”—by the insertion of foreign DNA into genes. Other reverse genetic approaches, for example, RNA interference, inhibit the expression of the genes but leave them structurally intact.

16.25 Insertional mutagenesis is a powerful tool in both plants and animals. However, when performing large-scale insertional mutagenesis, what major advantage do plants have over animals?

ANS: Plants have an advantage over animals in that once insertional mutations are induced they can be stored for long periods of time and distributed to researchers as dormant seeds.

16.26 We discussed the unfortunate effects of insertional mutagenesis in the four boys who developed leukemia after treatment of X-linked severe combined immunodeficiency disease by gene therapy. How might this consequence of gene therapy be avoided in the future? Do you believe that the use of somatic-cell gene therapy to treat human diseases can ever be made 100 percent risk-free?

Why? Why not?

ANS: The vectors used in somatic-cell gene therapy must not insert themselves preferentially into important genes, especially proto-oncogenes. Ideally, the vectors should insert themselves into unessential regions of the human genome. However, such vectors may not exist. At the minimum, however, vectors should be used that insert into the genome at random sites so that their chance of inserting into an important regulatory gene such as a proto-oncogene is very low. Gene therapy will probably never be completely risk-free, because anomalous insertion events will always occur at some low frequency. No biological process is 100 percent accurate. However, the potential benefits of gene therapy must outweigh the potential risks before the procedure will become an accepted tool for treating inherited diseases.

16.27 One strand of a gene in Arabidopsis thaliana has the following nucleotide sequence: atgagtgacgggaggaggaagaagagcgtgaacggaggt gcaccggcgcaaacaatcttggatgatcggagatctagtcttccgga agttgaagcttctccaccggctgggaaacgagctgttatcaagagtgc cgatatgaaagatgatatgcaaaaggaagctatcgaaatcgccatct ccgcgtttgagaagtacagtgtggagaaggatatagctgagaatataa agaaggagtttgacaagaaacatggtgctacttggcattgcattgttgg tcgcaactttggttcttatgtaacgcatgagacaaaccatttcgtttacttct acctcgaccagaaagctgtgctgctcttcaagtcgggttaa The function(s) of this gene is still uncertain. (a) How might insertional mutagenesis be used to investigate the function(s) of the gene? (b) Design an experiment using RNA interference to probe the function(s) of the gene.

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ANS: (a) You would first want to check the Salk Institute’s Genome Analysis Laboratory web site to see if a T-DNA or transposon insertion has already been identified in this gene (see Problem 16.28). If so, you can simply order seeds of the transgenic line from the Arabidopsis Biological Resource Center at Ohio State University. If no insertion is available in the gene, you can determine where it maps in the genome and use transposons that preferentially jump to nearby sites to identify a new insertional mutation (see http://www.arabidopsis.org/ abrc/ima/jsp). (b) You can construct a gene that has sense and antisense sequences transcribed to a single mRNA molecule (see Figure 16.23b), introduce it into Arabidopsis plants by A. tumefaciens-mediated transformation, and study its effect(s) on the expression of the gene and the phenotype of transgenic plants. The transcript will form a partially base-paired hairpin that will enter the RISC silencing pathway and block the expression of the gene (see Figure 16.23b).

16.28 Let’s check the Salk Institute’s Genome Analysis Laboratory web site (http://signal.salk.edu/cgi-bin/tdnaexpress) to see if any of their T-DNA lines have insertions in the gene shown in the previous question. At the SIGnAL web site, scroll down to “Blast” and paste or type the sequence in the box. The resulting map will show the location of mapped T-DNA insertions relative to the location of the gene (green rectangle at the top). The blue arrows at the top right will let you focus on just the short region containing the gene or relatively long regions of chromosome 4 of Arabidopsis. Are there any T-DNA insertions in the gene in question? Near the gene?

ANS: A search of the Salk Institute’s Genome Analysis Laboratory (SIGnAL) web site reveals that there are numerous T-DNA insertions (Salk T-DNA insertions and several others) located in the promoter region of this gene.

Moreover, there are two insertion lines (FLAG_177C02 and FLAG_216D01) in the collection at the Institute of Agronomic Research in Versailles, France, with inserts in the coding region of this gene. All of these insertion lines in these collections are available to researchers on request. Thus, the function of this gene can be studied by using these insertion lines.

16.29 The CRISPR/Cas9 anti-phage immunity system in Streptococcus pyogenes deploys a variety of crRNAs derived from the spacer and repeat sequences in the CRISPR array in the S. pyogenes genome. In combination with the transactivating RNA (tracrRNA), these crRNAs guide the Cas9 endonuclease to complementary sequences in infecting phage genomes, whereupon Cas9 cleaves the phage DNA. A requirement for cleavage is that the targeted phage DNA sequence be immediately upstream of a protospacer adjacent motif (PAM), which in the S. pyogenes system is 5′-NGG-3′. Why is it important that the CRISPR array in the S. pyogenes genome not contain this PAM?

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ANS: It is important that the CRISPR array in the S. pyogenes genome does not contain the PAM 5′NGG-3′ because if it did, crRNAs generated from the array could target the Cas9 endonuclease to the array and cleave it, resulting in breakage of the S. pyogenes chromosome.

16.30 The Streptococcus pyogenes Cas9 endonuclease can be targeted to a specific genomic DNA sequence by an sgRNA that at its 5′ end has 20 nucleotides complementary to the target sequence. If this target sequence is immediately upstream of the protospacer adjacent motif (PAM) 5′NGG-3′, Cas9 will cleave the target DNA. Suppose you have chosen a 20-nucleotide target sequence in the genome of Drosophila melanogaster and that this sequence is next to the required PAM. How could you determine if Cas9 will cleave only this sequence in the Drosophila genome?

effector molecule that usually, if not always, combines with the product of one or more regulator genes to turn off the synthesis of the enzymes in the biosynthetic pathway. Feedback inhibition occurs at the level of enzyme activity; it usually involved the first enzyme of the biosynthetic pathway. Feedback inhibition thus brings about an immediate arrest of the biosynthesis of the end-­ product. All together, feedback inhibition and repression rapidly and efficiently turn off the synthesis of both the enzymes and the end-products that no longer need to be synthesized by the cell.

17.3 In the lactose operon of E. coli, what is the function of each of the following genes or sites: (a) regulator, (b) operator, (c) promoter, (d) structural gene Z, and (e) structural gene Y?

ANS: You could use the 20-nucleotide sequence as a query in BLAST to scan the sequenced portion of the Drosophila genome to see if all or part of the target sequence is present anywhere else. If this sequence is present elsewhere, then you could check to see if the PAM 5′-NGG-3′ is immediately downstream of the sequence. If it is, then the Cas9 endonuclease will cleave at this site as well as at the intended target site.

ANS: Gene or Regulatory Element

16.31 How could the CRISPR/Cas9 system be used to create a translocation between two autosomes in cultured human cells?

ANS: Create two sgRNAs, one to target a sequence on a particular autosome and the other to target a sequence on a different autosome. Then introduce these sgRNAs and the Cas9 endonuclease into cultured cells to induce breakage at the two target sites. The broken DNA molecules may be repaired by the NHEJ pathway, and if they are, the broken pieces of different autosomes could be joined covalently, creating a reciprocal translocation.

Function

(a) Regulator gene

Encodes the repressor

(b) Operator

Binding site of repressor

(c) Promoter

Binding site of RNA polymerase and CAP-cAMP complex

(d) Structural gene Z

Encodes b-galactosidase

(e) Structural gene Y

Encodes b-galactoside permease

17.4 What would be the result of inactivation by mutation of the following genes or sites in the E. coli lactose operon: (a) regulator, (b) operator, (c) promoter, (d) structural gene Z, and (e) structural gene Y? ANS: (a) Constitutive synthesis of the lac enzymes. (b) Constitutive synthesis of the lac enzymes. (c) Uninducibility of the lac enzymes.

Chapter 17

(d) No b-galactosidase activity.

17.1 How can inducible and repressible enzymes of microorganisms be distinguished?

(e) No b-galactoside permease activity.

ANS: By studying the synthesis or lack of synthesis of the enzyme in cells grown on chemically defined media. If the enzyme is synthesized only in the presence of a certain metabolite or a particular set of metabolites, it is probably inducible. If it is synthesized in the absence but not in the presence of a particular metabolite or group of metabolites, it is probably repressible.

17.2 Distinguish between (a) repression and (b) feedback inhibition caused by the end-product of a biosynthetic pathway. How do these two regulatory phenomena complement each other to provide for the efficient regulation of metabolism?

ANS: Repression occurs at the level of transcription during enzyme synthesis. The end-product, or a derivative of the end-product, of a repressible system acts as an

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17.5 Groups of alleles associated with the lactose operon are as follows (in order of dominance for each allelic series): repressor, Is (superrepressor), I+ (inducible), and I− (constitutive); operator, Oc (constitutive, cis− dominant) and O+ (inducible, cis-dominant); structural, Z+ and Y+. (a) Which of the following genotypes will produce b-galactosidase and b-galactoside permease if lactose is present: (1) I+O+Z+Y+, (2) I−OcZ+Y+, (3) IsOcZ+Y+, (4) IsO+Z+Y+, and (5) I−O+Z+Y+? (b) Which of the above genotypes will produce b-galactosidase and b-galactoside permease if lactose is absent? Why? ANS: (a) 1, 2, 3, and 5. Genotype 1 is wild-type and inducible, whereas genotypes 2, 3, and 5 are constitutive; all except 4 will produce b-galactosidase and b-galactoside permease in the presence of lactose. However, genotype 4 has a superrepressor mutation (Is) and is uninducible with normal

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74-WC Answers to All Questions and Problems levels of lactose. (b) 2, 3, and 5. In genotypes 2 and 3, the repressor cannot bind to Oc and in genotype 5, no repressor is made; both situations render the operon constitutive.

17.6 Assume that you have discovered a new strain of E. coli that has a mutation in the lac operator region that causes the wild-type repressor protein to bind irreversibly to the operator. You have named this operator mutant Osb for “superbinding” operator. (a) What phenotype would a partial diploid of genotype I+OsbZ−Y+/I+O+Z+Y− have with respect to the synthesis of the enzymes b-galactosidase and b-galactoside permease? (b) Does your new Osb mutation exhibit cis or trans dominance in its effects on the regulation of the lac operon?

ANS: (a) b-Galactosidase will be produced only when lactose is present. Permease will not be produced at all. (b) cis dominance.

17.7 Why is the Oc mutation in the E. coli lac operon epistatic to the Is mutation?

ANS: The Oc mutant prevents the repressor from binding to the operator. The Is mutant repressor cannot bind to Oc. The Is mutant protein has a defect in the allosteric site that binds allolactose but has a normal operator binding site. Therefore, because the single Oc mutant would have the same phenotype as the Oc Is double mutant, the Oc mutation is, by definition, epistatic to Is. 17.8 For each of the following partial diploids indicate whether enzyme synthesis is constitutive or inducible (see Problem 17.5 for dominance relationships): (a) I+O+Z+Y+/I+O+Z+Y+, (b) I+O+Z+Y+/I+OcZ+Y+, (c) I+OcZ+Y+/I+OcZ+Y+,

(d) I+O+Z+Y+/I−O+Z+Y+,

(e) I−O+Z+Y+/I−O+Z+Y+.

Why? ANS: (a) Inducible, this is the wild-type genotype and phenotype. (b) Constitutive, the Oc mutation produces an operator that is not recognized by the lac repressor. (c) Constitutive, same as for (b). (d) Inducible, I+ is dominant to I−. (e) Constitutive, no active repressor is synthesized in this bacterium.

17.9 Write the partial diploid genotype for a strain that will (a) produce b-galactosidase constitutively and permease inducibly and (b) produce b-galactosidase constitutively but not permease either constitutively or inducibly, even though a Y+ gene is known to be present.

construct several E. coli strains that are partially diploid for the lac operon. You construct strains with the following genotypes: (1) I+OcZ+Y−/I+O+Z−Y+, (2) I+OcZ−Y+/ I+O+Z+Y−, (3) I−O+Z+Y+/I+O+Z−Y−, (4) IsO+Z−Y−/ I+O+Z+Y+, and (5) I+OcZ+Y+/IsO+Z−Y+. (a) Which of these strains will produce functional b-galactosidase in both the presence and absence of lactose? (b) Which of these strains will exhibit constitutive synthesis of functional b-galactoside permease? (c) Which of these strains will express both gene Z and gene Y constitutively and will produce functional products (b-galactosidase and b-galactoside permease) of both genes? (d) Which of these strains will show cis dominance of lac operon regulatory elements? (e) Which of these strains will exhibit trans dominance of lac operon regulatory elements?

ANS: (a) 1, 5. (b) 2, 5. (c) 5. (d) 1, 2, 5. (e) 3, 4.

17.11 Constitutive mutations produce elevated enzyme levels at all times; they may be of two types: Oc or I−. Assume that all other DNA present is wild-type. Outline how the two constitutive mutants can be distinguished with respect to (a) map position, (b) regulation of enzyme ­levels in Oc/O+ versus I−/I+ partial diploids, and (c) the position of the structural genes affected by an Oc mutation versus the genes affected by an I− mutation in a partial diploid.

ANS: (a) The Oc mutations map very close to the Z structural gene; I− mutations map slightly farther from the structural gene (but still very close by; see Figure 17.5). (b) An I+O+Z+Y+/I+OcZ+Y+ partial diploid would exhibit constitutive synthesis of b-galactosidase and b-galactoside permease, whereas an I+O+Z+Y+/I−O+Z+Y+ partial diploid would be inducible for the synthesis of these enzymes. (c) The Oc mutation is cis-dominant; the I− mutation is trans-recessive.

17.12 How could the tryptophan operon in E. coli have developed and been maintained by evolution?

ANS: The system could have developed from a series of tandem duplications of a single ancestral gene. Mutational changes that make the system more efficient and, therefore, favored by natural selection could have brought the system to its present level of efficiency.

ANS: (a) I+ Oc Z+ Y−/I+ O+ Z+ Y+ (b) I+ Oc A+ Y−/Is O+ Z+ Y+

17.13 Of what biological significance is the phenomenon of catabolite repression?

17.10 As a genetics historian, you are repeating some of the classic experiments conducted by Jacob and Monod with the lactose operon in E. coli. You use an F plasmid to

ANS: Catabolite repression has apparently evolved to assure the use of glucose as a carbon source when this carbohydrate is available, rather than less efficient energy sources.

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17.14 How might the concentration of glucose in the medium in which an E. coli cell is growing regulate the intracellular level of cyclic AMP? ANS: Possibly by directly or indirectly inhibiting the enzyme adenylcyclase, which catalyzes the synthesis of cyclic AMP from ATP. 17.15 Is the CAP–cAMP effect on the transcription of the lac operon an example of positive or negative regulation? Why? ANS: Positive regulation; the CAP–cAMP complex has a positive effect on the expression of the lac operon. It functions in turning on the transcription of the structural genes in the operon.

17.16 Would it be possible to isolate E. coli mutants in which the transcription of the lac operon is not sensitive to catabolite repression? If so, in what genes might the mutations be located?

ANS: Yes; in the gene encoding CAP. Some mutations in this gene might result in a CAP that binds to the promoter in the absence of cAMP. Also, mutations in the gene (or genes) coding for the protein (or proteins) that regulate the cAMP level as a function of glucose concentration.

17.17 Using examples, distinguish between negative regulatory mechanisms and positive regulatory mechanisms.

ANS: Negative regulatory mechanisms, such as that involving the repressor in the lactose operon, block the transcription of the structural genes of the operon, whereas positive mechanisms, such as the CAP–cAMP complex in the lac operon, promote the transcription of the structural genes of the operon.

17.18 The following table gives the relative activities of the enzymes b-galactosidase and b-galactoside permease in cells with different genotypes at the lac locus in E. coli.

The level of activity of each enzyme in wild-type E. coli not carrying F’s was arbitrarily set at 100; all other values are relative to the observed levels of activity in these wild-type bacteria. Based on the data given in the table for genotypes 1 through 4, fill in the levels of enzyme activity that would be expected for the fifth genotype. 𝛃-Galactoside Genotype

𝛃-Galactosidase Permease

- Inducer + Inducer - Inducer + Inducer +

1. I O Z Y

0.1

100

0.1

100

2. I−O+Z+Y+

100

100

100

100

3. I+OcZ+Y+

25

100

25

100

4. I−O+Z+Y−/ F′ I−O+Z+Y+

200

200

100

100

+

+

+

5. I−OcZ−Y+/ F′ I+O+Z+Y+ ANS: 0.1; 100; 25.1; 200.

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17.19 The rate of transcription of the trp operon in E. coli is controlled by both (1) repression/derepression and (2) attenuation. By what mechanisms do these two regulatory processes modulate trp operon transcript levels?

ANS: Repression/derepression of the trp operon occurs at the level of transcription initiation, modulating the frequency at which RNA polymerase initiates transcription from the trp operon promoters. Attenuation modulates trp transcript levels by altering the frequency of termination of transcription within the trp operon leader region (trpL).

17.20 What effect will deletion of the trpL region of the trp operon have on the rates of synthesis of the enzymes encoded by the five genes in the trp operon in E. coli cells growing in the presence of tryptophan?

ANS: Deletion of the trpL region would result in the levels of the tryptophan biosynthetic enzymes in cells growing in the presence of tryptophan being increased about 10-fold because attenuation would no longer occur if this region were absent.

17.21 By what mechanism does the presence of tryptophan in the medium in which E. coli cells are growing result in premature termination or attenuation of transcription of the trp operon?

ANS: First, remember that transcription and translation are coupled in prokaryotes. When tryptophan is present in cells, tryptophan-charged tRNATrp is produced. This allows translation of the trp leader sequence through the two UGG Trp codons to the trp leader sequence UGA termination codon. This translation of the trp leader region prevents base-pairing between the partially complementary mRNA leader sequences 75–83 and 110–121 (see Figure 17.15b), which in turn permits formation of the transcription–termination “hairpin” involving leader sequences 110–121 and 126–134 (see Figure 17.15c).

17.22 Suppose that you used site-specific mutagenesis to modify the trpL sequence such that the two UGG Trp codons at positions 54–56 and 57–60 (see Figure 17.14) in the mRNA leader sequence were changed to GGG Gly codons. Will attenuation of the trp operon still be regulated by the presence or absence of tryptophan in the medium in which the E. coli cells are growing?

ANS: No. Attenuation of the trp operon would now be controlled by the presence or absence of Gly-tRNAGly. 17.23 What do trp attenuation and the lysine riboswitch have in common?

ANS: Both trp attenuation and the lysine riboswitch turn off gene expression by terminating transcription upstream from the coding regions of the regulated genes. Both involve the formation of alternative mRNA secondary structures—switching between the formation of antiterminator and transcription–terminator hairpins—in response to the presence or absence of a specific

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76-WC Answers to All Questions and Problems metabolite (compare Figure 17.15 and Figure 2 in the Focus On The Lysine Riboswitch).

17.24 Would attenuation of the type that regulates the level of trp transcripts in E. coli be likely to occur in eukaryotic organisms?

ANS: No. Since transcription (nucleus) and translation (cytoplasm) are no coupled in eukaryotes, attenuation of the type occurring in prokaryotes would not be possible.

Chapter 18 18.1 Operons are common in bacteria but not in eukaryotes.

Suggest a reason why.

ANS: In multicellular eukaryotes, the environment of an individual cell is relatively stable. There is no need to respond quickly to changes in the external environment. In addition, the development of a multicellular organism involves complex regulatory hierarchies composed of hundreds of different genes. The expression of these genes is regulated spatially and temporally, often through intricate intercellular signaling processes.

18.2 In bacteria, translation of an mRNA begins before the synthesis of that mRNA is completed. Why is this “coupling” of transcription and translation not possible in eukaryotes?

ANS: Coupling of transcription and translation is not possible in eukaryotes because these two processes take place in different cellular compartments—transcription in the nucleus and translation in the cytoplasm.

18.3 Muscular dystrophy in humans is caused by mutations in an X-linked gene that encodes a protein called dystrophin. What techniques could you use to determine if this gene is active in different types of cells, say skin cells, nerve cells, and muscle cells?

ANS: Activity of the dystrophin gene could be assessed by blotting RNA extracted from the different types of cells and hybridizing it with a probe from the gene (northern blotting); or the RNA could be reverse transcribed into cDNA using one or more primers specific to the dystrophin gene and the resulting cDNA could be amplified by the polymerase chain reaction (RT-PCR). Another technique would be to hybridize dystrophin RNA in situ—that is, in the cells themselves—with a probe from the gene.

It would also be possible to check each cell type for production of dystrophin protein by using anti-dystrophin antibodies to analyze proteins from the different cell types on western blots or analyze the proteins in the cells themselves—that is, in situ.

18.4 Why do steroid hormones interact with receptors inside the cell, whereas peptide hormones interact with receptors on the cell surface?

ANS: Steroid hormones are small, lipid-soluble molecules that have little difficulty passing through the cell membrane.

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Peptide hormones are typically too large to pass through the cell membrane freely; rather, their influence must be mediated by a signaling system that begins with a membrane-bound receptor protein that binds the hormone.

18.5 In the polytene chromosomes of Drosophila larvae (Chapter 6), some bands form large “puffs” when the larvae are subjected to high temperatures. How could you show that these puffs contain genes that are vigorously transcribed in response to this heat shock treatment?

ANS: One procedure would be to provide larvae with radioactively labeled UTP, a building block of RNA, under different conditions—with and without heat shock. Then prepare samples of polytene cells from these larvae for autoradiography. If the heat shock-induced puffs contain genes that are vigorously transcribed, the radioactive signal should be abundant in the puffs.

18.6 How would you distinguish between an enhancer and a promoter?

ANS: An enhancer can be located upstream, downstream, or within a gene, and it functions independently of its orientation. A promoter is almost always immediately upstream of a gene and it functions only in one direction with respect to the gene.

18.7 Tropomyosins are proteins that mediate the interaction of actin and troponin, two proteins involved in muscle contractions. In higher animals, tropomyosins exist as a family of closely related proteins that share some amino acid sequences but differ in others. Explain how these proteins could be created from the transcript of a single gene.

ANS: By alternate splicing of the transcript.

18.8 A polypeptide consists of three separate segments of amino acids, A—B—C. Another polypeptide contains segments A and C but not segment B. How might you determine if these two polypeptides are produced by translating alternately spliced versions of RNA from a single gene or by translating mRNA from two different genes?

ANS: Southern blotting of genomic DNA digested with an appropriate restriction enzyme, followed by hybridization of the blot with a probe containing the DNA encoding segments A and B, or B and C, or at least parts of these adjacent segments. If one DNA fragment is detected on the blot, the two polypeptides are encoded by a single gene whose RNA is alternately spliced to produce two mRNAs. If two DNA fragments are detected, the two polypeptides are encoded by two different genes.

18.9 What techniques could be used to show that a plant gene is transcribed when the plant is illuminated with light?

ANS: Northern blotting of RNA extracted from plants grown with and without light, or PCR amplification of cDNA made by reverse transcribing these same RNA extracts.

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18.10 When introns were first discovered, they were thought to be genetic “junk”—that is, sequences without any useful function. In fact, they appeared to be worse than junk because they actually interrupted the coding sequences of genes. However, among eukaryotes, introns are pervasive and anything that is pervasive in biology usually has a function. What function might introns have? What benefit might they confer on an organism?

ANS: Introns make it possible for genes to encode different— but related—polypeptides by alternate splicing of their RNA transcripts.

18.11 The GAL4 transcription factor in yeast regulates two adjacent genes, GAL1 and GAL10, by binding to DNA sequences between them. These two genes are transcribed in opposite directions on the chromosome, one to the left of the GAL4 protein’s binding site and the other to the right of this site. What property of enhancers does this situation illustrate?

ANS: That enhancers can function in either orientation.

18.12 Using the techniques of genetic engineering, a researcher has constructed a fusion gene containing the heat-shock response elements from a Drosophila hsp70 gene and the coding region of a jellyfish gene (gfp) for green fluorescent protein. This fusion gene has been inserted into the chromosomes of living Drosophila by the technique of transposon-mediated transformation (Chapter 21 on the Instructor Companion site). Under what conditions will the green fluorescent protein be synthesized in these genetically transformed flies? Explain.

ANS: The green fluorescent protein will be made after the flies are heat shocked.

18.13 Suppose that the segment of the hsp70 gene that was used to make the hsp70/gfp fusion in the preceding problem had mutations in each of its heat-shock response elements. Would the green fluorescent protein encoded by this fusion gene be synthesized in genetically transformed flies?

ANS: Probably not unless the promoter of the gfp gene is recognized and transcribed by the Drosophila RNA polymerase independently of the heat-shock response elements.

18.14 The polypeptide products of two different genes, A and B, each function as transcription factors. These polypeptides interact to form dimers: AA homodimers, BB homodimers, and AB heterodimers. If the A and B polypeptides are equally abundant in cells, and if dimer formation is random, what is the expected ratio of homodimers to heterodimers in these cells?

ANS: With equal abundance of the A and B polypeptides, AA homodimers should constitute 1/4 of the total dimers formed, BB homodimers should constitute 1/4 of the total, and AB heterodimers should constitute 1/2 of the total. The expected ratio of homodimers to heterodimers is therefore (1/4 + 1/4):(1/2) = 1:1.

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18.15 A particular transcription factor binds to enhancers in 40 different genes. Predict the phenotype of individuals homozygous for a frameshift mutation in the coding sequence of the gene that specifies this transcription factor.

ANS: The mutation is likely to be lethal in homozygous condition because the transcription factor controls so many different genes, and a frameshift mutation in the coding sequence will almost certainly destroy the transcription factor’s function.

18.16 The alternately spliced forms of the RNA from the Drosophila doublesex gene encode proteins that are needed to block the development of one or the other set of sexual characteristics. The protein that is made in female animals blocks the development of male characteristics, and the protein that is made in male animals blocks the development of female characteristics. Predict the phenotype of XX and XY animals homozygous for a null mutation in the doublesex gene.

ANS: Both XX and XY animals would develop as intersexes because neither of the forms of the doublesex protein will be able to block sexual development. In these animals, both developmental pathways will be carried out, leading to tissues that have both male and female characteristics.

18.17 The RNA from the Drosophila Sex-lethal (Sxl) gene is alternately spliced. In males, the sequence of the mRNA derived from the primary transcript contains all eight exons of the Sxl gene. In females, the mRNA contains only seven of the exons because during splicing exon 3 is removed from the primary transcript along with its flanking introns. The coding region in the female’s mRNA is therefore shorter than it is in the male’s mRNA. However, the protein encoded by the female’s mRNA is longer than the one encoded by the male’s mRNA. How might you explain this paradox?

ANS: Exon 3 contains an in-frame stop codon. Thus, the protein translated from the Sxl mRNA in males will be shorter than the protein translated from the shorter Sxl mRNA in females.

18.18 In Drosophila, expression of the yellow gene is needed for the formation of dark pigment in many different tissues; without this expression, a tissue appears yellow in color.

In the wings, the expression of the yellow gene is controlled by an enhancer located upstream of the gene’s transcription initiation site. In the tarsal claws, expression is controlled by an enhancer located within the gene’s only intron. Suppose that by genetic engineering, the wing enhancer is placed within the intron and the claw enhancer is placed upstream of the transcription initiation site. Would a fly that carried this modified yellow gene in place of its natural yellow gene have darkly pigmented wings and claws? Explain.

ANS: Yes. Enhancers are able to function in different positions in and around a gene.

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78-WC Answers to All Questions and Problems 18.19 A researcher suspects that a 550-bp-long intron contains an enhancer that drives expression of an Arabidopsis gene specifically in root tip tissue. Outline an experiment to test this hypothesis.

ANS: The intron could be placed in a GUS expression vector, which could then be inserted into Arabidopsis plants. If the intron contains an enhancer that drives gene expression in root tips, transgenic plants should show GUS expression in their root tips. See the Problem-Solving Skills feature in Chapter 18 for an example of this type of analysis.

18.20 What is the nature of each of the following classes of enzymes? What does each type of enzyme do to chromatin? (a) HATs, (b) HDACs, (c) HMTs. ANS: (a) HATs, histone acetyl transferases, transfer acetyl groups to the amino acid lysine in histones; (b) HDACs, histone deacetylases, remove acetyl groups from these lysines; (c) HMTs, histone methyl transferases, transfer methyl groups to lysine, arginine, and histidine in histones.

18.21 In Drosophila larvae, the single X chromosome in males appears diffuse and bloated in the polytene cells of the salivary gland. Is this observation compatible with the idea that X-linked genes are hyperactivated in Drosophila males?

ANS: Yes. The diffuse, bloated appearance indicates that the genes on this chromosome are being transcribed vigorously—the chromatin is “open for business.”

18.22 Suppose that the LCR of the β-globin gene cluster was deleted from one of the two chromosomes 11 in a man.

What disease might this deletion cause?

ANS: The LCR regulates the expression of all the genes linked to it. Deletion of the LCR would ablate or impair globin gene expression from one of the two chromosomes 11.

With less b-globin being produced, the individual would likely suffer from anemia.

18.23 Would double-stranded RNA derived from an intron be able to induce RNA interference?

ANS: Short interfering RNAs target messenger RNA molecules, which are devoid of introns. Thus, if siRNA were made from double-stranded RNA derived from an intron, it would be ineffective against an mRNA target.

18.24 An RNA interference-like phenomenon has been implicated in the regulation of transposable elements. In Drosophila, two of the key proteins involved in this regulation are encoded by the genes aubergine and piwi. Flies that are homozygous for mutant alleles of these genes are lethal or sterile, but flies that are heterozygous for them are viable and fertile. Suppose that you have strains of Drosophila that are heterozygous for aubergine or piwi mutant alleles. Why might the genomic mutation rate in these mutant strains be greater than the genomic mutation rate in a wild-type strain?

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ANS: The aubergine and piwi gene products mediate the RNAilike response. Reduction in the amount of aubergine or piwi protein would likely impair the organism’s ability to mount this response, and without a vigorous capacity for regulation, transposable elements would be more likely to move in the genome. This movement would likely cause mutations because the transposons could insert into genes and inactivate them. Thus, Drosophila that are heterozygous for mutations in aubergine or piwi might experience higher mutation rates than Drosophila lacking these mutations.

18.25 Suppose that female mice homozygous for the a allele of the Igf2 gene are crossed to male mice homozygous for the b allele of this gene. Which of these two alleles will be expressed in the F1 progeny?

ANS: The paternally contributed allele (b) will be expressed in the F1 progeny.

18.26 Epigenetic states are transmitted clonally through cell division. What kinds of observations indicate that these states can be reversed or reset?

ANS: Here are two observations that show reversal or resetting of an epigenetic state: (1) For imprinted genes in mammals, the epigenetic state can be reset when the gene passes through the germ line of the opposite sex.

(2) Genes on the inactive X chromosome in mammals are reactivated in the female germ line.

18.27 A researcher hypothesizes that in mice gene A is actively transcribed in liver cells, whereas gene B is actively transcribed in brain cells. Describe procedures that would allow the researcher to test this hypothesis.

ANS: RNA could be isolated from liver and brain tissue.

Northern blotting or RT-PCR with this RNA could then establish which of the genes (A or B) is transcribed in which tissue. For northern blotting, the RNA samples would be fractionated in a denaturing gel and blotted to a membrane and then the RNA on the membrane would be hybridized with gene-specific probes, first for one gene, then for the other (or the researcher could prepare two separate blots and hybridize each one with a different probe). For RT-PCR, the RNA samples would be reverse transcribed into cDNA using primers specific for each gene; then the cDNA molecules would be amplified by standard PCR and the products of the amplifications would be fractionated by gel electrophoresis to determine which gene’s RNA was present in the original samples.

18.28 Suppose that the hypothesis mentioned in the previous question is correct and that gene A is actively transcribed in liver cells, whereas gene B is actively transcribed in brain cells. The researcher now extracts equivalent amounts of chromatin from liver and brain tissues and treats these extracts separately with DNase I for a limited period of time. If the DNA that remains after the treatments is then fractionated by gel electrophoresis,

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Answers to All Questions and Problems WC-79

transferred to a membrane by Southern blotting, and hybridized with a radioactively labeled probe specific for gene A, which sample (liver or brain) will be expected to show the greater signal on the autoradiogram? Explain your answer.

ANS: The sample of chromatin from brain tissue would be expected to show the greater signal on a Southern blot hybridized with a probe specific for gene A. The reason is that this gene is not so well transcribed in brain cells; consequently, it will be more resistant to digestion with DNase I in chromatin derived from brain cells than in chromatin derived from liver cells in which it is actively transcribed (and therefore more open to digestion with DNase I).

18.29 Why do null mutations in the msl gene in Drosophila have no effect in females?

ANS: The msl gene is not functional in females.

18.30 Suppose that a woman carries an X chromosome in which the XIST locus has been deleted. The woman’s other X chromosome has an intact XIST locus. What pattern of X-inactivation would be observed throughout the woman’s body?

ANS: The X chromosome containing the intact XIST locus will be silenced in all cases because that locus is located with the X inactivation center (XIC) and aids in silencing the inactive X. 18.31 In Drosophila, the variegated phenotype of the white mottled allele is suppressed by a dominant autosomal mutation that knocks out the function of the gene for heterochromatin protein 1 (HP1), an important factor in heterochromatin formation. Flies with the white mottled allele and the suppressor mutation have an almost uniform red color in their eyes; without the suppressor mutation, the eyes are mosaics of red and white tissue.

Can you suggest an explanation for the effect of the suppressor mutation?

ANS: HP1, the protein encoded by the wild-type allele of the suppressor gene, is involved in chromatin organization.

Perhaps this heterochromatic protein spreads from the region near the inversion breakpoint in the chromosome that carries the white mottled allele and brings about the “heterochromatization” of the white locus. When HP1 is depleted by knocking out one copy of the gene encoding it—that is, by putting the suppressor mutation into the fly’s genotype, the “heterochromatization” of the white locus would be less likely to occur, and perhaps not occur at all. The white locus would then function fully in all eye cells, producing a uniform red eye color.

18.32 The sheep Dolly (Chapter 2) was the first cloned mammal. Dolly was created by implanting a nucleus from a cell taken from the udder of a female sheep into an enucleated egg. This nucleus had two X chromosomes, and because it came from a differentiated cell, one of them must have been inactivated. If the udder cell was

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heterozygous for at least one X-linked gene whose expression you could assay, how could you determine if all of Dolly’s cells had the same X chromosome inactivated? If, upon testing, Dolly’s cells prove to be mosaic for X chromosome activity—that is, different X’s are active in different clones of cells—what must have happened during her embryological development?

ANS: Take samples of cells from Dolly and determine if the products of both alleles of the X-linked gene are present in them. If the products of both alleles are present, then Dolly must be a genetic mosaic for X chromosome activity. Thus, during her development, the pattern of X-inactivation that existed in the udder cell from which she was derived must have been reset. If only one of the gene’s products is detected—and if the sample of cells is representative of all Dolly’s cells—then Dolly must have maintained the pattern of X-inactivation that existed in the udder cell from which she was derived.

Chapter 19 19.1 If heart disease is considered to be a threshold trait, what genetic and environmental factors might contribute to the underlying liability for a person to develop this disease?

ANS: Some of the genes implicated in heart disease are listed in Table 19.2. Environmental factors might include diet, amount of exercise, and whether or not the person smokes.

19.2 A wheat variety with red kernels (genotype A′A′ B′B′) was crossed with a variety with white kernels (genotype AA BB). The F1 were intercrossed to produce an F2. If each primed allele increases the amount of pigment in the kernel by an equal amount, what phenotypes will be expected in the F2? Assuming that the A and B loci assort independently, what will the phenotypic frequencies be?

ANS: 1/16 red; 4/16 = 1/4 dark pink; 6/16 pink; 4/16 = 1/4 light pink; 1/16 white.

19.3 For alcoholism, the concordance rate for monozygotic twins is 55 percent, whereas for dizygotic twins, it is 28 percent. Do these data suggest that alcoholism has a genetic basis?

ANS: The concordance for monozygotic twins is almost twice as great as that for dizygotic twins. Monozygotic twins share twice as many genes as dizygotic twins. The data strongly suggest that alcoholism has a genetic basis.

19.4 The height of the seed head in wheat at maturity is determined by several genes. In one variety, the head is just 9 inches above the ground; in another, it is 33 inches above the ground. Plants from the 9-inch variety were crossed to plants from the 33-inch variety. Among the F1, the seed head was 21 inches above the ground. After selffertilization, the F1 plants produced an F2 population in which 9-inch and 33-inch plants each appeared with a frequency of 1/256. (a) How many genes are involved in

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80-WC Answers to All Questions and Problems the determination of seed head height in these strains of wheat? (b) How much does each allele of these genes contribute to seed head height? (c) If a 21-inch F1 plant were crossed to a 9-inch plant, how often would you expect 18-inch wheat to occur in the progeny?

ANS: (a) 4; (b) 3 inches; (c) frequency of 1/4.

19.5 Assume that size in rabbits is determined by genes with equal and additive effects. From a total of 2012 F2 progeny from crosses between true-breeding large and small varieties, eight rabbits were as small as the small variety and eight were as large as the large variety. How many size-determining genes were segregating in these crosses?

ANS: Because 8/2012 is approximately 1/256 = (1/4)4, it appears that four size-determining genes were segregating in the crosses.

19.6 A sample of 20 plants from a population was measured in inches as follows: 18, 21, 20, 23, 20, 21, 20, 22, 19, 20, 17, 21, 20, 22, 20, 21, 20, 22, 19, and 23. Calculate (a) the mean, (b) the variance, and (c) the standard deviation.

ANS: (a) The mean is 20.45 inches. (b) The variance is 2.37 inches2. (c) The standard deviation is 1.54 inches.

19.7 Quantitative geneticists use the variance as a measure of scatter in a sample of data; they calculate this statistic by averaging the squared deviations between each measurement and the sample mean. Why don’t they simply measure the scatter by computing the average of the deviations without bothering to square them?

ANS: Because S(Xi − mean) = 0.

19.8 Two inbred strains of corn were crossed to produce an F1, which was then intercrossed to produce an F2. Data on ear length from a sample of F1 and F2 individuals gave phenotypic variances of 15.2 cm2 and 27.6 cm2, respectively. Why was the phenotypic variance greater for the F2 than for the F1?

ANS: For the F1, Vg = 0 because they are all genetically identical and heterozygous; for the F2, Vg > 0 because genetic differences result from the segregation and independent assortment of genes. Thus, in the F2, the phenotypic variance has a pronounced genetic component.

19.9 A study of quantitative variation for abdominal bristle number in female Drosophila yielded estimates of VT = 6.08, Vg = 3.17, and Ve = 2.91. What was the broad-sense heritability?

ANS: 3.17/6.08 = 0.52 19.10 A researcher has been studying kernel number on ears of corn. In one highly inbred strain, the variance for kernel number is 426. Within this strain, what is the broadsense heritability for kernel number?

ANS: The broad-sense heritability within a highly inbred strain is expected to be zero because there is no genetic variability.

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19.11 Measurements on ear length were obtained from three populations of corn—two inbred varieties and a randomly pollinated population derived from a cross between the two inbred strains. The phenotypic variances were 9.2 cm2 and 9.6 cm2 for the two inbred varieties and 26.4 cm2 for the randomly pollinated population.

Estimate the broad-sense heritability of ear length for these populations.

ANS: Ve is estimated by the average of the variances of the inbreds: 9.4 cm2. Vg is estimated by the difference between the variances of the randomly pollinated population and the inbreds: (26.4 − 9.4) = 17.0 cm2. The broad-sense heritability is H2 = Vg /VT = 17.0/26.4 = 0.64.

19.12 Figure 19.4 summarizes data on maturation time in populations of wheat. Do these data provide any insight as to whether or not this trait is influenced by dominance?

Explain.

ANS: Because the F1 plants have maturation times midway between those of the parental strains, there seems to be little or no dominance for this trait.

19.13 A quantitative geneticist claims that the narrow-sense heritability for body mass in human beings is 0.7, while the broad-sense heritability is only 0.3. Why must there be an error?

ANS: Broad-sense heritability must be greater than narrowsense heritability because H2 = Vg /VT > Va /VT = h2.

19.14 The mean value of a trait is 100 units, and the narrowsense heritability is 0.4. A male and a female measuring 124 and 126 units, respectively, mate and produce a large number of offspring, which are reared in an average environment. What is the expected value of the trait among these offspring?

ANS: (125 − 100)(0.4) + 100 = 110 units.

19.15 The narrow-sense heritability for abdominal bristle number in a population of Drosophila is 0.3. The mean bristle number is 12. A male with 10 bristles is mated to a female with 20 bristles, and a large number of progeny are scored for bristle number. What is the expected number of bristles among these progeny?

ANS: (15 − 12)(0.3) + 12 = 12.9 bristles.

19.16 A breeder is trying to decrease the maturation time in a population of sunflowers. In this population, the mean time to flowering is 100 days. Plants with a mean flowering time of only 90 days were used to produce the next generation. If the narrow-sense heritability for flowering time is 0.2, what will the average time to flowering be in the next generation?

ANS: (90 − 100)(0.2) + 100 = 98 days.

19.17 A fish breeder wishes to increase the rate of growth in a stock by selecting for increased length at 6 weeks after hatching. The mean length of 6-week-old fingerlings is currently 10 cm. Adult fish that had a mean length of

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Answers to All Questions and Problems WC-81

15 cm at 6 weeks of age were used to produce a new generation of fingerlings. Among these, the mean length was 12.5 cm. Estimate the narrow-sense heritability of fingerling length at 6 weeks of age and advise the breeder about the feasibility of the plan to increase growth rate.

ANS: h = R/S = (12.5 − 10)/(15 − 10) = 0.5; selection for increased growth rate should be effective.

2

19.18 Leo’s IQ is 86 and Julie’s IQ is 110. The mean IQ in the population is 100. Assume that the narrow-sense heritability for IQ is 0.4. What is the expected IQ of Leo and Julie’s first child?

to environmental factors, and eg is the deviation due to epigenetic factors arising from the interaction of genetic and environmental factors.

Chapter 20 20.1 The following data for the M–N blood types were obtained from native villages in Central and North America: Sample Size

M

MN

N

Central American

86

53

29

4

ANS: (98 − 100)(0.4) + 100 = 99.2.

North American

278

78

61

139

19.19 One way to estimate a maximum value for the narrowsense heritability is to calculate the correlation between half-siblings that have been reared apart and divide it by the fraction of genes that half-siblings share by virtue of common ancestry. A study of human half-siblings found that the correlation coefficient for height was 0.14. From this result, what is the maximum value of the narrowsense heritability for height in this population?

Calculate the frequencies of the LM and LN alleles for the two groups.

ANS: Half-siblings share 25 percent of their genes. The maximum value for h2 is therefore 0.14/0.25 = 0.56.

19.20 A selection differential of 40 mg per generation was used in an experiment to select for increased pupa weight in Tribolium. The narrow-sense heritability for pupa weight was estimated to be 0.3. If the mean pupa weight was initially 2000 mg and selection was practiced for 10 generations, what was the mean pupa weight expected to become?

ANS: The response to selection in one generation is R = h S = (0.3)(40 mg) = 12 mg. The cumulative effect over 10 generations is therefore 10 × 12 mg = 120 mg. Thus, the mean pupa weight should become 2120 mg.

2

Group

ANS: Frequency of LM in Central American population: p = (2 × 53 + 29)/(2 × 86) = 0.78; q = 0.22. Frequency of LM in North American population: p = (2 × 78 + 61)/(2 × 278) = 0.39; q = 0.61.

20.2 The frequency of an allele in a large randomly mating population is 0.2. What is the frequency of heterozygous carriers?

ANS: 2pq = 2(0.2)(0.8) = 0.32.

20.3 The incidence of recessive albinism is 0.0004 in a human population. If mating for this trait is random in the population, what is the frequency of the recessive allele?

ANS: q2 = 0.0004; q = 0.02.

20.4 In a sample from an African population, the frequencies of the LM and LN alleles were 0.78 and 0.22, respectively. If the population mates randomly with respect to the M–N blood types, what are the expected frequencies of the M, MN, and N phenotypes?

19.21 On the basis of the observed correlations for personality traits shown in Table 19.5, what can you say about the value of the environmentality (C2 in Table 19.3)?

ANS: Phenotype M

(0.78)2 = 0.61

ANS: The correlations for MZT are not much different from those for MZA. Evidently, for these personality traits, the environmentality (C2 in Table 19.3) is negligible.

MN

2(0.78)(0.22) = 0.34

N

(0.22)2 = 0.05

19.22 Correlations between relatives provide estimates of the broad and narrow-sense heritabilities on the assumption that the genetic and environmental factors influencing quantitative traits are independent of each other and that they do not interact in some peculiar way. In Chapter 18, we considered epigenetic modifications of chromatin that regulate genes and noted the possibility that some of these modifications might be induced by environmental factors. How could epigenetic influences on complex traits be incorporated into the basic theory of quantitative genetics?

ANS: We might represent the value of a quantitative trait, T, as m + g + e + eg, where m is the mean of the population, g is the deviation due to genetic factors, e is the deviation due

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Hardy–Weinberg Frequency

20.5 Human beings carrying the dominant allele T can taste the substance phenylthiocarbamide (PTC). In a population in which the frequency of this allele is 0.4, what is the probability that a particular taster is homozygous?

ANS: Frequency of tasters (genotypes TT and Tt): (0.4)2 + 2 (0.4)(0.6) = 0.64. Frequency of TT tasters among all tasters: (0.4)2/(0.64) = 0.25.

20.6 A gene has three alleles, A1, A2, and A3, with frequencies 0.6, 0.3, and 0.1, respectively. If mating is random, predict the combined frequency of all the heterozygotes in the population.

ANS: Frequency of heterozygotes combined = 2[(0.6)(0.3) + (0.3)(0.1) + ((0.6)(0.1)] = 0.54.

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82-WC Answers to All Questions and Problems 20.7 Hemophilia is caused by an X-linked recessive allele. In a particular population, the frequency of males with hemophilia is 1/4000. What is the expected frequency of females with hemophilia?

ANS: (0.00025)2 = 6.25 × 10−8.

20.8 In Drosophila, the ruby eye phenotype is caused by a recessive, X-linked mutant allele. The wild-type eye color is red. A laboratory population of Drosophila is started with 25 percent ruby-eyed females, 25 percent homozygous red-eyed females, 5 percent ruby-eyed males, and 45 percent red-eyed males. (a) If this population mates randomly for one generation, what is the expected frequency of ruby-eyed males and females? (b) What is the frequency of the recessive allele in each of the sexes?

ANS: (a) Half the males will be ruby-eyed and 5 percent (0.50 × 0.10 × 100 percent) of the females will be ruby-eyed. (b) Among males, the frequency of the recessive allele will be 0.5, which was its frequency among females in the previous generation; among females, the frequency of the recessive allele will be (0.5 + 0.1)/2 = 0.3, which is the average of the frequencies of this allele in males and females in the previous generation.

20.9 A trait determined by an X-linked dominant allele shows 100 percent penetrance and is expressed in 36 percent of the females in a population. Assuming that the population is in Hardy–Weinberg equilibrium, what proportion of the males in this population express the trait?

ANS: In females, the frequency of the dominant phenotype is 0.36. The frequency of the recessive phenotype is 0.64 = q2; thus, q = 0.8 and p = 0.2. The frequency of the dominant phenotype in males is therefore p = 0.2.

20.10 A phenotypically normal couple has had one normal child and a child with cystic fibrosis, an autosomal recessive disease. The incidence of cystic fibrosis in the population from which this couple came is 1/500. If their normal child eventually marries a phenotypically normal person from the same population, what is the risk that the newlyweds will produce a child with cystic fibrosis?

ANS: The probability that the unaffected child of the couple is a carrier of the mutant allele for cystic fibrosis is 2/3. The probability that the mate of this individual is a carrier can be determined by using the population incidence of the disease. The mutant allele frequency is the square root of the incidence—0.045—and the frequency of heterozygotes under the assumption of random mating is 2 × 0.045 × (1 − 0.045) = 0.086. If both individuals are carriers, the chance that they will have an affected child is 1/4. Putting all this analysis together, the risk for the child to have cystic fibrosis is therefore 2/3 × 0.086 × 1/4 = 0.014, which is seven times the incidence in the population at large.

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20.11 What frequencies of alleles A and a in a randomly mating population maximize the frequency of heterozygotes?

ANS: Frequency of heterozygotes = H = 2pq = 2p(1 − p).

Using calculus, take the derivative of H and set the result to zero to solve for the value of p that maximizes H: dH/ dp = 2 − 4p = 0 implies that p = 2/4 = 0.5.

20.12 In an isolated population, the frequencies of the IA, IB, and i alleles of the A–B–O blood type gene are, respectively, 0.15, 0.25, and 0.60. If the genotypes of the A–B–O blood type gene are in Hardy–Weinberg proportions, what fraction of the people who have type A blood in this population is expected to be homozygous for the IA allele?

ANS: In a Hardy–Weinberg population, the frequency of IAIA homozygotes is (0.15)2 = 0.0335 and the frequency of IAi heterozygotes is 2 × 0.15 × 0.60 = 0.18. The sum of these frequencies—0.2025—is the frequency of individuals with type A blood. Thus, the frequency of IAIA homozygotes among all individuals with type A blood is 0.0225/0.2025 = 0.1111.

20.13 In a survey of moths collected from a natural population, a researcher found 51 dark specimens and 49 light specimens. The dark moths carry a dominant allele, and the light moths are homozygous for a recessive allele. If the population is in Hardy–Weinberg equilibrium, what is the estimated frequency of the recessive allele in the population? How many of the dark moths in the sample are likely to be homozygous for the dominant allele?

ANS: Under the assumption that the population is in Hardy– Weinberg equilibrium, the frequency of the allele for light coloration is the square root of the frequency of recessive homozygotes. Thus, q = √0.49 = 0.7, and the frequency of the allele for dark color is 1 − q = p = 0.3.

From p2 = 0.09, we estimate that 0.09 × 100 = 9 of the dark moths in the sample are homozygous for the dominant allele.

20.14 A population of Hawaiian Drosophila is segregating two alleles, P1 and P2, of the phosphoglucose isomerase (PGI) gene. In a sample of 100 flies from this population, 30 were P1P1 homozygotes, 60 were P1P2 heterozygotes, and 10 were P2P2 homozygotes. (a) What are the frequencies of the P1 and P2 alleles in this sample? (b) Perform a chi-square test to determine if the genotypes in the sample are in Hardy–Weinberg proportions. (c) Assuming that the sample is representative of the population, how many generations of random mating would be required to establish Hardy–Weinberg proportions in the population?

ANS: (a) P1 = 0.6 and P2 = 0.4. (b) Predict H–W proportions by multiplying the expected genotype frequencies by the sample size and compare these values with the observed genotype frequencies using a chi-square test:

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Answers to All Questions and Problems WC-83

Genotype

H–W Frequency

Predicted Number

P1P1

0.62 = 0.36

0.36 × 100 = 36

P1P2

2(0.6)(0.4) = 0.48

0.28 × 100 = 48

PP

0.42 = 0.16

0.16 × 100 = 16

2

2

c2 = (30 − 36)2/36 + (60 − 48)2/48 + (10 − 16)2/16 = 6.25, df = 1; 6.25 is greater than 3.841; therefore, the observed genotypes are not in agreement with H–W. 20.15 In a large population that reproduces by random mating, the frequencies of the genotypes GG, Gg, and gg are 0.04, 0.32, and 0.64, respectively. Assume that a change in the climate induces the population to reproduce exclusively by self-fertilization. Predict the frequencies of the genotypes in this population after many generations of self-fertilization.

ANS: Ultimate frequency of GG is 0.2; ultimate frequency of gg is 0.8.

20.16 The frequencies of the alleles A and a are 0.6 and 0.4, respectively, in a particular plant population. After many generations of random mating, the population goes through one cycle of self-fertilization. What is the expected frequency of heterozygotes in the progeny of the self-fertilized plants?

ANS: 2pq(1 − F) = 2(0.6)(0.4)(1 − 0.5) = 0.24.

20.17 Each of two isolated populations is in Hardy–Weinberg equilibrium with the genotype frequencies shown below:

(genotype Tt). Predict the ultimate phenotypic and genotypic composition of the population if, generation after generation, mating is strictly assortative (i.e., tall individuals mate with tall individuals, short individuals mate with short individuals, and intermediate individuals mate with intermediate individuals).

ANS: Ultimately, all members of the population will either be TT or tt, each 50% of the total.

20.19 In controlled experiments with different genotypes of an insect, a researcher has measured the probability of survival from fertilized eggs to mature, breeding adults. The survival probabilities of the three genotypes tested are 0.92 (for GG), 0.90 (for Gg), and 0.56 (for gg). If all breeding adults are equally fertile, what are the relative fitnesses of the three genotypes? What are the selection coefficients for the two least fit genotypes?

ANS: The relative fitnesses can be obtained by dividing each of the survival probabilities by the largest probability (0.92).

Thus, the relative fitnesses are 1 for GG, 0.98 = 1 − 0.02 for Gg, and 0.61 = 1 − 0.39 for gg. The selection coefficients are s1 = 0.02 for Gg and s2 = 0.39 for gg.

20.20 In a large randomly mating population, 0.84 of the individuals express the phenotype of the dominant allele A and 0.16 express the phenotype of the recessive allele a. (a) What is the frequency of the dominant allele? (b) If the aa homozygotes are 5 percent less fit than the other two genotypes, what will the frequency of A be in the next generation?

Genotype:

AA

Aa

Aa

Frequency in Population 1:

0.04

0.32

0.64

ANS: Frequency of a is q = 0.4; (a) thus p = 1 − q = 0.6; (b) use the following scheme:

Frequency in Population 2:

0.64

0.32

0.04

Genotype

AA

Aa

Aa

Hardy–Weinberg frequency

0.36

0.48

0.16

Relative fitness

1

1

0.95

Relative contribution to next generation

0.36

0.48

0.152

Proportional contribution to next generation

0.363

0.484

0.153

(a) If the populations are equal in size and they merge to form a single large population, predict the allele and genotype frequencies in the large population immediately after merger. (b) If the merged population reproduces by random mating, predict the genotype frequencies in the next generation. (c) If the merged population continues to reproduce by random mating, will these genotype frequencies remain constant?

ANS: (a) Frequency of A in merged population is 0.5 and that of a is also 0.5; (b) 0.25 (AA), 0.50 (Aa), and 0.25 (aa); (c) frequencies in (b) will persist.

20.18 A population consists of 25 percent tall individuals (genotype TT), 25 percent short individuals (genotype tt), and 50 percent individuals of intermediate height

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Thus, the frequency of the A allele in the next generation will be (0.363 + 0.484/2) = 0.605.

20.21 Because individuals with cystic fibrosis die before they can reproduce, the coefficient of selection against them is s = 1. Assume that heterozygous carriers of the ­recessive mutant allele responsible for this disease are as fit as wild-type homozygotes and that the population ­frequency of the mutant allele is 0.02. (a) Predict the incidence of cystic fibrosis in the population after one generation of selection. (b) Explain why the incidence of cystic fibrosis hardly changes even with s = 1.

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84-WC Answers to All Questions and Problems ANS: (a) Use the following scheme: Genotype

CC

Cc

cc

Hardy–Weinberg frequency

(0.98) = 0.9604

2(0.98)(0.02) = (0.02) = 0.0392 0.0004

Relative fitness Relative contribution to next generation Proportional contribution

1 (0.9604) × 1

1 (0.0392) × 1

2

2

0.9604/0.9996

= 0.9608

0 0

0.0392/0.9996 0

= 0.0392

The new frequency of the allele for cystic fibrosis is (0.5) (0.0392) = 0.0196; thus, the incidence of the disease will be (0.0196)2 = 0.00038, which is very slightly less than the incidence in the previous generation. (b) The incidence of cystic fibrosis does not change much because selection can only act against the recessive allele when it is in homozygotes, which are rare in the population.

20.22 For each set of relative fitnesses for the genotypes AA, Aa, and aa, explain how selection is operating. Assume that 0 < t < s < 1.

AA

Aa

aa

Case 1

1

1

1-s

Case 2

1-s

1-s

1

Case 3

1

1-t

1-s

Case 4

1-s

1

1-t

ANS: Case 1: selection is operating against a deleterious recessive allele. Case 2: selection is operating against a deleterious dominant allele. Case 3: selection is operating against a deleterious allele that has some expression in heterozygotes, that is, it is partially dominant. Case 4: selection is operating against both alleles in homozygous condition; this is a case of balancing selection.

20.23 The frequency of newborn infants homozygous for a recessive lethal allele is about 1 in 25,000. What is the expected frequency of carriers of this allele in the population?

ANS: q2 = 4 × 10−5; thus q = 6.3 × 10−3 and 2pq = 0.0126.

20.24 A population of size 50 reproduces in such a way that the population size remains constant. If mating is random, how rapidly will genetic variability, as measured by the frequency of heterozygotes, be lost from this population?

ANS: The frequency of heterozygotes will decrease by a 1/(2N) = 1/100 = 0.01 per generation.

20.25 A population is segregating three alleles, A1, A2, and A3, with frequencies 0.2, 0.5, and 0.3, respectively. If these alleles are selectively neutral, what is the probability that A2 will ultimately be fixed by genetic drift? What is the probability that A3 will ultimately be lost by genetic drift?

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ANS: Probability of ultimate fixation of A2 is 0.5; probability of ultimate loss of A3 is 1 − 0.3 = 0.7.

20.26 A small island population of mice consists of roughly equal numbers of males and females. The Y chromosome in one-fourth of the males is twice as long as the Y chromosome in the other males because of an expansion of heterochromatin. If mice with the large Y chromosome have the same fitness as mice with the small Y chromosome, what is the probability that the large Y chromosome will ultimately be fixed in the mouse population?

ANS: 0.25.

20.27 In some regions of west Africa, the frequency of the HBBS allele is 0.2. If this frequency is the result of a dynamic equilibrium due to the superior fitness of HBBSHBBA heterozygotes, and if HBBSHBBS homozygotes are essentially lethal, what is the intensity of selection against the HBBAHBBA homozygotes?

ANS: p = 0.2; at equilibrium, p = t/(s + t). Because s = 1, we can solve for t; t = 0.25.

20.28 Mice with the genotype Hh are twice as fit as either of the homozygotes HH and hh. With random mating, what is the expected frequency of the h allele when the mouse population reaches a dynamic equilibrium because of balancing selection?

ANS: The relative fitnesses of the genotypes HH, Hh, and hh are 0.5, 1, and 0.5, respectively. At equilibrium, the frequency of h will be s/(t + s) = (0.5)/(0.5 + 0.5) = 0.5.

20.29 A completely recessive allele g is lethal in homozygous condition. If the dominant allele G mutates to g at a rate of 10−6 per generation, what is the expected frequency of the lethal allele when the population reaches mutation– selection equilibrium?

ANS: At mutation–selection equilibrium q = u / s = 10−6 / 1 = 0.001.

20.30 Individuals with the genotype bb are 20 percent less fit than individuals with the genotypes BB or Bb. If B mutates to b at a rate of 10−6 per generation, what is the expected frequency of the allele b when the population reaches mutation–selection equilibrium?

ANS: q = u / s = 10−6 / (0.2) = 2.2 × 10−3.

Chapter 21 21.1 Which of the following pairs of DNA sequences could qualify as the terminal repeats of a bacterial IS element.

Explain. (a) 5′-GAATCCGCA-3′ and 5′-ACGCCTAAG-3′ (b) 5′-GAATCCGCA-3′ and 5′-CTTAGGCGT-3′ (c)5′-GAATCCGCA-3′ and 5′-GAATCCGCA-3′ (d) 5′-GAATCCGCA-3′ and 5′-TGCGGATTC-3′.

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Answers to All Questions and Problems WC-85

ANS: The pair in (d) are inverted repeats and could therefore qualify.

21.2 Which of the following pairs of DNA sequences could qualify as target site duplications at the point of an IS50 insertion? Explain. (a) 5′-AATTCGCGT-3′ and 5′-AATTCGCGT-3′ (b) 5′-AATTCGCGT-3′ and 5′-TGCGCTTAA-3′ (c) 5′-AATTCGCGT-3′ and 5′-TTAAGCGCA-3′ (d) 5′-AATTCGCGT-3′ and 5′-ACGCGAATT-3′.

ANS: The pair in (a) are direct repeats and could therefore qualify.

21.3 One strain of E. coli is resistant to the antibiotic streptomycin, and another strain is resistant to the antibiotic ampicillin. The two strains were cultured together and then plated on selective medium containing streptomycin and ampicillin. Several colonies appeared, indicating that cells had acquired resistance to both antibiotics.

Suggest a mechanism to explain the acquisition of double resistance.

ANS: Resistance for the second antibiotic was acquired by conjugative gene transfer between the two types of cells.

21.4 What distinguishes IS and Tn3 elements in bacteria?

transposon, only IS50R produces the catalytically active transposase. Would you expect IS50R to be able to be excised from the Tn5 composite transposon and insert elsewhere in the chromosome? Would you expect IS50L to be able to do this?

ANS: Both IS50 elements should be able to excise from the transposon and insert elsewhere in the chromosome, because even though IS50L does not produce its own transposase, IS50R provides a source of this enzyme.

21.7 By chance, an IS1 element has inserted near an IS2 element in the E. coli chromosome. The gene between them, sug+, confers the ability to metabolize certain sugars.

Will the unit IS1 sug+ IS2 behave as a composite transposon? Explain.

ANS: No. IS1 and IS2 are mobilized by different transposases.

21.8 A researcher has found a new Tn5 element with the structure IS50L strr bler kanr IS50L. What is the most likely origin of this element?

ANS: IS50L inserted on each side of the cluster of antibiotic resistance genes.

21.9 Would a Tn3 element with a frameshift mutation early in the tnpA gene be able to form a cointegrate? Would a Tn3 element with a frameshift mutation early in the tnpR gene be able to form a cointegrate?

ANS: Tn3 elements carry a gene that is not essential for transposition.

ANS: The tnpA mutation: no; the tnpR mutation: yes.

21.5 The circular order of genes on the E. coli chromosome is *A B C D E F G H*, with the * indicating that the ends of the chromosome are attached to each other. Two copies of an IS element are located in this chromosome: one between genes C and D, and the other between genes D and E. A single copy of this element is also present in the F plasmid. Two Hfr strains were obtained by selecting for integration of the F plasmid into the chromosome.

During conjugation, one strain transfers the chromosomal genes in the order D E F G H A B C, whereas the other transfers them in the order D C B A H G F E. Explain the origin of these two Hfr strains. Why do they transfer genes in different orders? Does the order of transfer reveal anything about the orientation of the IS elements in the E. coli chromosome?

ANS: Two enzymes, transposase and resolvase, are needed for replicative transposition. These enzymes are encoded by genes of Tn3. Transposase catalyzes formation of a cointegrate between donor and recipient plasmids. During this process, Tn3 is replicated so that there is a copy of it at each junction in the cointegrate. Resolvase catalyzes the site-specific recombination between the two Tn3 elements and thereby resolves the cointegrate, generating two molecules each with a copy of the transposon.

Resolvase also represses the synthesis of both the transposase and resolvase enzymes.

ANS: In the first strain, the F factor integrated into the chromosome by recombination with the IS element between genes C and D. In the second strain, it integrated by recombination with the IS element between genes D and E. The two strains transfer their genes in different orders because the two chromosomal IS elements are in opposite orientation.

21.6 The composite transposon Tn5 consists of two IS50 elements, one on either side of a group of three genes for antibiotic resistance. The entire unit IS50L kanr bler strr IS50R can transpose to a new location in the E. coli chromosome. However, of the two IS50 elements in this

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21.10 What enzymes are necessary for replicative transposition of Tn3? What are their respective functions?

21.11 What is the medical significance of bacterial transposons?

ANS: Many bacterial transposons carry genes for antibiotic resistance, and it is relatively simple for these genes to move from one DNA molecule to another. DNA molecules that acquire resistance genes can be passed to other cells in a bacterial population, both vertically (by descent) and horizontally (by conjugative transfer). Over time, continued exposure to an antibiotic will select for cells that have acquired a gene for resistance to that antibiotic.

The antibiotic will therefore no longer be useful in combating these bacteria.

21.12 Describe the structure of the Ac transposon in maize. In what ways do the Ds transposons differ structurally and functionally from the Ac transposon?

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86-WC Answers to All Questions and Problems ANS: The Ac element consists of 4563 nucleotide pairs bounded by inverted repeats that are 11 nucleotide pairs long. The Ac element is flanked by direct repeats of eight nucleotide pairs long; however, these repeats are created at the time the element is inserted into a chromosome (target site duplications) and are therefore not considered to be integral parts of the element itself. Ds elements possess the same terminal inverted repeats as Ac, but their internal sequences vary. Some residue of the Ac sequence may be present, or non-Ac sequences may be present; sometimes, one Ds element is contained within another Ds element.

21.13 In homozygous condition, a deletion mutation of the c locus, cn, produces colorless (white) kernels in maize; the dominant wild-type allele, C, causes the kernels to be purple. A newly identified recessive mutation of the c locus, cm, has the same phenotype as the deletion mutation (white kernels), but when cmcm and cncn plants are crossed, they produce white kernels with purple stripes.

If it is known that the cncn plants harbor Ac elements, what is the most likely explanation for the cm mutation?

ANS: The cm mutation is due to a Ds or an Ac insertion.

21.14 In maize, the O2 gene, located on chromosome 7, controls the texture of the endosperm, and the C gene, located on chromosome 9, controls its color. The gene on chromosome 7 has two alleles, a recessive, o2, which causes the endosperm to be soft, and a dominant, O2, which causes it to be hard. The gene on chromosome 9 also has two alleles, a recessive, c, which allows the endosperm to be colored, and a dominant, CI, which inhibits coloration. In one homozygous CI strain, a Ds element is inserted on chromosome 9 between the C gene and the centromere. This element can be activated by introducing an Ac element by appropriate crosses. Activation of Ds causes the CI allele to be lost by chromosome breakage. In CI/c/c kernels, such loss produces patches of colored tissue in an otherwise colorless background.­ A geneticist crosses a strain with the genotype o2/o2; CI Ds/CI Ds to a strain with the genotype O2/o2; c/c. The latter strain also carries an Ac element somewhere in the genome. Among the offspring, only those with hard endosperm show patches of colored tissue. What does this tell you about the location of the Ac element in the O2/o2; c/c strain?

ANS: The Ac element must be tightly linked to the O2 allele.

21.15 In maize, the recessive allele bz (bronze) produces a lighter color in the aleurone than does the dominant allele, Bz. Ears on a homozygous bz/bz plant were fertilized by pollen from a homozygous Bz/Bz plant. The resulting cobs contained kernels that were uniformly dark except for a few on which light spots occurred. Suggest an explanation.

ANS: The paternally inherited Bz allele was inactivated by a transposable element insertion.

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21.16 The X-linked singed locus is one of several in Drosophila that controls the formation of bristles on the adult cuticle. Males that are hemizygous for a mutant singed allele have bent, twisted bristles that are often much reduced in size. Several P element insertion mutations of the singed locus have been characterized, and some have been shown to revert to the wild-type allele by excision of the inserted element. What conditions must be present to allow such reversions to occur?

ANS: The P transposase to catalyze excision and the absence of P-specific piRNAs that would repress excision.

21.17 Dysgenic hybrids in Drosophila have elevated mutation rates as a result of P element transposition. How could you take advantage of this situation to obtain P element insertion mutations on the X chromosome?

ANS: Cross dysgenic (highly mutable) males carrying a wildtype X chromosome to females homozygous for a balancer X chromosome; then cross the heterozygous F1 daughters individually to their brothers and screen the F2 males that lack the balancer chromosome for mutant phenotypes, including failure to survive (lethality). Mutations identified in this screen are probably due to P element insertions in X-linked genes.

21.18 If DNA from a P element insertion mutation of the Drosophila white gene and DNA from a wild-type white gene were purified, denatured, mixed with each other, renatured, and then viewed with an electron microscope, what would the hybrid DNA molecules look like?

ANS:

P element DNA

White gene DNA

(Should see a single-stranded DNA loop corresponding to the insertion.)

21.19 When complete P elements are injected into embryos from an M strain, they transpose into the chromosomes of the germ line, and progeny reared from these embryos can be used to establish new P strains. However, when complete P elements are injected into embryos from insects that lack these elements, such as mosquitoes, they do not transpose into the chromosomes of the germ line.

What does this failure to insert in the chromosomes of other insects indicate about the nature of P element transposition?

ANS: Factors made by the fly’s genome are required for transposition; other insects apparently lack the ability to provide these factors.

21.20 (a) What are retroviruslike elements? (b) Give examples of retroviruslike elements in yeast and Drosophila.

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(c) Describe how retroviruslike elements transpose. (d) After a retroviruslike element has been inserted into a chromosome, is it ever expected to be excised?

ANS: (a) Retroviruslike elements resemble integrated retroviruses in overall structure and behavior. (b) Examples include the Ty1 element in yeast and the copia element in Drosophila. (c) Retroviruslike elements transpose using an RNA intermediate. The element DNA is transcribed into single-stranded RNA, which is reverse-transcribed into double-stranded DNA (cDNA). The doublestranded cDNA is then inserted into a site in the genome. (d) No. However, the LTRs could pair and recombine to excise all but one LTR. 21.21 Sometimes, solitary copies of the LTR of Ty1 elements are found in yeast chromosomes. How might these solitary LTRs originate?

ANS: Through crossing over between the LTRs of a Ty1 element.

21.22 Would you ever expect the genes in a retrotransposon to possess introns? Explain.

ANS: No. The intron sequences would be removed by RNA processing prior to reverse transcription into DNA. 21.23 Suggest a method to determine whether the TART retroposon is situated at the telomeres of each of the chromosomes in the Drosophila genome.

ANS: In situ hybridization to polytene chromosomes using a TART probe.

21.24 It has been proposed that the hobo transposable elements in Drosophila mediate intrachromosomal recombination— that is, two hobo elements on the same chromosome pair and recombine with each other. What would such a recombination event produce if the hobo elements were oriented in the same direction on the chromosome?

What if they were oriented in opposite directions?

ANS: Same orientation: a deletion; opposite orientation: an inversion.

21.25 What evidence suggests that some transposable elements are not simply genetic parasites?

ANS: TART and HeT-A replenish the ends of Drosophila chromosomes.

21.26 Approximately half of all spontaneous mutations in Drosophila are caused by transposable element insertions. In human beings, however, the accumulated evidence suggests that the vast majority of spontaneous mutations are not caused by transposon insertions. Propose a hypothesis to explain this difference.

ANS: The transposition rate in humans may be very much less than it is in Drosophila.

21.27 Z. Ivics, Z. Izsvák, and P. B. Hackett have “resurrected” a nonmobile member of the Tc1/mariner family of transposable elements isolated from the DNA of salmon.

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These researchers altered 12 codons within the coding sequence of the transposase gene of the salmon element to restore the catalytic function of its transposase. The altered element, called Sleeping Beauty, is being tested as an agent for the genetic transformation of vertebrates such as mice and zebra fish (and possibly humans).

Suppose that you have a bacterial plasmid containing the gene for green fluorescent protein (gfp) inserted between the ends of a Sleeping Beauty element. How would you go about obtaining mice or zebra fish that express the gfp gene?

ANS: The Sleeping Beauty element could be used as a transformation vector in vertebrates much like the P element has been used in Drosophila. The gfp gene could be inserted between the ends of the Sleeping Beauty element and injected into eggs or embryos along with an intact Sleeping Beauty element capable of encoding the element’s transposase. If the transposase that is produced in the injected egg or embryo acts on the element that contains the gfp gene, it might cause the latter to be inserted into genomic DNA. Then, if the egg or embryo develops into an adult, that adult can be bred to determine if a Sleeping Beauty/gfp transgene is transmitted to the next generation. In this way, it would be possible to obtain strains of mice or zebra fish that express the gfp gene.

21.28 The human genome contains about 5000 “processed pseudogenes,” which are derived from the insertion of DNA copies of mRNA molecules derived from many different genes. Predict the structure of these pseudogenes. Would each type of processed pseudogene be expected to found a new family of retrotransposons within the human genome? Would the copy number of each type of processed pseudogene be expected to increase significantly over evolutionary time, as the copy number of the Alu family has? Explain your answers.

ANS: The processed pseudogenes will, in the best of cases, contain sequences from the transcription start site to the poly-A tail of the transcript (if there is one); however, they will not contain the gene’s promoter or any of its introns. Because these pseudogenes will not have a promoter, they are not likely to found new retrotransposon families. Without a promoter, they will not be transcribed; hence, they will not produce RNA to be reverse transcribed into DNA for insertion into other sites in the genome. Most likely, the copy number of each of these processed pseudogenes will not increase as the copy number of the Alu element has. The Alu element contains an “internal” promoter recognized by RNA polymerase III. Each insertion of the Alu element contains this promoter and can therefore be transcribed into RNA, which can subsequently be reverse transcribed into DNA. The L1 element also contains an “internal” promoter, but this promoter is recognized by RNA polymerase II. Most protein-coding genes contain an “external” promoter—that is, one that is not transcribed—and this promoter is recognized by RNA polymerase II.

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88-WC Answers to All Questions and Problems Chapter 22 22.1 During oogenesis, what mechanisms enrich the cytoplasm of animal eggs with nutritive and determinative materials?

ANS: Unequal division of the cytoplasm during the meiotic divisions; transport of substances into the oocyte from surrounding cells such as the nurse cells in Drosophila.

22.2 Predict the phenotype of a fruit fly that develops from an embryo in which the posterior pole cells had been destroyed by a laser beam.

ANS: The fly will be sterile because the posterior pole cells form the germ line in adults of both sexes.

22.3 Outline the main steps in the genetic analysis of development in a model organism such as Drosophila.

ANS: Collect mutations with diagnostic phenotypes; map the mutations and test them for allelism with one another; perform epistasis tests with mutations in different genes; clone individual genes and analyze their function at the molecular level.

22.4 Why is the early Drosophila embryo a syncytium?

ANS: Mitotic division is so rapid that there is not enough time for membranes to form between cells.

22.5 In Drosophila, what larval tissues produce the external organs of the adult?

ANS: Imaginal discs.

22.6 Like dorsal, bicoid is a strict maternal-effect gene in Drosophila; that is, it has no zygotic expression. Recessive mutations in bicoid (bcd) cause embryonic death by preventing the formation of anterior structures. Predict the phenotypes of (a) bcd/bcd animals produced by mating heterozygous males and females; (b) bcd/bcd animals produced by mating bcd/bcd females with bcd/+ males; (c) bcd/+ animals produced by mating bcd/bcd females with bcd/+ males; (d) bcd/bcd animals produced by mating bcd/+ females with bcd/bcd males; (e) bcd/+ animals produced by mating bcd/+ females with bcd/bcd males.

ANS: (a) Wild-type; (b) embryonic lethal; (c) embryonic lethal; (d) wild-type; (e) wild-type.

22.7 Why do women, but not men, who are homozygous for the mutant allele that causes phenylketonuria produce children that are physically and mentally retarded?

ANS: In homozygous condition, the mutation that causes phenylketonuria has a maternal effect. Women homozygous for this mutation influence the development of their children in utero.

22.8 In Drosophila, recessive mutations in the dorsal–ventral axis gene dorsal (dl) cause a dorsalized phenotype in embryos produced by dl/dl mothers; that is, no ventral structures develop. Predict the phenotype of embryos produced by females homozygous for a recessive mutation in the anterior–posterior axis gene nanos.

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ANS: Some structures fail to develop in the posterior portion of the embryo.

22.9 A researcher is planning to collect mutations in maternal-effect genes that control the earliest events in ­Drosophila development. What phenotype should the researcher look for in this search for maternal-effect mutations?

ANS: Female sterility. Females affected by these mutations will lay abnormal eggs that will not develop into viable embryos.

22.10 A researcher is planning to collect mutations in the gap genes, which control the first steps in the segmentation of Drosophila embryos. What phenotype should the researcher look for in this search for gap gene mutations?

ANS: Screen for lethal mutations that prevent regions of the embryo from developing normally.

22.11 How do the somatic cells that surround a developing Drosophila egg in the ovary influence the formation of the dorsal–ventral axis in the embryo that will be produced after the egg is fertilized?

ANS: The somatic cells surrounding a developing Drosophila egg in the ovary determine where the spätzle protein, which is the ligand for the Toll receptor protein, will be cleaved. This cleavage will eventually occur on the ventral side of the developing embryo.

22.12 What events lead to a high concentration of hunchback protein in the anterior of Drosophila embryos?

ANS: The hunchback mRNA is translated into protein only in the anterior region of the developing embryo. This RNA is supplied to the egg by the nurse cells and it is also synthesized after fertilization by transcription of the hunchback gene. This zygotic transcription is stimulated by a transcription factor encoded by maternally supplied bicoid mRNA, which is located in the anterior of the egg. Thus, hunchback mRNA is concentrated in the anterior of the embryo. In addition, the hunchback mRNA that is located in the posterior of the embryo is bound by nanos protein and then degraded. The nanos protein is concentrated in the posterior of the embryo because maternally supplied nanos mRNA is preferentially localized there.

22.13 Diagram a pathway that shows the contributions of the sevenless (sev) and bride of sevenless (boss) genes to the differentiation of the R7 photoreceptor in the ommatidia of Drosophila eyes. Where would eyeless (ey) fit in this pathway?

ANS: ey → boss → sev → R7 differentiation 22.14 The sevB4 allele is temperature sensitive; at 22.7°C, flies that are homozygous for it develop normal R7 photoreceptors, but at 24.3°C, they fail to develop these photoreceptors. sos2A is a recessive, loss-of-function mutation in the son of sevenless (sos) gene. Flies with the genotype

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sevB4/sevB4; sos2A/+ fail to develop R7 photoreceptors if they are raised at 22.7°C. Therefore, sos2A acts as a dominant enhancer of the sevB4 mutant phenotype at this temperature. Based on this observation, where is the protein product of the wild-type sos gene— called SOS—likely to act in the pathway for R7 differentiation?

ANS: If the SEV protein is activated—either by the BOSS ligand or by a gain-of-function mutation in the sev gene, a faulty effector protein could stop it from inducing the R7 cell to differentiate. The SOS protein is likely to be a downstream effector in the pathway for R7 differentiation because when it is depleted by mutating one copy of the sos gene, flies that have a partially functional SEV protein show a mutant phenotype—that is, transmission of the developmental signal through SEV and its downstream effector proteins is weakened.

20.15 When the mouse Pax6 gene, which is homologous to the Drosophila eyeless gene, is expressed in Drosophila, it produces extra compound eyes with ommatidia, just like normal Drosophila eyes. If the Drosophila eyeless gene were introduced into mice and expressed there, what effect would you expect? Explain.

ANS: Because the Pax6 gave the same phenotype in flies as overexpression of the eyeless gene, the genes must be functionally homologous, as well as structurally homologous. Therefore, expect extra mouse eyes or eye primorida when expressing eyeless in the mouse.

22.16 Would you expect to find homologues of Drosophila’s BX-C and ANT-C genes in animals with radial symmetry such as sea urchins and starfish? How could you address this question experimentally?

ANS: Maybe these organisms would not have homologues of the BX-C and ANT-C genes because they do not have segmented bodies with bilateral symmetry as Drosophila does. To see if homologues to these genes are present, use Drosophila BX-C and ANT-C DNA as probes to hybridize with starfish or sea urchin genomic DNA on a Southern blot. The hybridization would have to be done under conditions that allow DNA that is not a perfect match to form a duplex—that is, under conditions of low stringency. Usually, hybridizations of this type are carried out at lower temperatures than typical Southern hybridizations. If the probes stick to the DNA on the blot, there is evidence for homologues to the BX-C and ANT-C genes in the genomic DNA. Follow-up experiments might endeavor to clone this DNA and, ultimately, to sequence it to determine just how close a match it is to the Drosophila probe DNA. 22.17 How might you show that two mouse Hox genes are expressed in different tissues and at different times during development?

ANS: Northern blotting of RNA extracted from the tissues at different times during development. Hybridize the blot with gene-specific probes.

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22.18 Distinguish between therapeutic and reproductive cloning.

ANS: Therapeutic cloning involves the creation of an embryo by implanting the nucleus of a somatic cell into an enucleated egg and stimulating the egg to divide. Stem cells are then taken from the embryo to differentiate into specific tissues in the individual from which the somatic cell was taken. These tissues will be genetically identical to the other tissues of the individual—thus, they are unlikely to be rejected by the individual’s immune system. Reproductive cloning involves the creation of an embryo by implanting the nucleus of a somatic cell into an enucleated egg and then allowing the egg to develop into an entire individual.

22.19 What is the scientific significance of reproductive cloning?

ANS: Reproductive cloning of mammals such as sheep, mice, and cats indicates that somatic cell nuclei have all the genetic information to direct the development of a complete, viable organism. It also shows that epigenetic modifications of chromatin, such as X chromosome inactivation, can be reset.

22.20 The methylation of DNA, the acetylation of histones, and the packaging of DNA into chromatin by certain kinds of proteins are sometimes referred to as epigenetic modifications of the DNA. These modifications portend difficulties for reproductive cloning. Do they also portend difficulties for therapeutic cloning and for the use of stem cells to treat diseases or injuries that involve the loss of specific cell types?

ANS: Methylation of DNA, acetylation of histones, and packaging of DNA into chromatin by certain kinds of proteins all portend difficulties for therapeutic cloning as well as for reproductive cloning. These epigenetic modifications of somatic cell DNA would have to be “reprogrammed” in the oocyte or they could affect how the stem cells derived from the oocyte would develop.

22.21 Assume that an animal is capable of producing 100 million different antibodies and that each antibody contains a light chain of 220 amino acids long and a heavy chain of 450 amino acids. How much genomic DNA would be needed to accommodate the coding sequences of these genes?

ANS: If each antibody consists of one kind of light chain and one kind of heavy chain, and if light and heavy chains can combine freely, the potential to produce 100 million different antibodies implies the existence of 10,000 light chain genes and 10,000 heavy chain genes (10,000 × 10,000 = 100 million). If each light chain is 220 amino acids long, each light chain gene must comprise 3 × 220 = 660 nucleotides because each amino acid is specified by a triplet of nucleotides; similarly, each heavy chain gene must comprise 3 × 450 = 1350 nucleotides. Therefore, the genome must contain 10,000 × 660 = 6.6 million nucleotides devoted to light chain production and

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90-WC Answers to All Questions and Problems 10,000 × 1,350 = 13.5 million nucleotides devoted to heavy chain production. Altogether, then, the genome must contain 19.5 million nucleotides dedicated to encoding the amino acids of the various antibody chains.

22.22 Each LkVk gene segment in the kappa light chain locus on chromosome 2 consists of two coding exons, one for the leader peptide and one for the variable portion of the kappa light chain. Would you expect to find a stop codon at the end of the coding sequence in the second (Vk) exon?

ANS: No, because the Vk coding sequence must be joined to the coding sequence of the constant region to encode a complete kappa light chain.

Chapter 23 23.1 Many cancers seem to involve environmental factors.

Why, then, is cancer called a genetic disease?

ANS: Cancer has been called a genetic disease because it results from mutations of genes that regulate cell growth and division. Nonhereditary forms of cancer result from mutations in somatic cells. These mutations, however, can be induced by environmental factors including tobacco smoke, chemical pollutants, ionizing radiation, and UV light. Hereditary forms of cancer also frequently involve the occurrence of environmentally induced somatic mutations.

23.2 Both embryonic cells and cancer cells divide quickly.

How can these two types of cells be distinguished from each other?

ANS: Cancer cells do not display contact inhibition—they pile up on top of each other—whereas embryonic cells spread out in flat sheets. Cancer cells are frequently aneuploid; embryonic cells are euploid.

23.3 Most cancer cells are aneuploid. Suggest how aneuploidy might contribute to deregulation of the cell cycle.

ANS: Aneuploidy might involve the loss of functional copies of tumor suppressor genes, or it might involve the inappropriate duplication of proto-oncogenes. Loss of tumor suppressor genes would remove natural brakes on cell division, and duplication of proto-oncogenes would increase the abundance of factors that promote cell division.

23.4 Would you ever expect to find a tumor-inducing retrovirus that carried a processed cellular tumor suppressor gene in its genome?

23.6 How might the absence of introns in a retroviral oncogene explain that gene’s overexpression in the tissues of an infected animal?

ANS: The absence of introns might speed up the expression of the gene’s protein product because there would be no need for splicing. In addition, some introns contain sequences called silencers that negatively regulate transcription. Removal of these sequences might cause transcription to occur when it otherwise would not.

23.7 When cellular oncogenes are isolated from different animals and compared, the amino acid sequences of the polypeptides they encode are found to be very similar.

What does this suggest about the functions of these polypeptides?

ANS: The products of these genes play important roles in cell activities.

23.8 The majority of the c-ras oncogenes obtained from cancerous tissues have mutations in codon 12, 59, or 61 in the coding sequence. Suggest an explanation.

ANS: Mutations in these codons cause amino acid changes that activate the Ras protein.

23.9 When a mutant c-H-ras oncogene with a valine for glycine substitution in codon 12 is transfected into cultured NIH 3T3 cells, it transforms those cells into cancer cells.

When the same mutant oncogene is transfected into cultured embryonic cells, it does not transform them. Why? ANS: The cultured NIH 3T3 cells probably carry other mutations that predispose them to become cancerous; transfection of such cells with a mutant c-H-ras oncogene may be the last step in the process of transforming the cells into cancer cells. Cultured embryonic cells probably do not carry the predisposing mutations needed for them to become cancerous; thus, when they are transfected with the mutant c-H-ras oncogene, they continue to divide normally.

23.10 A mutation in the ras cellular oncogene can cause cancer when it is in heterozygous condition, but a mutation in the RB tumor suppressor gene can cause cancer only when it is in homozygous condition. What does this difference between dominant and recessive mutations imply about the roles that the ras and RB gene products play in normal cellular activities?

ANS: Ras protein is an activator of cell division, whereas RB protein is a suppressor of cell division.

ANS: No. A virus that carried a processed copy of a tumor suppressor gene would not be expected to induce tumor formation because the product of the tumor suppressor gene would help to restrain cell growth and division.

23.11 Explain why individuals who develop nonhereditary retinoblastoma usually have tumors in only one eye, whereas individuals with hereditary retinoblastoma usually develop tumors in both eyes.

23.5 How do we know that normal cellular oncogenes are not simply integrated retroviral oncogenes that have acquired the proper regulation?

ANS: Retinoblastoma results from homozygosity for a loss-offunction (recessive) allele. The sporadic occurrence of retinoblastoma requires two mutations of this gene in the same cell or cell lineage. Therefore, retinoblastoma is rare among individuals who, at conception, are

ANS: They possess introns.

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homozygous for the wild-type allele of the RB gene. For such individuals, we would expect the frequency of tumors in both eyes to be the square of the frequency of tumors in one eye. Individuals who are heterozygous for a mutant RB allele require only one somatic mutation to occur for them to develop retinoblastoma. Because there are millions of cells in each retina, there is a high probability that this somatic mutation will occur in at least one cell in each eye, causing both eyes to develop tumors.

23.12 Approximately 5 percent of the individuals who inherit an inactivated RB gene do not develop retinoblastoma.

Use this statistic to estimate the number of cell divisions that form the retinal tissues of the eye. Assume that the rate at which somatic mutations inactivate the RB gene is one mutation per 106 cell divisions.

ANS: The probability that a carrier does not develop retinoblastoma is 0.05, which is equal to the probability that the wild-type RB allele is not mutationally inactivated during the cell divisions that form the retinas of the eyes.

If the rate of mutational inactivation is u = 10−6 per cell division, and n is the number of cell divisions, then 0.05 = (1 − u)n. If we take logarithms of both sides, log(0.05) = nlog(1 − u), which to a good approximation, is equal to nlog (u). After substituting 10−6 for u and solving, we find that n = 1.3 × 106.

23.13 Inherited cancers like retinoblastoma show a dominant pattern of inheritance. However, the underlying genetic defect is a recessive loss-of-function mutation—often the result of a deletion. How can the dominant pattern of inheritance be reconciled with the recessive nature of the mutation?

ANS: At the cellular level, loss-of-function mutations in the RB gene are recessive; a cell that is heterozygous for such a mutation divides normally. However, when a second mutation occurs, that cell becomes cancerous. If the first RB mutation was inherited, there is a high probability that the individual carrying this mutation will develop retinoblastoma because a second mutation can occur any time during the formation of the retinas in either eye.

Thus, the individual is predisposed to develop retinoblastoma, and it is this predisposition that shows a dominant pattern of inheritance.

23.14 The following pedigree shows the inheritance of familial ovarian cancer caused by a mutation in the BRCA1 gene.

Should II-1 be tested for the presence of the predisposing mutation? Discuss the advantages and disadvantages of testing.

I II

Ovarian cancer Normal

III

ANS: II-1 should be tested for the BRCA1 mutation that apparently was involved in the ovarian cancer that developed in her mother and sister. If she is found to carry this

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mutation, a prophylactic oophorectomy can be prescribed to reduce the chance that she will develop cancer. If she is found to be free of the mutation carried by her mother and sister, then she is not more likely to develop ovarian cancer than a woman in the general population.

23.15 In what sense is pRB a negative regulator of E2F transcription factors?

ANS: By binding to E2F transcription factors, pRB prevents those transcription factors from activating their target genes—which encode proteins involved in progression of the cell cycle; pRB is therefore a negative regulator of transcription factors that stimulate cell division.

23.16 A particular E2F transcription factor recognizes the sequence TTTCGCGC in the promoter of its target gene. A temperature-sensitive mutation in the gene encoding this E2F transcription factor alters the ability of its protein product to activate transcription; at 25°C, the mutant protein activates transcription normally, but at 35°C, it fails to activate transcription at all. However, the ability of the protein to recognize its target DNA sequence is not impaired at either temperature. Would cells heterozygous for this temperature-sensitive mutation be expected to divide normally at 25°C? at 35°C? Would your answers change if the E2F protein functions as a homodimer?

ANS: At 25°, cell division should be normal—the same as for cells homozygous for a wild-type allele of the E2F gene.

At 35°, division would be expected to be impaired either because the mutant E2F protein binds unproductively to the sequence in its target gene or because a mutant E2F polypeptide dimerizes with a wild-type E2F polypeptide and abolishes the activation function of the wild-type polypeptide.

23.17 During the cell cycle, the p16 protein is an inhibitor of cyclin/CDK activity. Predict the phenotype of cells homozygous for a loss-of-function mutation in the gene that encodes p16. Would this gene be classified as a proto-oncogene or as a tumor suppressor gene?

ANS: Cells homozygous for a loss-of-function mutation in the p16 gene might be expected to divide in an uncontrolled manner because the p16 protein would not be able to inhibit cyclin-CDK activity during the cell cycle. The p16 gene would therefore be classified as a tumor suppressor gene.

23.18 The BCL-2 gene encodes a protein that represses the pathway for programmed cell death. Predict the phenotype of cells heterozygous for a dominant activating mutation in this gene. Would the BCL-2 gene be classified as a proto-oncogene or as a tumor suppressor gene?

ANS: Cells heterozygous for a dominant activating mutation in the BCL-2 gene would be expected to be unable to execute the programmed cell death pathway in response to DNA damage induced by radiation treatment.

Such cells would continue to divide and accumulate

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92-WC Answers to All Questions and Problems mutations; ultimately they would have a good chance of becoming cancerous. The BCL-2 gene would therefore be classified as a proto-oncogene.

23.23 Would cancer-causing mutations of the APC gene be expected to increase or decrease the ability of pAPC to bind b-catenin?

23.19 The protein product of the BAX gene negatively regulates the protein product of the BCL-2 gene—that is, BAX protein interferes with the function of the BCL-2 protein. Predict the phenotype of cells homozygous for a loss-of-function mutation in the BAX gene. Would this gene be classified as a proto-oncogene or as a tumor suppressor gene?

ANS: They would probably decrease the ability of pAPC to bind b-catenin.

ANS: Cells homozygous for a loss-of-function mutation in the BAX gene would be unable to prevent repression of the programmed cell death pathway by the BCL-2 gene product. Consequently, these cells would be unable to execute that pathway in response to DNA damage induced by radiation treatment. Such cells would ­continue to divide and accumulate mutations; ultimately, they would have a good chance of becoming cancerous.

The BAX gene would therefore be classified as a tumor suppressor gene.

23.20 Cancer cells frequently are homozygous for loss-offunction mutations in the TP53 gene, and many of these mutations map in the portion of TP53 that encodes the DNA-binding domain of p53. Explain how these mutations contribute to the cancerous phenotype of the cells.

ANS: Loss-of-function mutations in the DNA-binding domain of p53 abolish the ability of that protein to activate transcription of target genes whose products are involved in the restraint of cell division or in the promotion of programmed cell death. Without restraint of cell division or promotion of programmed cell death, cells accumulate damage to their DNA and ultimately become cancerous.

23.21 Suppose that a cell is heterozygous for a mutation that caused p53 to bind tightly and constitutively to the DNA of its target genes. How would this mutation affect the cell cycle? Would such a cell be expected to be more or less sensitive to the effects of ionizing radiation?

ANS: If a cell were heterozygous for a mutation that caused p53 to bind tightly and constitutively to the DNA of its target genes, its growth and division might be retarded, or it might be induced to undergo apoptosis. Such a cell would be expected to be more sensitive to the effects of ionizing radiation because radiation increases the expression of p53, and in this case, the p53 would be predisposed to activate its target genes, causing the cell to respond vigorously to the radiation treatment.

23.22 Mice homozygous for a knockout mutation of the TP53 gene are viable. Would they be expected to be more or less sensitive to the killing effects of ionizing radiation?

ANS: Homozygous TP53 knockout mice might actually be less sensitive to the killing effects of ionizing radiation because p53 would be unable to mediate the apoptotic response to the radiation treatment.

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23.24 Mice that are heterozygous for a knockout mutation in the RB gene develop pituitary and thyroid tumors. Mice that are homozygous for this mutation die during embryonic development. Mice that are homozygous for a knockout mutation in the gene encoding the p130 homologue of RB and heterozygous for a knockout mutation in the gene encoding the p107 homologue of RB do not have a tendency to develop tumors. However, homozygotes for knockout mutations in both of these genes die during embryonic development. What do these findings suggest about the roles of the RB, p139, and p107 genes in embryos and adults?

ANS: All three genes (RB, p130, and p107) are essential for embryonic development, although by themselves, p130 and p107 are dispensable, possibly because their products are functionally redundant. (Both gene products must be inactivated before any deleterious effect is seen.) In adults, only pRB appears to play a role in suppressing tumor formation.

23.25 It has been demonstrated that individuals with diets poor in fiber and rich in fatty foods have an increased risk to develop colorectal cancer. Fiber-poor, fat-rich diets may irritate the epithelial lining of the large intestine. How could such irritation contribute to the increased risk for colorectal cancer?

ANS: The increased irritation to the intestinal epithelium caused by a fiber-poor, fat-rich diet would be expected to increase the need for cell division in this tissue (to replace the cells that were lost because of the irritation), with a corresponding increase in the opportunity for the occurrence of cancer-causing mutations.

23.26 Messenger RNA from the KAI1 gene is strongly expressed in normal prostate tissues but weakly expressed in cell lines derived from metastatic prostate cancers.

What does this finding suggest about the role of the KAI1 gene product in the etiology of prostate cancer?

ANS: The KAI1 gene is a prostate tumor suppressor gene.

Functional inactivation of this gene allows prostate tumors to develop.

23.27 The p21 protein is strongly expressed in cells that have been irradiated. Researchers have thought that this strong expression is elicited by transcriptional activation of the p21 gene by the p53 protein acting as a transcription factor.

Does this hypothesis fit with the observation that p21 expression is induced by radiation treatment in mice homozygous for a knockout mutation in the TP53 gene? Explain.

ANS: No. Apparently there is another pathway—one not mediated by p53—that leads to the activation of the p21 gene.

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Answers to All Questions and Problems WC-93

Chapter 24 24.1 What was some of the evidence that led Charles Darwin to argue that species change over time?

ANS: Among other things, Darwin observed species on islands that were different from each other and from continental species but were still similar enough to indicate that they were related. He also observed variation within species, especially within domesticated breeds, and saw how the characteristics of an organism could be changed by selective breeding. His observations of fossilized organisms indicated that some species have become extinct.

24.2 Darwin stressed that species evolve by natural selection.

What was the main gap in his theory?

ANS: Darwin did not understand the mechanism of inheritance; he did not know of Mendel’s principles.

24.3 Using the data in Table 24.1, and assuming that mating is random with respect to the blood type, predict the frequencies of the three genotypes of the Duffy blood-type locus in a South African and an English population.

ANS: The frequency of the a allele is 0.06 in the South African population and 0.42 in the English population. The predicted genotype frequencies under the assumption of random mating are as follows: Genotype

South Africa

England

aa

(0.06)2 = 0.004

(0.42)2 = 0.18

ab

2(0.06)(0.94) = 0.11

2(0.42)(0.58) = 0.49

bb

(0.94)2 = 0.88

(0.58)2 = 0.33

24.4 Theodosius Dobzhansky and his collaborators studied chromosomal polymorphisms in Drosophila pseudoobscura and its sister species in the western United States. In one study of polymorphisms in chromosome III of D. pseudoobscura sampled from populations at different locations in the Yosemite region of the Sierra Nevada, Dobzhansky (1948, Genetics 33: 158–176) recorded the following frequencies of the Standard (ST) banding pattern:

ANS: The frequency of the ST banding pattern declines with increasing altitude. Thus, the data indicate that this chromosomal polymorphism exhibits an altitudinal cline.

24.5 In a survey of electrophoretically detectable genetic variation in the alcohol dehydrogenase gene of Drosophila melanogaster, a researcher found two allozymes, denoted F (fast) and S (slow) in a population; 32 individuals were homozygous for the F allele of the gene, 22 were homozygous for the S allele, and 46 were heterozygous for the F and S alleles. Are the observed frequencies of the three genotypes consistent with the assumption that the population is in Hardy–Weinberg equilibrium?

ANS: In the sample, the frequency of the F allele is (2 × 32 + 46)/(2 × 100) = 0.55 and the frequency of the S allele is 1 − 0.55 = 0.45. The predicted and observed genotype frequencies are as follows: Genotype

Observed

Hardy–Weinberg Predicted

FF

32

100 × (0.55)2 = 30.25

FS

46

100 × 2(0.55)(0.45) = 49.5

SS

22

100 × (0.45)2 = 20.25

To test for agreement between the observed and predicted values, we compute a chi-square statistic with 1 degree of freedom: c2 = S(obs. − pred.)2/pred. = 0.50, which is not significant at the 5 percent level. Thus, the population appears to be in Hardy–Weinberg equilibrium for the alcohol dehydrogenase locus.

24.6 A researcher has been studying genetic variation in fish populations by using PCR to amplify microsatellite repeats at a particular site on a chromosome (see Chapter 16). The diagram below shows the gel-fractionated products of amplifications with DNA samples from 10 different fish. How many distinct alleles of this microsatellite locus are evident in the gel?

Elevation (in feet)

Location

Frequency ST

Jacksonville

0.46

850

Lost Claim

0.41

3,000

Mather

0.32

4,600

ANS: There are four alleles.

Aspen

0.26

6,200

Porcupine

0.14

8,000

24.7 Within the coding region of a gene, where would you most likely find silent polymorphisms?

Tuolumne

0.11

8,600

Timberline

0.10

9,900

Lyell Base

0.10

10,500

What is interesting about these data?

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ANS: In the third position of some of the codons. Due to the degeneracy of the genetic code, different codons can specify the same amino acid. The degeneracy is most pronounced in the third position of many codons, where different nucleotides can be present without changing the amino acid that is specified.

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94-WC Answers to All Questions and Problems 24.8 Why are the nucleotide sequences of introns more polymorphic than the nucleotide sequences of exons?

ANS: Introns do not encode amino acids; most exons do.

Many—perhaps most—of the nucleotides within an intron can be changed without impairing the expression of the gene or the integrity of its polypeptide product.

By contrast, many of the nucleotides within exons— especially the first and second positions of codons—are functionally constrained by the amino acids specified.

24.9 DNA and protein molecules are “documents of evolutionary history.” Why aren’t complex carbohydrate molecules such as starch, cellulose, and glycogen considered “documents of evolutionary history”?

ANS: Complex carbohydrates are not “documents of evolutionary history” because, although they are polymers, they are typically made of one subunit incorporated repetitiously into a chain. Such a polymer has little or no “information content.” Thus, there is little or no opportunity to distinguish a complex carbohydrate obtained from two different organisms. Moreover, complex carbohydrates are not part of the genetic machinery; their formation is ultimately specified by the action of enzymes, which are gene products, but they themselves are not genetic material or the products of the genetic material.

24.10 A geneticist analyzed the sequences of a gene cloned from four different individuals. The four clones were identical except for a few base pair differences, a deletion (gap), and a transposable element (TE) insertion: Sequences TE

G C

1

TE

A T

2 C G

3

T A C G

A T

4

G C

Using this information, compute the minimum number of mutations required to explain the derivation of the four sequences (1, 2, 3, and 4) in the following phylogenetic trees: 1

2

Tree A

3

4

1

2

4

Tree B

3

3

4

2

1

Tree C

Which of these trees provides the most parsimonious explanation for the evolutionary history of the four DNA sequences?

ANS: The tree on the right would require nine mutations.

These mutations are a deletion in the branch leading to

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the common ancestor of sequences 1, 2, and 3; a TE insertion in the branch leading to the common ancestor of sequences 1 and 2; and seven base-pair changes: one leading to sequence 1, another leading to sequence 2, two leading to sequence 3 and three leading to sequence 4.

The tree in the middle would also require nine ­mutations.

In this tree, we regard the gap as ancestral—which means that sequence 4 has acquired an “insertion” at the position of the gap. The other mutations are a TE ­insertion in the branch leading to the common ancestor of sequences 1 and 2, and seven base-pair changes: one ­leading to sequence 1, another to sequence 2, two leading to sequence 3, and three leading to sequence 4.

The tree on the right would also require nine mutations.

Here again we regard the gap as ancestral; evidently, an insertion occurred at the position of the gap in the branch leading to sequence 4. The TE insertion can also be regarded as ancestral, with loss occurring in the branch leading to the common ancestor of sequences 3 and 4. There are also seven base-pair changes in this tree: one leading to sequence 1, another leading to sequence 2, two leading to sequence 3 and three leading to sequence 4. These interpretations of the data assume that insertions and deletions (gaps) are reversible and that there is no way of telling which way the sequence evolved—that is, through insertion or deletion. However, if we know, for example, that the TE is a retrotransposon incapable of excision, then we would need more than nine mutations to explain the tree on the right. The TE insertions must have occurred independently in the branches leading to sequences 1 and 2, and there is no need to postulate e­ xcision of the TE in the branch leading to the common ancestor of sequences 3 and 4. Thus, on the assumption that the TE is not excisable, 10 mutations are needed to explain the tree on the right. Given this reservation, the left and middle trees provide the most parsimonious explanations for the evolution of the four sequences.

24.11 The heme group in hemoglobin is held in place by histidines in the globin polypeptides. All vertebrate globins possess these histidines. Explain this observation in terms of the Neutral Theory of Molecular Evolution.

ANS: The histidines are rigorously conserved because they perform an important function—anchoring the heme group in hemoglobin. Because these amino acids are strongly constrained by natural selection, they do not evolve by mutation and random genetic drift.

24.12 During the early evolutionary history of the vertebrates, a primordial globin gene was duplicated to form the α- and b-globin genes. The rate of evolution of the polypeptides encoded by these duplicate genes has been ­ estimated to be about 0.9 amino acid substitutions per site every billion years. By comparing the human α- and b-globins, the average number of amino acid substitutions per site has been estimated to be 0.800. From this estimate, calculate when the duplication event that ­produced the α- and b-globin genes must have occurred.

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Answers to All Questions and Problems WC-95

ANS: The total elapsed evolutionary time is 0.800/0.9 = 880 million years, which must be apportioned equally to the α and β gene lineages by dividing by 2; thus, the time since the duplication event is estimated to be 440 m ­ illion years.

24.17 How might you explain the 1000-fold difference in the evolutionary rates of fibrinopeptide and histone 3?

24.13 Ribonuclease, a protein that degrades RNA, is 124 amino acids long. A comparison between the amino acid sequences of cow and rat ribonucleases reveals 40 differences. What is the average number of amino acid substitutions that have occurred per site in these two evolutionary lineages? If the cow and the rat lineages diverged from a common ancestor 80 million years ago, what is the rate of ribonuclease evolution?

24.18 A geneticist has studied the sequence of a gene in each of three species, A, B, and C. Species A and species B are sister species; species C is more distantly related. The geneticist has calculated the ratio of nonsynonymous (NS) to synonymous (S) nucleotide substitutions in the coding region of the gene in two ways—first, by comparing the gene sequences of species A and C, and second, by comparing the gene sequences of species B and C. The NS:S ratio for the comparison of species B and C is five times greater than it is for the comparison of species A and C. What might this difference in the NS:S ratios suggest?

ANS: Estimate the average number of substitutions per site in the ribonuclease molecule as −ln(S), where S = (124 − 40)/124 = 0.68, the proportion of amino acids that are the same in the rat and cow molecules. The average number of substitutions per site since the cow and rat lineages diverged from a common ancestor is therefore 0.39. The evolutionary rate in the cow and rat lineages is 0.39/(2 × 80 million years) = 2.4 substitutions per site every billion years.

24.14 If a randomly mating population is segregating n selectively neutral alleles of a gene and each allele has the same frequency, what is the frequency of all the homozygotes in the population?

ANS: With n alleles having equal frequency, the frequency of any one allele is 1/n. Under random mating, the frequency of homozygotes for a particular allele is (1/n)2.

The frequency of all the homozygotes is therefore ∑(1/n)2 = n(1/n)2 = 1/n.

24.15 If the evolutionary rate of amino acid substitution in a protein is K, what is the average length of time between successive amino acid substitutions in this protein?

ANS: The reciprocal of the rate, that is, 1/K. 24.16 The coding sequence of the alcohol dehydrogenase (Adh) gene of D. melanogaster consists of 765 nucleotides (255 codons); 192 of these nucleotides are functionally silent—that is, they can be changed without changing an amino acid in the Adh polypeptide. In a study of genetic variation in the Adh gene, Martin Kreitman observed that 13 of the 192 silent nucleotides were polymorphic. If the same level of polymorphisms existed among the nonsilent nucleotides of the Adh gene, how many amino acid polymorphisms would Kreitman have observed in the populations he studied?

ANS: The fraction of polymorphic sites among the silent nucleotides is 13/192 = 0.068. If the same level of polymorphism existed among the nonsilent sites, the number of amino acid polymorphisms would be 225 × 0.067 = 17.2. Kreitman actually observed only one amino acid polymorphism—evidence that amino acid changes in alcohol dehydrogenase are deleterious.

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ANS: The protein with the higher evolutionary rate is not as constrained by natural selection as the protein with the lower evolutionary rate.

ANS: The difference in the NS:S ratios suggests that in at least one lineage, positive selection has been operating to change nucleotides in the gene.

24.19 Dispersed, repetitive sequences such as transposable elements may have played a role in duplicating short regions in a genome. Can you suggest a mechanism? (Hint: See Chapter 21 on the Instructor Companion site.)

ANS: Repetitive sequences that are near each other can mediate displaced pairing during meiosis. Exchange involving the displaced sequences can duplicate the region between them.

24.20 Exon shuffling is a mechanism that combines exons from different sources into a coherent sequence that can encode a composite protein—one that contains peptides from each of the contributing exons. Alternate splicing is a mechanism that allows exons to be deleted during the expression of a gene; the mRNAs produced by alternate splicing may encode different, but related, polypeptides (see Chapter 18). What bearing do these two mechanisms have on the number of genes in a eukaryotic genome? Do these mechanisms help to explain why the gene number in the nematode Caenorabditis elegans is not too different from the gene number in Homo sapiens?

ANS: Exon shuffling is a way of creating genes that have pieces from disparate sources. Through exon shuffling, the number of genes in a genome could be increased without having to evolve the genes “from scratch.” However, the genome sequencing projects indicate that the gene number in complex, multicellular vertebrates is not much different from the gene number in simple, multicellular invertebrates or in plants. If, judging from their phenotypes, multicellular vertebrates need more gene products than phenotypically simpler invertebrates such as C. elegans, alternate splicing could provide some of these gene products without increasing the number of genes.

24.21 Drosophila mauritiana inhabits the island of Mauritius in the Indian Ocean. Drosophila simulans, a close relative, is

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96-WC Answers to All Questions and Problems widely distributed throughout the world. What experimental tests would you perform to determine if D. mauritiana and D. simulans are genetically different species?

ANS: Cross D. mauritiana with D. simulans and determine if these two species are reproductively isolated. For instance, can they produce offspring? If they can, are the offspring fertile?

24.22 Distinguish between allopatric and sympatric modes of speciation.

ANS: Allopatric speciation occurs when populations diverge genetically while they are geographically separated.

Sympatric speciation occurs when populations diverge genetically while they inhabit the same territory.

24.23 The prune gene (symbol pn) is X-linked in Drosophila melanogaster. Mutant alleles of this gene cause the eyes to be brown instead of red. A dominant mutant allele of another gene located on a large autosome causes ­hemizygous or homozygous pn flies to die; this dominant mutant allele is therefore called Killer of prune (symbol Kpn). How could mutants such as these play a role in the evolution of reproductive isolation between populations?

ANS: The Kpn–pn interaction is an example of the kind of negative epistasis that might prevent populations that have evolved separately from merging into one panmictic population. The Kpn mutation would have evolved in one population and the pn mutation in another, geographically separate population. When the populations merge, the two mutations can be brought into the same

ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 96

fly by interbreeding. If the combination of these mutations is lethal, then the previously separate populations will not be able to exchange genes, that is, they will be reproductively isolated.

24.24 A segment of DNA in an individual may differ at several nucleotide positions from a corresponding DNA segment in another individual. For instance, one individual may have the sequence …A…G…C… and another individual may have the sequence …T…A…A…. These two DNA segments differ in three nucleotide positions.

Because the nucleotides within each segment are tightly linked, they will tend to be inherited together as a unit, that is, without being scrambled by recombination.

We call such heritable units DNA haplotypes. Through sampling and DNA sequencing, researchers can determine which DNA haplotypes are present in a particular population. When this kind of analysis is performed on human populations by sequencing, for example, a segment of mitochondrial DNA, it is found that samples from Africa exhibit more haplotype diversity than samples from other continents. What does this observation tell us about human evolution?

ANS: More haplotype diversity refers to the number of different haplotypes that are found in a population. If African populations of humans have the greatest haplotype diversity, then these populations appear to have had a longer time to accumulate different haplotypes—that is, they are older than other populations. Greater haplotype diversity in African populations of humans is therefore evidence that African populations were at the root of the modern human evolutionary tree.

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