Principles of Genetics, 6th Edition Test Bank

Principles of Genetics, 6th Edition

By Snustad, Simmons

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

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

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

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

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c) The discovery of the structure of DNA d) All of these e) None of these Answer: d

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

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

7) Different forms of the same gene are known as: a) Peptides b) Amino acids c) Proteins d) Alleles e) Gene differences Anwer: d

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 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

10) The building blocks of genes are: a) Proteins b) Amino acids c) Nucleic acids d) Lipids e) Carbohydrates Answer: c

11) Nucleic acids are composed of a series of: a) Amino acids b) Nucleotides c) Nucleic chromosomes d) Peptide chains e) Sugars Answer: b

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

13) Which of the following is the sugar found in RNA? a) Deoxyribose b) Ribose c) Sucrose d) Fructose e) Glucose Answer: b

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

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

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

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 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

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

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

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

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

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

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

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

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

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

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

29) What substance is produced via the process of transcription? a) ssDNA b) dsDNA c) proteins d) mRNA e) enzymes Answer: d

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:

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: b

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

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 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

35) Established historical relationships between organisms are known as a: a) Phylogeny b) Genealogy c) Genomics d) Proteomics e) None of these Answer: a

36) If a geneticist pursues their 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

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

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

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

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

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

42) An example of a commonly produced agricultural GMO is: a) Philodendron b) Corn c) Bacillus thuringiensis d) Canker worm e) Butterfly

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

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

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

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.

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.

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

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

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.

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

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

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

4) The internal liquid portion of the cell is known as: a) Cytoplasmic membrane b) Nucleus c) Cytoplasm d) Mitochondria e) Lysosome Answer: c

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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

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

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

8) Proteins that catalyze biochemical reactions are known as: a) Proteosomes b) Lysosomes c) Chemosomes d) Enzymes e) Anions Answer: d

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

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

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

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

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

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

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

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 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

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

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

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

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

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

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

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

25)Eukaryotic cells divide during which phase of the cell cycle? a) G1 b) S c) G2 d) M e) K Answer: d

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

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

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

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

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

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

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

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

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

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

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 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

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

39) The points where chromosomes have physically crossed over are known as: a) Centromeres b) Kinetochores c) Chiasmata d) Diplonema e) Pachynema Answer: c

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 AnswerL d

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

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

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

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

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

46)

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

47) 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.

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 can not be used for fertilization e) The organism can not be used for genetic study Answer: a

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

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

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

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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

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

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 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

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

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

11) In a monohybrid cross, if the P generation consists of plants that are true breeding (one is homozygous dominant, the other homozygous recessive) then 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

12) In a monohybrid cross, if the P generation consists of plants that are true breeding (one is homozygous dominant, the other homozygous recessive) 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

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

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

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

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

17) In a dihybrid cross, if the P generation consists of plants that are true breeding (one is homozygous dominant, the other homozygous recessive) 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

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

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

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) Seven pairs of alleles determining these traits behave as if they are on the same chromosome e) Formation of gametes in plants is by mitosis only

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

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, assuming that the trait exhibits complete penetrance? a) 10% b) 25% c) 50% d) 100% e) 0% Answer: c

23) What is a genetic cross between one homozygous recessive individual and one of unknown genotype?

a) Self cross b) Hybrid cross c) Dihybrid cross d) Test cross e) F1 cross Answer: d

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

25) In a trihybrid cross, 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

26) In a cross AABBCC and aabbcc (P generation), what would be the phenotypic frequency of A_bbC_ in the F2 (produced by an F1 self cross), where ___= dominant or recessive allele? a) 9/16 b) 9/64 c) 3/16 d) 3/64 e) 1/64 Answer: b

27)

27)

Which of the following is a method for analyzing a cross involving 2 genes? a) b) c) d) e)

Punnett Square Method Forked-Line Method Probability Method All of these None of these

Answer: d

28) In the Chi-Square test a critical value is: a) The point where the discrepancies between observed and expected numbers are not b) likely to be due 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

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

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

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

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%

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

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

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

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

Question Type: Essay

37) In a cross between individuals of the genotype AaBbCc x AaBbCc, what is the probability of producing the genotype AABBCC? Answer: 1/64 (¼AA x ¼ BB x ¼ CC)

38) A normally pigmented man (dominant) marries an albino woman (recessive). They have 3 children, one of whom is an albino. What is the genotype of the man?

Answer: Heterozygote 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

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

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 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

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%.

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%.

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. Given that one gene pair is involved, is the inheritance pattern recessive or dominant

Answer: Recessive

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

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 x ¾ normal pigmentation child #2)

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, if the curly haired individual was a heterozygote?

Answer: 1/2

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)

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 x 0.25 CF)

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

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

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

4) Which of the following is the most appropriate representation of a codominant allele? a) M b) m c) LM d) LM

e) None of these Answer: d

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

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

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

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

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

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

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

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

13) Non-functional alleles are said to be: a) Polymorphic b) Amorphic c) Hypomorphic d) Heteromorphic e) Homomorphic Answer: b

14) Alleles that are partially functional are said to be: a) Polymorphic b) Amorphic c) Hypomorphic d) Heteromorphic e) Homomorphic Answer: c

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

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

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

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

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 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 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

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

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

24) Polypeptides are macromolecules comprised of what monomer subunit?

a) Nucleic acids b) Amino acids c) Micromolecules d) Enzymes e) Proteins Answer: b

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 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%

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

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.

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

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

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 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

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

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

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

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

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

10) X and Y chromosomes are known as: a) Autosomes b) Somosomes c) Sex chromosomes d) Non-sex chromosomes e) Somatosomes Answer: c

11) All chromosomes except X and Y are known as: a) Autosomes b) Somosomes c) Sex chromosomes d) Somatosomes e) Tetrasomes Answer: a

12) In a certain species, a somatic cell has 48 chromsomes. How many chromosomes are in a sex cell? a) 12 b) 24 c) 96 d) 144 e) 200 Answer: b

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

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

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) Sweetpeas Answer: c

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

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

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

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

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

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

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

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

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 mae b) The individual is sterile c) The individual is male and sterile d) The individual is female and sterile

Answer: d 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

26) Down Syndrome occurs when there is an extra chromosome #21. Such individuals therefore have 47 chromosomes. Assume that a mating occurs between a female with Down Syndrome and a normal male. Assuming normal disjunction during meiosis, what percentage of offspring would be expected to have Down Syndrome? a) 25% b) 50% c) 75% d) 100% e) 0% Answer: b

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

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

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

30) Which of the following is an X-linked disorder in human beings? a) Hemophilia b) Color blindness c) HIV d) Hemophiila and Color blindness e) All of these Answer: d

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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 anaphase1 or anaphase2 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.

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

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.

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.

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

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

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

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

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

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

7) Human diploid cells contain _____ chromosomes. a) 12 b) 24 c) 46 d) 92 e) 23 Answer: c

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

9) The human diploid cell has ______ pairs of chromosomes. a) 12 b) 23 c) 24 d) 46 e) 48 Answer: b

10) The long arm of the chromosome is designated: a) P b) Q c) L d) M e) N Answer: b

11) The short arm of the chromosome is designated: a) P b) Q c) L d) M e) N Answer: a

12) The pattern of bands within a chromosome is termed: a) Karyotype b) Chromogram c) Ideogram d) Sonogram e) None of these

Answer: c

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

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

15) An organism that carries extra sets of chromosomes is termed: a) Aneuploid b) Euploid c) Polyploid d) Diploid e) Multiploid Answer: c

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

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

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

19) a) Wheat b) Coffee c) Strawberries d) All of these e) None of these Answer: d

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

38) Cells with triplicate chromosomes are said to exhibit: a) Down syndrome b) Trisomy c) Autosomy d) Triploidy e) Polyploidy Answer: b

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

40) An organism in which one chromosome or chromosome segment is overrepresented is referred to as a: a) Hypoploid b) Hyperploid c) Polyploid d) Autopolyploid

e) Allopolyploid Answer: b

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) Triplo-X Syndrome

Answer: c

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

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

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

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

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

47) An individual who possesses one inverted chromosome and one that is not inverted is said to be a/an: a) Inversion heterozygote b) Inversion mutant c) Inversion homozygote d) All of these e) None of these Answer: a

48) When a segment from one chromosome is detached and reattached to a nonhomologous chromosome this is known as:

a) Inversion b) Translocation c) Polyploidy d) Trisomy e) Monosomy Answer: b

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

Question Type: Essay

50) 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.

51) 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.

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

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.

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.

Question Type: Multiple Choice

1) Which of the following most accurately represents how Alfred H 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

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

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

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

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

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

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

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

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

10) Typically, each crossover event produces how many recombinants? a) 1 b) 2 c) 3 d) 4 e) 5 Answer: b 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

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

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 1 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 1 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

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

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

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

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

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

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 blu100 white, short e, long What is the recombination frequency, as shown in the F2 generation? a) 0.25

b) .20 c) .08 d) .10 e) .05 Answer: b

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

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

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

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

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

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

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

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

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

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

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

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

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

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

34) Saccharomyces cerevisiae is a good experimental organism for the study of meiosis because: a) It is haploid with a sexual cycle. b) It produces 4 ascospores. c) You can analyze all 4 products of a single meiosis. d) There are many mutations that have been identified. e) All of these Answer: e

35) If ascospores are analyzed in the order in which they were created, the fungus must produce spore pairs which are _____________. a) Unordered tetrads b) Ordered tetrads c) Random tetrads d) Labeled tetrads e) Allof these Answer: b

36) In one type of ascus, two of the ascospores are genotypically like one of the parents and two are like the other parent. An ascus with this pattern is called a: a) Parental ditype ascus b) Nonparental ditype ascus c) Original phenotype ascus d) Recombinant phenotype ascus e) None of these Answer: a

37) In one type of ascus neither of the two genotypes is like the parental genotypes; thus, the ascus is called a: a) Parental ditype ascus b) Nonparental ditype ascus c) Original phenotype ascus d) Recombinant phenotype ascus e) None of these

Answer: b

38) In this type of ascus there are four kinds of ascospores, one that is like each of the parents and two that are recombinant. a) Parental ditype ascus b) Nonparental ditype ascus c) Tetratype d) Quadratype e) Recombinant type ascus Answer: c

39) Two yeast strains are crossed and, it is observed that there is a very low frequency of nonparental ditype asci. What can you conclude about the genes being studied?

1. The genes are most likely not linked 2. The genes are most likely linked 3. The genes are most likely far apart from each other a) 1 b) 2 c) 3 d) 2 and 3 e) All of these Answer: b

40) In ordered tetrad analysis, what type of pattern occurs when there is a crossover between a locus and the centromere in the first meiotic division? a) First division segregation b) Second division segregation c) Third division segregation d) Independent assortment e) Linkage Answer: b

41) 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

42) 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

43) 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

44) 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) 1 and 2 e) All of these

Answer: e 45) 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

Question Type: Essay

46) 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

47) 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 cross?

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

48) 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 + xt cm : 39 + xt +: 474 + + cm: 158

z + cm : 460 z xt +: 164 z + +: 35 + + +: 40

What is the order of the three genes on chromosome 3?

Answer: z cm xt

49) 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 + xt cm : 39 + xt +: 474 + + cm: 158

z + cm : 460 z xt +: 164 z + +: 35 + + +: 40

What is the map distance between z and cm ?

Answer: 28 map units or centimorgans

50) 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 + xt cm : 39 + xt +: 474 + + cm: 158

z + cm : 460 z xt +: 164 z + +: 35 + + +: 40

Answer: 11 map units or centimorgans

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

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

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

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

5) Two of the most important bacteriophages that have played a role in they study of genetics are: a) E. coli and T4 b) T4 and TMZ c) T4 and Lambda d) Lambda and E. coli e) Lambda and TMZ Answer: c

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

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

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

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

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

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 f these Answer: b

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 lifecycle 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

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 14) When bacteriophage  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

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

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

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

18) Plaque morphology is a function of: a) A physiological interaction between bacteria and phage. b) Phage genotype c) Bacterial strain d) All of these e) None of these Answer: d

19) In a mixed infection of T4 of genotype h+r X hr+ (where h is the mutant that produces clear plaques and r is the mutant that produces sharp edges) on a mixture of E. coli B and B2 cells, wild type recombinants form: a) Turbid plaques with sharp edges b) Clear plaques with fuzzy edges c) Turbid plaques with fuzzy edges d) Clear plaques with sharp edges e) Clear and turbid plaques with jagged edges Answer: c 20) What is the difference in the plaque morphology of r+ and r phage on E. coli B? a) r+ is larger than r b) r is larger than r+ c) r+ has sharper edges than r d) r has sharper edges than r+ e) None of these Answer: d

21) Which of the following sequences are terminally redundant and circularly permuted? a) (ABCD...WXYZ, BCDE...XYZA, CDEF...YZAB) b) (ABCD...YZAB, CDEF...ABCD, EFGH...CDEF) c) (ABCD...YZAB, EFGH...CDEF, IJKL...GHJI) d) (ABCD...WXYZ, ABCD....WXYZ, ABCD...WXYZ) e) None of these Answer: b

22) Terminally redundant DNA molecules: a) contain the same nucleotide sequence at both ends of a linear molecule b) contain different nucleotide sequences at the ends of a linear molecule c) contain the same nucleotide sequence in the middle segments of a linear molecule d) All of these e) None of these Answer: a

23) The head-full mechanism of DNA packaging incorporates _____________ of T4 DNA per phage head. a) more than one genome b) less than one genome c) one genome d) all of these e) none of these Answer: a

24) The genetic map of phage T4 is circular because: 1. The sequence is terminally redundant 2. The sequence is circularly permuted 3. The sequence is 50kbp long a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: d

25) 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

26) 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

27) 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

28) 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

29) Wild type bacteria that can synthesize all necessary metabolites are known as: a) Prototrophs b) Auxotrophs c) Phototrophs d) Heterotrophs e) Chemoheterotrophs Answer: a

30)

30) Bacteria that can not synthesize all the necessary amino acids are known as: a) b) c) d) e)

Prototrophs Auxotrophs Phototrophs Heterotrophs Chemoheterotrophs

Answer: b 31) A mutant of E. coli is designated Strr and arg-. Which of the following best describes this genotype? a) b) c) d) e)

antibiotic resistant, prototrophic antibiotic resistant, auxotrophic antibiotic sensitive, prototrophic antibiotic sensitive, auxotrophic antibiotic resistant, carbon-source mutant

Answer: b

32) With rare exceptions, after gene transfer occurs, the recipient cells become: a) b) c) d) e)

Full diploids Partial diploids Full haploids Partial haploids Triploids

Answer: b

33) Recombination between bacteria occurs between: a) b) c) d) e)

The main chromosome of the donor and a fragment of the recipient.

The main chromosome of the recipient and a fragment of the donor.

The main chromosome of the donor and the main chromosome of the recipient.

The main chromosome of the recipient and the main chromosome of the donor.

None of these

Answer: b

34) Which process relies on phage as a vehicle of DNA transport? a) b) c) d) e)

Conjugation Transformation Transduction Conjugation and Transformation Transformation and Transduction

Answer: c

35) Which process requires cell-to-cell contact for DNA transfer? a) b) c) d) e)

Conjugation Transformation Transduction Conjugation and Transformation Transformation and Transduction

Answer: a

36) 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) b) c) d) e)

Transduction was taking place before the addition of DNAse Transformation was taking place before the addition of DNAse Conjugation was taking place before the addition of DNAse Binary Fission was taking place before the addition of DNAse There is not enough information provided to determine what method of genetic exchange was occurring prior to the addition of DNAse

Answer: b

37) Cells that are capable of DNA uptake are known as: a) b) c) d) e)

Capable Competent Conjugative Universal recipient None of these

Answer: b

38) During B. subtilis transformation, which of the following statements is incorrect? a) b) c) d)

Large, double-stranded DNA fragments are bound to the cell surface Bound DNA is attached to specific receptor sites DNA moves across the membrane in an energy independent manner 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

39) A recombinant double helix that has one allele on one strand and the other allele on the second strand is known as a: a) b) c) d) e)

Heteroduplex Homoduplex Heterotriplex Homotriplex Diploid

Answer: a

40) 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

41) 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) b) c) d) e)

1 2 3 1 and 2 2 and 3

Answer: a

42) A cell that carries an integrated F factor is best known as a/an: a) b) c) d) e)

Episome Hfr cell F' cell F+ cell None of these

Answer: b

43) Hfr mapping: a) b) c) d) e)

Is carried out using interrupted mating experiments Maps gene distance as a unit of base pairs rather than a crossover frequency.

Requires F+ to integrate into a multiple positions to map chromosomal genes.

Only maps genes that are more than 4 minutes apart.

None of these

Answer: a

44) Which form of replication mediates the transfer of chromosomes from Hfr cells to Fcells? a) b) c) d) e)

Rolling circle replication Bidirectional replication Semi-conservative replication Conservative replication Dispersive replication

Answer: a

45) 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

46) 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

47) A Hfr strain is produced by: a) b) c) d) e)

The excision of a lambda phage The integration of the F factor into a bacterial chromosome The integration of lambda phage DNA into a bacterial chromosome All of these None of these

Answer: b

48) Modified F factors that also carry extra bacterial genes are known as: a) b) c) d) e)

F+ Hfr FF' None of these

Answer: d

49) Transfer of F' factors into recipient cells is a process known as: a) b) c) d) e)

Conjugation Sexjugtion Transduction Sexduction Transformation

Answer: d

Question Type: Essay

50) 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.

51) 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.

52) 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.

53) Four Hfr strains of E. coli transferred a series of genes in the order of left to right as follows Hfr1: BAKJI Hfr2: EFGHI Hfr3: ABC 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 54) 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 x+y z+ 0.02 Answer: They are linked and the order places y in the middle (xyz or zyx)

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

2) Which of the following are essential functions that the genetic material must perform? a) Replication, mutation, gene expression b) Replication, adaptation, mutation c) Transcription, translation, storage d) Transcription, translation, replication e) None of these Answer: a

3) 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

4) Which of the following is a genotypic function that the genetic material must be able to perform? a) Replication b) Mutation c) Adaptation d) Gene expression e) Translation Answer: a

5) Which of the following is a phenotypic function that genetic material must be able to perform? a) Replication b) Mutation c) Adaptation d) Gene expression e) Translation Answer: d

6) How does mutation allow an organism to evolve?

1. Changes in the genetic material allow organisms to adapt to modifications in the environment 2. Changes in the genetic material allow organisms to survive in limited types of environment 3. Mutations do not allow an organism to evolve a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and Answer: a

7) Which of the following is a type of nucleic acid? a) DNA b) RNA c) DNAse d) RNAse e) DNA and RNA Answer: e

8) The genetic information in most organisms, excluding viruses, viroids, and prions, is encoded in what molecule?

A) RNA b) Protein c) DNA d) Lactose e) Lactase

Answer: c

9) 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

10) 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

11) 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

12) 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

13) The first direct evidence indicating that DNA, rather than RNA or protein, is the genetic material in bacteriophages was performed by: a) Griffith b) Avery, McLeod, and McCarty c) Hershey and Chase d) Watson and Crick e) Sia and Dawson Answer: c

14) 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

15) Hershey and Chase used radiolabeled T2 phages to identify which component of the phage entered E. coli during infection. Which of the following matches 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 16) The work carried out by Hershey and Chase had one flaw: a high degree of 35S was found inside the cells. This flaw was later addressed by:

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 a) c

17) 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

18) Infectious agents that contain small circular pieces of RNA that are not packaged in a protein coat are known as: a) Prions b) Viruses c) Bacteria d) Viroids e) Protisits Answer: d

19) The pathogenic effects of viroids are thought to result from: a) The ability to alter normal patterns of gene expression b) The ability to directly mutate DNA c) The ability to indirectly mutate RNA d) The ability to stop gene expression e) The ability to enhance gene expression Answer: a

20) Infectious agents that contain only altered forms of protein and no nucleic acid are known as:

a) Prions b) Viruses c) Bacteria d) Viroids e) Protists Answer: a

21) The pathogenic effects of prions are thought to result from: a) Viroids acting as a template for the synthesis of the prion proteins b) Aberrant forms of mammalian cellular proteins that convert more mammalian cellular proteins to aberrant forms, leading to a clumping of the proteins, eventually killing the cell. c) Prions converting viroids into abberant proteins that kill the cell d) All of these e) None of these Answer: c

22) 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

23) Which of the following is true regarding the structure of RNA? 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: e

24) 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

25) 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

26) 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

27) 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

28) 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

29) 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

30) 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 degrees) would the helix make? a) 1 b) 1.1 c) 1.5 d) 2 e) 2.1 Answer: b

31) DNA extracted from a bacterium was identified as having a guanine content of 14%. From this information determine the percentage of A. a) 14% b) 28% c) 36% d) 78% e) There is not enough information to determine this.

Answer: c

32) 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

33) Underwound DNA is said to exhibit: a) Positive supercoiling b) Negative supercoiling c) Positive relaxation d) Negative relaxation e) None of these Answer: b

34) 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

35) 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

36) 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

37) 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

38) According to the unineme model of chromosomal structure, how many double stranded DNA molecules run parallel throughout a eukaryotic chromosome? a) One b) Two c) Three d) Four e) Five Answer: a 39) The technique for visualizing large chromosomes using [3H] thymidine-labeled DNA on protein coated glass slides is called: a) Pulse-field gel electrophoresis b) Autoradiography c) Viscoelastometry d) DNA sequencing e) RFLP Answer: b

40) DNA migrates toward which pole during gel electrophoresis?

a) Positive because DNA is negatively charged b) Negative because DNA is positively charged c) Positive because DNA is positively charged d) Negative because DNA is negatively charged e) Neither positive and negative because DNA is neutral in charge Answer: a

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

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

43) Satellite DNA is a) Repetitive b) Randomly located throughout the genome c) Prokaryotic d) Expressed e) None of these Answer: a

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

45) For dispersed moderately repetitive DNA in the human genome, which of the following is incorrect? a) LINEs are the longest and L1 is the only functional element in the genome b) SINEs are typically 100-400 bp in length and the most abundant c) Transposable elements with LTRs are rarely observed d) DNA transposons comprise about 3% of the genome e) None of these Answer: c

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.

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.

48) How does pulsed field gel electrophoresis differ from standard gel electrophoresis?

Answer: Pulsed-field gel electrophoresis, which is used to separate large DNA molecules, differs from standard gel electrophoresis in that instead of a single (one-dimensional), constant electric field, two electric fields offset by about 90° are applied across the gel in an alternating or pulsed manner. In standard gel electrophoresis, the DNA molecules pass through the gel in an end-first or snake-like fashion. In pulsed-field gel electrophoresis, the application of intermittent and alternating electric fields requires the molecules to reorient themselves before continuing to migrate through the gel. Larger molecules take longer to undergo these reorientation events and move more slowly. As a result, pulsed-field gel electrophoresis yields superior separation of very large DNA molecules.

49) 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.

50) 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.

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

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

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

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

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

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

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

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

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

10) E. coli chromosomes: a) Replicate in a unidirectional pattern from a single, unique origin, called oriC. b) Adopt a -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

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 -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

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

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

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

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

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

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

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 -shaped replicative structure of the lambda chromosome d) All of these e) None of these Answer: a

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

36) Which of the following prokaryotic replication events is incorrect? a) Interaction of prepriming 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

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

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

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

40) In eukaryotes, two DNA polymerases were found to be essential to leading and lagging strand synthesis of the SV40 virus. They are: a) , b) , c) , d) , e) , Answer: a

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

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

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)  b)  c)  d)  e)  Answer: e

44) Telomere length has not been correlated with: a) Aging b) Sex determination c) Progeria d) Cancer e) All of these Answer: b

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

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.

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.

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.

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 (Figure 10.28). 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 (Figure 10.28). 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 -type intermediates.

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.

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

2) Which process transfers information from RNA molecules to proteins? a) Replication b) Transcription c) Translation d) Splicing e) None of these Answer: c

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

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

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

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

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

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

9) Which part of the eukaryotic cell is the location of transcription? a) Nucleus b) Mitochondria c) Ribosome d) Cytosol e) Lysosome Answer: a

10) Which part of the cell is the location for translation? a) Nucleus b) Mitochondria c) Ribosome d) Cytosol e) Lysosome Answer: c

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

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

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

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

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

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

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

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

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

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

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

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

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

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

25) Which of the following E. coli RNA polymerase subunits are responsible only for promoter specific transcription in vitro? a)  b)  c) ' d)  e)  Answer: d

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) 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 27) During a research event, the  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

28) A basepair 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

29) Which statement is correct 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 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

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

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

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

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

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

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

37) Which of the following RNA polymerases catalyze the initiation of transcription in most eukaryotic genes? a) RNA Polymerase 1 b) RNA Polymerase 2 c) RNA Polymerase 3 d) RNA Polymerase 4 e) RNA Polymerase 5 Answer: b

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

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

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

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

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

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

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

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

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

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.

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.

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.

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.

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.

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.

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

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

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

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

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

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

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

8) Amino acids in polypeptides are joined by what type of bond? a) Hydrogen b) Peptide c) Covalent d) Ionic e) Weak Answer: b

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

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

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

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

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

14) Which of the following is a secondary structure often observed in proteins? a)  helix b)  sheet c)  subunit d)  helix and  sheet e) All of these Answer: d

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  helices and  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

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

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

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

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

20) Which of the following statements is correct about RNA involvement in translation? 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

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

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

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 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

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

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

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

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 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

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

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

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

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

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

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

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

37) The genetic code is: a) Composed of triplet codons b) Non-overlapping c) Degenerate d) Universal e) All of these Answer: e

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

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

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

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

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

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

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

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

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

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

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.

49) Explain how chain termination occurs during translation, including the differences between chain 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.

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

Question Type: Multiple Choice

1) Living cells contain numerous enzymes that constantly scan DN, 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

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

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

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 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

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

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

8) Which cell type experiences germinal mutations? a) Squamous Epithelial b) Chondrocytes c) Gametes d) Cuboidal Epithelial

e) Columnar Epithelial Answer: c

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

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

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

12) Mutation is a random and non-adaptive process. This was first proven by: a) Lysenko and Lamarck, using the removal of mouse tails as an environmental stress. b) Joshua and Esther Lederberg, using the removal of mouse tails as an environmental stress. c) Lysenko and Lamarck, using replica plated E. coli and streptomycin as an environmental stress. d) Joshua and Esther Lederberg, using replica-plated E. coli and streptomycin as an environmental stress

e) Mutation is not a random, nonadaptive process.

Answer: d

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

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

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

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

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

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

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

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

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

22) 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

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

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

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

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

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

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

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

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

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

32) 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

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

34) 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

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

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

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

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

39) Which of the following molecules carries out single-strand assimilation? a) XPA b) MutD c) RecA d) RecBCD e) UvrD Answer: c

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

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

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

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

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

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

Question Type: Essay

46) 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.

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.

48) 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.

49) 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.

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

14) Why is it possible to join a piece of DNA from a human that has been digested using EcoR1 and a piece of DNA from a bacterium that has been digested using EcoR1? 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 can not 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

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

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

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

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

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

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

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

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

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

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

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

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

27) The intragenic complementation between the amino terminal fragment of -galactosidase and the defective M15 polypeptide which permits the Xgal color test to be utilized without placing the entire lacZ gene in the pUC vectors is called: a) Forced cloning b) -complementation c) Blue/white color selection d) In situ hybridization e) Complementation screening Answer: b

28) 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

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 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a) AATGTCAGCCTT b) TTCCGACTGTAA c) AAATTTTCCCGG d) GGCCCTTTTAAA e) TTTAAAAGGGCC Answer: a

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:

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.

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.

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.

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.

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.

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

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

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

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 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

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e) All of these Answer: e

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

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

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

8) Overlapping genomic clones found in physical chromosome maps are also known as:

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a) STSs b) Contigs c) ORI sites d) Polylinkers e) All of these Answer:b

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

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

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

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Answer: a

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

13) Within the euchromatic regions of human chromosomes one CentiMorgan is equivalent, on average, to what physical distance of DNA? a) 1mb b) 1kb c) 1mm d) 1cm e) 10mb Answer: a

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

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

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c) PCR d) FISH e) VNTR Answer: e

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

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

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

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

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3. A large amount of dispersed repetitive DNA a) 1 b) 2 c) 3 d) 1 and 2 e) 1 and 3 Answer: d

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

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

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

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Answer: d

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

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

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

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

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e) Francis Collins and James Watson Answer: d

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

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 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 individuals genetic makeup a) 1 b) 2 c) 3 d) 1 and 2 e) All of these Answer: e

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:

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a) NSPs b) VNTRs c) STRs d) SNPs e) ESTs Answer: d

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

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

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

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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

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

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

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

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d) Lactase and GFP e) Lactase and Kinase Answer: b

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

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

40) Homologous genes within a species are known as: a) Paralogues b) Orthologues c) Heterologues d) Similogues e) None of these Answer: a

41) Homologous genes within a species are known as: a) Paralogues b) Orthologues c) Heterologues

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d) Similogues e) None of these Answer: a

41) Homologous genes present within different species are known as: a) Paralogues b) Orthologues c) Heterologues d) Similogues e) None of these Answer: b

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

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

44) Which of the following is a bioinformatics tool? a) BLAST search b) GENBANK c) Microarray hybridization d) BLAST search and GENBANK

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e) All of these Answer: d

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

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.

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

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near the gene, and then “walking” or “jumping” along the chromosome until the gene is reached.

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.

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.

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

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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.

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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

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

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

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

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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

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

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

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

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Answer: a

9) Seventy percent of all Cystic Fibrosis cases are caused by which of the following mutations? a) F508 trinucleotide deletion b) F508 trinucleotide repeat c) CAG trinucleotide repeat d) CAG trinuclotide deletion e) None of these Answer: a

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

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

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

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Answer: c

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

14) When testing for the presence of the sickle cell anemia Hb allele, a Southern blot exhibits two small bands for the HbA allele and one band for the Hb 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

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

16) A gene that has been introduced into a cell or an organism is known as a: a) Polygene b) Cisgene c) Transgene

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d) Xenogene e) Autogene Answer: c

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

18) Which of the following statements true 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

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

20) Which of the following requirements must be fulfilled before a gene therapy procedure will

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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

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

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

23) Presently, somatic cell gene therapy for ADA-SCID presents challenges. What are those challenges? a) The therapy is too expensive

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b) The therapy is transient an 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

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

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

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

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

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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

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

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 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

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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

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

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

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.

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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

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

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

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

37) Which of the following was the first pharmaceutical protein produced by bacteria approved for human use? a) Insulin

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b) hGH c) Erythropoietin d) Rennin e) Interferon Answer: a

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

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

40) Which of the following is used to make cheese? a) Insulin b) hGH c) Erythropoietin d) Renning e) Interferon Answer: d

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.

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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

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

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

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

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

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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

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.

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.

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

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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.

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.

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.

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.

51) How is hGH produced for pharmaceutical use by using E. coli as a producer?

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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.

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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

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

3) Which enzyme catalyzes the excision and insertion events conducted by transposons? a) Transformase b) Transposase c) Ligase d) Helicase e) Polymerase Answer: b

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 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

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

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

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

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

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

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

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

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

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

34) What is the hobo element found in Drosophila? a) Ac/Ds element b) Replicative transposon c) Retrotransposon d) Tn3 e) IS Answer: a

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

36) Why do Drosophila, from a P male x 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 x M female crosses do not generate viable offspring.

Answer: c

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

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

39) Which of the following crosses would result in the movement of P elements and subsequent hybrid dysgenesis? a) P cytotype females x P cytotype males b) M cytotype females x M cytotype males c) P cytotype females x M cytotype males d) M cytotype females x P cytotype males e) Hybrid dysgenesis is not induced by cytotype.

Answer: d

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

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

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

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

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

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

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

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

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

49) What is the function of HeT-A and TART transposons in flies? a) Provide genetic variation by randomly hopping throughout the genome b) To counterbalance the loss of chromosomal ends c) To promote crossing-over between homologous chromosomes d) To aid in the segregation of chromosomes e) They have no known function.

Answer: b

Question Type: Essay

50) 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.

51) 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.

52) 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.

53) 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.

54) 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.

55) 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.

56) 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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 26) The enzyme -galactosidase is encoded by a) lacA. b) lacy. c) lacZ. d) lacI. e) both lacA and lacY. Answer: c 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 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

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

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

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

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

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

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

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

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

37) Which of the following is the best example of derepression? a) Switching on of the trp operon in the presence of tryptophan b) Switching off of the trp operon in the presence of tryptophan c) Switching off of the trp operon in the absence of tryptophan d) Switching off of the lac operon by the presence of glucose in the medium e) Switching on of the lac operon in the presence of lactose Answer: a

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

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 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 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 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 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 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 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

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

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

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

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

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

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

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.

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.

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).

55) Draw a sketch of a simplistic operon with 4 structural genes.

Answer: --Promoter--Operator---Gene 1—Gene 2—Gene 3—Gene 4

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.

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.

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

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

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

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

5) Where does RNA transcript modification occur in eukaryotic cells? a) Nucleus b) Mitochondria c) Rough ER d) Lysosome e) Ribosome Answer: a

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

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

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

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

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

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

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

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

14) Which of the following would not induce eukaryotic gene expression? a) Heat b) Light c) Hormones d) siRNAs e) Growth factors Answer: d

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

31) RISC-associated RNAs that result in mRNA cleavage are usually termed a) siRNAs. b) miRNAs. c) SNRPs. d) mtDNAs. e) STRs. Answer: a

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

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 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

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

Answer: a

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

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 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

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

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

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

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

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

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

45) Which of the following is a mechanism of dosage compensation? a) Inactivation b) Hyperactivation c) Hypoactivation d) All of these are correct. e) None of these are correct.

Answer: d

46) 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

Question Type: Essay

47) 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.

48) 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).

49) 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.

50) 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).

51) 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.

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) Chickens d) Drosophila and C. elegans e) All of these are correct.

Answer: e

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

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

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

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

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

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) None of these are correct.

Answer: a

8) Which of the following is true regarding the model organism C. elegans? a) C. elegans completes its life cycle in three days. b) C. elegans is a hermaphroditic species.

c) Self fertilization is possible. d) All of these are true. e) None of these are true.

Answer: d

9) Which of the following is not a reason why C. elegans is an excellent model system for studying development? a) The animal has a life cycle that is completed in three days. b) It is a hermaphrodite species. c) It is a transparent animal. d) It has invariant cell lineages that form tissue of the adult. e) XO mutants produced by meiotic disjunction are sterile.

Answer: e

10) Which statement is false with regard to development in C. elegans? a) Hermaphrodites have two X chromosomes and five autosomes. b) XO animals can be crossed with XX hermaphrodites. c) It is possible to trace the lineage of all of the cells in the adult from the zygote. d) All adult tissue arises from six founder cells. e) Self-incompatibility ensures that sperm cannot fertilize eggs from the same animal.

Answer: e 11) How many “founder” cells are produced by the series of asymmetric cell divisions in C. elegans? a) 1 b) 2 c) 4 d) 6 e) 8 Answer: d

12) 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

13) 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

14) What is encoded by the somatic sex-determination pathway in Drosophila? a) Proteins that regulate RNA splicing b) Signaling molecules c) Signaling receptors d) Signaling transcription factors e) All of these are correct.

Answer: a

15) What is encoded by the somatic sex determination pathway in C. elegans?

1. Proteins that regulate RNA splicing 2. Signaling molecules 3. Signaling receptors a) 1 b) 2 c) 3 d) 1 and 2 e) 2 and 3 Answer: e

16) Which of the following signals is used in the sex determination pathway in both Drosophila and C. elegans. a) Number of X chromosomes b) Number of Y chromosomes c) Ratio of X and Y chromosomes d) Ratio of X chromosomes to autosomes e) Number of autosomes Answer: d

17) The system to ascertain the X:A ratio involves interactions between which of the following? a) Maternally synthesized proteins that have been deposited in the egg cytoplasm, and embryonically synthesized proteins that are encoded by several X-linked genes b) Embryonically synthesized proteins that have been deposited in the egg cytoplasm and maternally synthesized proteins that are encoded by several X-linked genes c) Paternally synthesized proteins that have been deposited in the sperm cytoplasm and embryonically synthesized proteins that are encoded by several X-linked genes d) Embryonically synthesized proteins that have been deposited in the sperm cytoplasm and paternally synthesized proteins that are encoded by several Y-linked genes e) None of these are correct.

Answer: a

18) In Drosophila with two X chromosomes and three pairs of autosomes (genotype XX; AAA), how do flies develop? a) They become males. b) They become females. c) They become intersex flies. d) They do not develop as this is lethal. e) They become hermaphrodites.

Answer: c

19) Which of the following is the master regulator of the Drosophila sex-determination pathway? a) Numerator elements b) Denominator elements c) Transformer gene

d) Sex-lethal gene e) Doublesex gene Answer: d

20) Which statement does not apply to sex determination in Drosophila? a) Sxl gene is turned on early in development only in XX embryos. b) Sxl gene is never turned on in XY embryos. c) Sxl gene is the master regulator of sex determination in Drosophila. d) SXL protein is a regulator of splicing of Sxl transcripts. e) Sxl transcript can be differentially spliced.

Answer: b

21) What is the phenotype of homozygous, loss-of-function mutations in the Sxl gene in XY animals? a) They develop into normal males. b) They develop into males, but the mutation is lethal. c) They develop into normal females. d) They develop into females, but the mutation is lethal. e) The animals develop as intersexes.

Answer: a

22) Translation of the Sxl mRNA in XY embryos results in a) functional SXL protein. b) a polypeptide without regulatory functions. c) synthesis of functional transformer protein. d) repression of the genes required for male development. e) repression of the genes required for female development.

Answer: b

23) Which of the following genes is involved in sex-determination and in dosage compensation in Drosophila? a) Numerator elements b) Denominator elements c) Sxl gene

d) Transformer gene e) Doublesex gene Answer: c

24) The following mutations would result in phenotypic males except for loss-of-function mutations in a) the transformer gene. b) the transformer2 gene. c) Sxl in XY animals. d) Dsx in XX animals. e) Sxl in XX animals.

Answer: d

25) In C. elegans loss-of-function mutations in the two transformer genes tra-1 and tra-2 cause XX animals a) to develop as normal females. b) to develop as males. c) to develop as hermaphrodites. d) to develop as intersex individuals. e) None of these are correct.

Answer: b

26) 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

27) 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

28) 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 29) 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

30) 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

31) 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

32) 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

33) 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

34) 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

35) 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

36) 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

37) 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

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

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

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

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

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

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

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

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

Question Type: Essay

46) Why is C. elegans an excellent model system to study development?

Answer: C. elegans is a small, free-living soil nematode that can be reared easily on agar plates and has a short life cycle. Because it is hermaphroditic and there is no self-incompatibility, researchers can make recessive mutations homozygous very conveniently. XO males can be crossed with XX hermaphrodites to carry out standard genetic procedures such as mapping and complementation testing. It is transparent and each cell can be observed in the course of development. The events of development are essentially invariant and the invariant cell lineages

that form the tissues of the adult make it an excellent system for studying development.

47) Explain how the developmental signal controls sex determination in Drosophila?

Answer: Once the X:A ratio has been ascertained, it is converted into a molecular signal that controls the expression of the X-linked Sex-lethal (Sxl) gene, the master regulator of the sex-determination pathway (Figure 21.4). Early in development, this signal activates transcription of the Sxl gene from PE, the gene's “early” promoter, but only in XX embryos. The early transcripts from this promoter are processed and translated to produce functional Sex-lethal protein, denoted SXL. After only a few division cycles, transcription from the PE promoter is replaced by transcription from another promoter, PM, the so-called maintenance promoter of the Sxl gene. Curiously, transcription from the PM promoter is also initiated in XY embryos.

However, the transcripts from PM are correctly processed only if SXL protein is present.

Consequently, in XY embryos, where this protein has not yet been made, the Sxl transcripts are alternately spliced to include an exon with a stop codon, and when these alternately spliced transcripts are translated, they generate a short polypeptide without regulatory function. Thus, alternate splicing of the Sxl transcripts in XY embryos does not lead to the production of functional SXL protein, and in the absence of this protein, these embryos develop as males. In XX embryos, where SXL protein was initially made in response to the X:A signal, Sxl transcripts from the PM promoter are spliced to produce more SXL protein. This expression pattern is maintained because the SXL protein regulates the splicing of transcripts from the Sxl gene in such a way that they encode the full-length SXL protein. In XX embryos, this protein is therefore a positive regulator of its own synthesis—a curious feedback mechanism that maintains the expression of the SXL protein in XX embryos and prevents its expression in XY embryos. The SXL protein also regulates the splicing of transcripts from another gene in the sex-determination pathway, transformer (tra) (Figure 21.5) These transcripts can be processed in two different ways. In chromosomal males, where the SXL protein is absent, the splicing apparatus always leaves a stop codon in the second exon of the tra RNA. Thus, when spliced tra RNA is translated, it generates a truncated (and nonfunctional) polypeptide. In females, where the SXL protein is present, this premature stop codon is removed by alternate splicing in at least some of the transcripts. Thus, when they are translated, some functional transformer protein (denoted TRA) is produced. The SXL protein therefore allows the synthesis of functional TRA protein in XX embryos but not in XY embryos.

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.

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.

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.

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

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

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

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

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

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

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

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

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

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

11) Programmed cell death is called a) cell lysis b) apoptosis c) cell popping d) hemolysis e) None of these are correct.

Answer: b

12) Which enzyme family plays a crucial role in apoptosis? a) Caspases b) Lactases c) Maltases d) Kinases e) Polymerases Answer: a

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

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

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

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

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

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

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

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

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

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

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

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

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

26) Burkitts 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

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

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

29) Which of the following is not a tumor suppressor gene? a) p16 b) c-myc c) RB d) NF2 e) TP53 Answer: b

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 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 (Figure 22.2). 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.

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.

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?

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.

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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) X2 value.

Answer: a

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

36) Which statistic quantitatively summarizes a pattern of resemblance? a) Regression analysis b) Correlation coefficient c) Standard deviation d) Mean e) Mode Answer: b

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

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

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

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

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

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

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

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

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

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.

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:

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

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.

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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 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 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 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

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

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 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 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 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

40) What general effect 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

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

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

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

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

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

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

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

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

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

50) What type of selection is demonstrated by the following chart?

Answer: Directional Selection

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

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 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

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

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 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 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 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

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

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

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

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

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

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

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

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

36) A distinct basic structure in a phylogenetic tree is known as a) heterology. b) topography. c) topology. d) parsimony. e) homology.

Answer: c

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

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

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

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

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

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

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

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

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 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 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

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

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.

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.

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.

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.

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.