Test Bank for Genetic Analysis, 4th edition Mark F. Sanders John L. Bowman Chapters 1-22

Test Bank For Genetic Analysis, 4th Edition Mark F.Sanders John L. Bowman Chapter 1-22

Chapter 1 The Molecular Basis of Heredity, Variation, and Evolution 1.1 Multiple-Choice Questions 1) Bacteria are single-celled organisms with a single chromosome found in which cell space?

A) nucleus B) nucleolus C) nucleoid D) nuclear envelope E) mitochondria Answer: C Section: 1.1 Skill: Knowledge/Comprehension 2) Sexual reproduction uses meiosis to generate A) two identical haploid gametes.

B) two identical diploid gametes.

C) four identical haploid gametes.

D) four unique diploid gametes.

E) four unique haploid gametes.

Answer: E Section: 1.1 Skill: Application/Analysis 3) When a diploid cell divides by mitosis, the result is A) two identical haploid cells.

B) two identical diploid cells.

C) four identical haploid cells.

D) four unique diploid cells.

E) four unique haploid cells.

Answer: B Section: 1.1 Skill: Application/Analysis 4) Modern genetics consists of three major branches. Which of these branches, also known as "transmission genetics," involves the study of the transmission of traits and characteristics in successive generations?

A) evolutionary B) Mendelian C) molecular D) population E) reproductive

Answer: B Section: 1.1 Skill: Knowledge/Comprehension 5) You identify a new unicellular organism with multiple chromosomes organized by proteins within the cell's nucleus. Into which of the three domains of life might this organism fit?

A) Archaea B) Bacteria C) Eukarya D) Archaea or Bacteria E) Archaea or Eukarya Answer: C Section: 1.1 Skill: Application/Analysis 6) Watson and Crick used evidence from several studies to determine the structure of DNA. What conclusion were they able to draw from Rosalind Franklin's X-ray diffraction data, specifically?

A) DNA consists of four types of nucleotide bases: A, T, C, and G. B) DNA nucleotides form complementary base pairs.

C) Adenine pairs with thymine and cytosine pairs with guanine when they are on opposite DNA strands.

D) DNA is a duplex, with two strands forming a double helix.

E) The DNA strands are antiparallel, and the strands are held together by hydrogen bonds.

Answer: D Section: 1.2 Skill: Knowledge/Comprehension 7) What kind of bond is formed between the 5′ phosphate group of one nucleotide and the 3′ hydroxyl (OH) group of the adjacent nucleotide?

A) ionic bond B) phosphodiester bond C) hydrogen bond D) disulfide bond E) hydroxyl bond Answer: B Section: 1.3 Skill: Knowledge/Comprehension 8) What kind of bond is formed between complementary base pairs to join the two DNA strands into a double helix?

A) ionic bond B) phosphodiester bond C) hydrogen bond D) disulfide bond E) peptide bond Answer: C Section: 1.3 Skill: Knowledge/Comprehension

9) Which type(s) of RNA is/are translated into proteins?

A) mRNA B) rRNA C) tRNA D) mRNA and rRNA E) mRNA, rRNA, and tRNA Answer: A Section: 1.4 Skill: Knowledge/Comprehension 10) Which enzyme is active in DNA replication?

A) DNA polymerase B) RNA polymerase C) reverse transcriptase D) aminoacyl-tRNA synthetase E) RNase Answer: A Section: 1.4 Skill: Knowledge/Comprehension 11) The promoter is bound by the transcription factors and which enzyme?

A) DNA polymerase B) RNA polymerase C) reverse transcriptase D) DNase E) RNase Answer: B Section: 1.4 Skill: Knowledge/Comprehension 12) What is the process of synthesizing proteins from mRNA sequences?

A) replication B) transcription C) translation D) transformation E) transduction Answer: C Section: 1.4 Skill: Knowledge/Comprehension 13) What is the process of synthesizing single-stranded RNA from template DNA? A) replication B) transcription C) translation D) transformation E) transduction Answer: B Section: 1.4 Skill: Knowledge/Comprehension

14) What kind of bond is formed between successive amino acids during translation?

A) ionic bond B) phosphodiester bond C) hydrogen bond D) disulfide bond E) peptide bond Answer: E Section: 1.4 Skill: Knowledge/Comprehension 15) Which evolutionary process describes the movement of members of a species from one population to another?

A) natural selection B) migration C) mutation D) random genetic drift E) population genetics Answer: B Section: 1.5 Skill: Knowledge/Comprehension 16) Which evolutionary process is most pronounced in small populations where statistical fluctuations in allele frequencies can be significant from one generation to the next?

A) natural selection B) migration C) mutation D) random genetic drift E) population genetics Answer: D Section: 1.5 Skill: Application/Analysis 17) Which evolutionary process involves the slow addition of allelic variation that increases the hereditary diversity of populations, ultimately leading to evolutionary change?

A) natural selection B) migration C) mutation D) random genetic drift E) population genetics Answer: C Section: 1.5 Skill: Application/Analysis

18) Which evolutionary process relies on the premise that individuals with the best adaptations are most successful at reproducing and leave more offspring than those with less adaptive forms?

A) natural selection B) migration C) mutation D) random genetic drift E) population genetics Answer: A Section: 1.5 Skill: Application/Analysis 19) Which term describes a set of organisms that descended from a single common ancestor and are more closely related to other members of the group than to organisms outside the group?

A) paraphyletic group B) phylogeny C) species D) monophyletic group E) parsimony Answer: D Section: 1.5 Skill: Knowledge/Comprehension 20) Which concept describes the idea that the most closely related molecular sequences are those with the smallest number of differences between them?

A) cladistics B) phylogeny C) species D) monophyletic group E) parsimony Answer: E Section: 1.5 Skill: Knowledge/Comprehension 1.2 Short-Answer Questions 1) What are the three domains of life?

Answer: Bacteria, Archaea, and Eukarya Section: 1.1 Skill: Knowledge/Comprehension 2) With the assistance of William Bateson, Archibald Garrod produced the first documented example of a human hereditary disorder. Which disorder were they describing?

Answer: alkaptonuria Section: 1.1 Skill: Knowledge/Comprehension

3) After replication, chromosomes consist of how many molecules of DNA (assuming one chromatid is equal to one molecule of DNA)?

Answer: two Section: 1.2 Skill: Application/Analysis 4) The physical units of heredity composed of defined DNA sequences that collectively control gene transcription and contain the information to produce RNA molecules or proteins are better known as what?

Answer: genes Section: 1.2 Skill: Knowledge/Comprehension 5) A complete set of chromosomes is transmitted to produce identical daughter cells in which cell division process?

Answer: mitosis Section: 1.2 Skill: Application/Analysis 6) In eukaryotes, most of the cells' DNA is found in the form of chromosomes in the nucleus. Which organelles contain their own genomes (descended from ancient endosymbiotic bacteria)?

Answer: mitochondria and chloroplasts Section: 1.2 Skill: Application/Analysis 7) What are the three principal components of a DNA nucleotide?

Answer: sugar (deoxyribose), phosphate, and nitrogenous base (nucleotide) Section: 1.2 Skill: Knowledge/Comprehension 8) In double-stranded DNA, the percentages of adenine and thymine are approximately equal to each other, as are the percentages of cytosine and guanine. This relationship is better known as what (named for the person who made this discovery)?

Answer: Chargaff's rule Section: 1.3 Skill: Knowledge/Comprehension 9) The polarities of the complementary nucleic acid strands run in opposite directions 5′-to-3′ and 3′-to5′. The 5′ designation refers to the placement of what chemical group on the 5′ carbon of the deoxyribose sugar?

Answer: phosphate group Section: 1.3 Skill: Application/Analysis 10) During DNA replication, nascent DNA strands are synthesized in only one direction. Nucleotides are added only to which end of the nascent strand?

Answer: the 3′ hydroxyl end Section: 1.3 Skill: Knowledge/Comprehension

11) In retroviruses, DNA is synthesized from an RNA template using which enzyme?

Answer: reverse transcriptase Section: 1.4 Skill: Application/Analysis 12) Messenger RNA codons pair with tRNA anticodons at which cell structure?

Answer: the ribosome Section: 1.4 Skill: Knowledge/Comprehension 13) Peptidyl transferase and other proteins power the continuous progression of the ribosome along mRNA and catalyze what type of bond formation in the growing chain?

Answer: peptide bonds Section: 1.4 Skill: Knowledge/Comprehension 14) Only 61 of the 64 codons specify an amino acid. What do the other three codons encode?

Answer: stop codons Section: 1.4 Skill: Knowledge/Comprehension 15) What process proposed by Wallace and Darwin describes the higher rates of survival and reproduction of certain forms of a species over alternative forms?

Answer: natural selection Section: 1.5 Skill: Application/Analysis 16) As natural selection increases the frequency of one morphological form over another in the population, what changes at the genotypic level?

Answer: allele frequency Section: 1.5 Skill: Application/Analysis 17) Theodosius Dobzhansky and Ernst Mayr drew on ideas from Darwin, Fisher, Wright, Haldane, and others to demonstrate how evolution operates in real populations. These observations lead to which evolutionary genetics theory?

Answer: modern synthesis of evolution Section: 1.5 Skill: Knowledge/Comprehension 18) What type of diagram would you use to depict morphological or molecular similarities and differences that identify evolutionary relationships?

Answer: phylogenetic tree Section: 1.5 Skill: Knowledge/Comprehension

19) Both sugar gliders and flying squirrels have evolved characteristics that allow them to glide, despite being geographically separated. Similar traits that have independent origins arise as a result of what phenomenon?

Answer: convergent evolution Section: 1.5 Skill: Application/Analysis 20) Phylogenetic trees are constructed based on morphological characteristics, but molecular phylogenetic trees are constructed based on which feature?

Answer: nucleic or amino acid sequence Section: 1.5 Skill: Knowledge/Comprehension 1.3 Fill-in-the-Blank Questions 1) The work of Walter Sutton and Theodor Boveri suggested that the hereditary units, or genes, described by Mendel are located on ________.

Answer: chromosomes Section: 1.1 Skill: Knowledge/Comprehension 2) Genetic experiments have revealed the relationship between the observable traits of an organism, or ________, and the genetic constitution of an organism, or ________.

Answer: phenotype; genotype Section: 1.1 Skill: Knowledge/Comprehension 3) DNA replication is called ________ because the newly replicated DNA consist of a parental strand (from the original DNA) and a newly synthesized daughter strand.

Answer: semiconservative Section: 1.3 Skill: Knowledge/Comprehension 4) The ________, first proposed by Francis Crick, summarizes the relationships between DNA, RNA, and protein.

Answer: central dogma of biology Section: 1.3 Skill: Knowledge/Comprehension 5) There are 64 codons and 20 amino acids, meaning that the genetic code is ________.

Answer: redundant Section: 1.4 Skill: Knowledge/Comprehension

1.4 Essay Questions 1) What are the five essential characteristics of hereditary material, and why was protein originally believed to be a better candidate for the hereditary material than DNA? Answer: The hereditary material must be 1. a component of chromosomes.

2. present in a stable form in cells.

3. sufficiently complex to contain the genetic information that directs the structure, function, development, and reproduction of organisms.

4. able to accurately replicate itself so that daughter cells contain the same information as parental cells.

5. mutable, undergoing mutation at a low rate that introduces genetic variation and serves as a foundation for evolutionary change.

Proteins were thought to be a better candidate because they were found in high concentrations in the chromosomes, and the "20-letter" alphabet of amino acids was thought to have more potential for transmitting information than the "4-letter" alphabet of nucleotides of DNA. Section: 1.2 Skill: Synthesis/Evaluation 2) What are the three major types of RNA and their functions? What would happen to translation if each of these RNAs were mutated or degraded?

Answer: 1. Messenger RNA (or mRNA) is transcribed from the DNA template and translated into proteins.

2. Ribosomal RNA (rRNA) forms part of the ribosomes, the plentiful cellular structures where protein assembly takes place.

3. Transfer RNA (tRNA) carries amino acids, the building blocks of proteins, to ribosomes.

If any of these types of RNA were mutated, translation would not occur. Mutating mRNA would prevent translation of that particular gene. Mutating rRNA or tRNA would prevent translation of any mRNAs because the ribosome would not form properly, and the transfer RNA would not bring the correct amino acid to the growing polypeptide chain.

Section: 1.2 Skill: Synthesis/Evaluation 3) DNA strands can be pulled apart by adding heat and "melting" the double-stranded DNA. The temperature required to melt a region of DNA changes based on the base-pair composition. Based on the structure of the A-T and C-G bonds in Figure 1.11, which bonds would require more energy (heat) to break them? How might this help you predict which regions of the DNA helix may be the most stable and harder to break apart? (This can be used to predict regions that are not transcribed as often, as you'll see in later chapters.)

Answer: C-G bonds contain three hydrogen bonds, whereas A-T bonds have only two hydrogen bonds.

The more hydrogen bonds in a particular region of DNA, the more energy required to break those bonds apart. Thus, regions of DNA with large numbers of C and G residues will be more heat resistant (and probably transcribed less often) than A-T rich regions.

Section: 1.3 Skill: Synthesis/Evaluation

4) Describe what is meant by adaptive and nonadaptive evolution. Which type of evolution might be represented by the presence of the sickle cell phenotype in regions with high rates of malaria? By the presence of both attached and detached earlobes in a given population?

Answer: Adaptive evolution implies that one form reproduces in greater numbers than others in a population because of being better adapted to the conditions driving natural selection. Sickle cell genotype is an example of adaptive evolution. Nonadaptive evolution describes the evolution of characteristics that are reproductively equivalent to other forms in the population. Nonadaptive traits are neutral with respect to natural selection, conferring neither a selective advantage nor a selective disadvantage to their bearer (e.g., attached versus detached earlobes).

Section: 1.5 Skill: Synthesis/Evaluation Organisms are constantly evolving and phylogenetic relationships can be reevaluated. Refer to the phylogenetic tree to answer the following questions.

5) Which groups are more closely related: lancelets and tunicates, or lancelets and hagfishes?

Answer: Lancelets and tunicates, because there are fewer branches separating them on the phylogenetic tree.

Section: 1.5 Skill: Synthesis/Evaluation 6) What feature distinguishes snakes and mammals from frogs and salamanders?

Answer: Snakes and mammals both have an amniotic egg, whereas amphibians do not.

Section: 1.5 Skill: Synthesis/Evaluation

7) How might you explain the fact that the common ancestor of snakes had legs?

Answer: Snakes had legs and evolved a morphological structure that minimized their legs to the point where they were eventually "lost." So, just as traits can be gained by evolution, they can be lost if there is an evolutionary advantage to that change.

Section: 1.5 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 2 Transmission Genetics 2.1 Multiple-Choice Questions 1) Mendel performed many types of crosses, including those in which the same genotypes are crossed but the sexes of the parents are switched. These are known as A) replicate crosses.

B) reciprocal crosses.

C) test crosses.

D) monohybrid crosses.

E) dihybrid crosses.

Answer: B Section: 2.1 Skill: Knowledge/Comprehension 2) Crosses in which F1 plants heterozygous for a given allele are crossed to generate a 3:1 phenotypic ratio in the F2 generation are known as A) replicate crosses.

B) reciprocal crosses.

C) test crosses.

D) monohybrid crosses.

E) dihybrid crosses.

Answer: D Section: 2.2 Skill: Knowledge/Comprehension 3) In peas, the round allele is dominant over the wrinkled allele. If a plant with round peas is crossed to a plant with wrinkled peas, all of the resulting plants have round peas. What is the genotype of the parents in this cross?

A) RR × rr B) RR × Rr C) Rr × rr D) Rr × Rr E) rr × rr Answer: A Section: 2.2 Skill: Application/Analysis

4) In peas, the yellow allele is dominant over the green allele. If a plant with yellow peas is crossed to a plant with green peas, the resulting plants are 50% yellow and 50% green. What is the genotype of the parents in this cross?

A) YY × yy B) YY × Yy C) Yy × yy D) Yy × Yy E) yy × yy Answer: C Section: 2.2 Skill: Application/Analysis 5) Assuming independent assortment, what phenotypic ratio would you expect to see if an individual with the genotype RrGg is self-crossed?

A) 1:3 B) 9:3:3:1 C) 1:2:1 D) 1:3:2:1 E) 3:1 Answer: B Section: 2.3 Skill: Application/Analysis 6) What genotypic ratio would you expect to see among the progeny of a monohybrid cross?

A) 1:3 B) 9:3:3:1 C) 1:2:1 D) 1:3:2:1 E) 3:1 Answer: C Section: 2.2 Skill: Application/Analysis 7) A couple has four children. What is the probability that they would have four boys?

A) 1/2 B) 1/4 C) 1/8 D) 1/16 E) 1/32 Answer: D Section: 2.4 Skill: Application/Analysis

8) Humans have a gene, "T," that is involved in muscle formation of the tongue. Individuals with the one allele can roll their tongues, while individuals with the other allele cannot. If both parents can roll their tongues, but their child cannot, what can be said about the mode of inheritance?

A) Tongue rolling is dominant.

B) Tongue rolling is recessive.

C) The parents were both homozygous, but the child was heterozygous.

D) Tongue rolling is dominant, and both parents were heterozygous (Tt).

E) Tongue rolling is recessive, and both parents were heterozygous (Tt).

Answer: D Section: 2.2 Skill: Synthesis/Evaluation 9) In peas, axial (A) flower position is dominant to terminal (a), tall (L) is dominant to short (l), and yellow (Y) is dominant to green (y). If a plant that is heterozygous for all three traits is allowed to selffertilize, how many of the offspring would be dominant for all three traits?

A) 3/64 B) 9/64 C) 27/64 D) 32/64 E) 64/64 Answer: C Section: 2.3 Skill: Application/Analysis 10) In peas, axial (A) flower position is dominant to terminal (a), and tall (L) is dominant to short (l). If a plant that is heterozygous for both traits is allowed to self-fertilize, how many of the offspring would also be heterozygous for both traits?

A) 9/16 B) 4/16 C) 3/16 D) 2/16 E) 1/16 Answer: B Section: 2.3 Skill: Application/Analysis 11) The law of segregation states that the F2 progeny of F1 heterozygous plants will exhibit A) a 3:1 phenotypic ratio.

B) a 9:3:3:1 phenotypic ratio.

C) a 1:2:1 genotypic ratio.

D) a 9:3:3:1 phenotypic ratio and 1:2:1 genotypic ratio.

E) both a 3:1 phenotypic ratio and a 1:2:1 genotypic ratio.

Answer: E Section: 2.3 Skill: Knowledge/Comprehension

12) The law of independent assortment would predict that the F2 progeny of F1 heterozygous plants will exhibit a A) 3:1 phenotypic ratio.

B) 9:3:3:1 phenotypic ratio.

C) 1:2:1 genotypic ratio.

D) 9:3:3:1 phenotypic ratio and 1:2:1 genotypic ratio.

E) 3:1 phenotypic ratio and 1:2:1 genotypic ratio.

Answer: B Section: 2.3 Skill: Knowledge/Comprehension 13) What phenotypic ratio would you expect as a result of a test cross between a dihybrid organism and one that is homozygous recessive for alleles at two independent loci?

A) 3:1 B) 1:2:1 C) 1:1:1:1 D) 9:3:3:1 E) 9:4:2:1 Answer: C Section: 2.3 Skill: Knowledge/Comprehension 14) How many possible gametes can be produced by a short plant with yellow, round peas with a heterozygous genotype (YyRrSs)?

A) 3 B) 6 C) 8 D) 10 E) 12 Answer: C Section: 2.3 Skill: Application/Analysis 15) By convention, when the difference between the observed experimental outcomes and the expected outcome is less than 5 percent (< 0.05), the experimental results are considered to be A) within normal expected range.

B) statistically significantly and different from the expected outcome.

C) not significant.

D) less than one standard deviation from the mean.

E) equal to the mean.

Answer: B Section: 2.4 Skill: Knowledge/Comprehension

16) The statistical interpretation of a chi-square value is determined by identifying the A) mean.

B) degrees of freedom.

C) average.

D) P value.

E) joint probability.

Answer: D Section: 2.5 Skill: Knowledge/Comprehension 17) The P value is a quantitative expression of the probability that the results of another experiment of the same size and structure will deviate as much or more from expected results by chance. The greater the difference between observed and expected results of an experiment, A) the lower the χ2 value and the lower the P value.

B) the greater the χ2 value and the greater the P value.

C) the greater the χ2 value and the lower the P value.

D) the lower the χ2 value and the greater the P value.

E) the greater the χ2 value; but the P value is unaffected.

Answer: C Section: 2.5 Skill: Knowledge/Comprehension 18) The statistical interpretation of a chi-square value is determined by identifying the P value for each experiment, and the P value is dependent on the number of degrees of freedom (df) in the experiment being examined. For the coin flip experiment, the df value is equal to 1. What would be the df value for the example about rolling a 1 in a single roll of a die?

A) 1 B) 2 C) 3 D) 4 E) 5 Answer: E Section: 2.5 Skill: Application/Analysis 19) The genes responsible for some of the traits that Mendel observed have been recently identified and have helped in determining how molecular variation produces morphologic variation in pea plants.

Allelic variation in the Sbe1 gene, which produces starch-branching enzyme 1, is responsible for which trait in peas?

A) round and wrinkled pea shape B) yellow and green pea color C) purple and white flowers D) tall and short plant height E) axial and terminal flower position Answer: A Section: 2.6 Skill: Knowledge/Comprehension

20) In 1997, a gene called Le was discovered by two research groups led by David Martin and Diane Lester. Allelic variation in the Le gene, which controls elongation of the plant stem between branches, is responsible for which trait in peas?

A) inflated and constricted pod shape B) yellow and green pod color C) purple and white flowers D) tall and short plant height E) axial and terminal flower position Answer: D Section: 2.6 Skill: Knowledge/Comprehension 2.2 Short-Answer Questions 1) Mendel performed numerous controlled genetic crosses to obtain strains that consistently produced a single phenotype without variation. What are these strains that consistently produce the same phenotype called?

Answer: pure-breeding or true-breeding strains Section: 2.1 Skill: Knowledge/Comprehension 2) In a test cross, a pure-breeding plant is crossed with a plant suspected to be heterozygous (Aa). What is the genotype of the pure-breeding plant?

Answer: aa Section: 2.1 Skill: Application/Analysis 3) Why did Mendel cut off the nascent anthers during the process of artificial cross-fertilization?

Answer: to prevent self-fertilization or to prevent uncontrolled crosses Section: 2.2 Skill: Knowledge/Comprehension 4) What simple type of cross that looks at the inheritance of only one trait could be used to illustrate Mendel's law of segregation?

Answer: monohybrid cross Section: 2.2 Skill: Knowledge/Comprehension 5) If an affected individual is born to parents who are unaffected, what is the likely mode of inheritance?

Answer: autosomal recessive Section: 2.2 Skill: Knowledge/Comprehension 6) What type of cross would be used to illustrate Mendel's law of independent assortment?

Answer: dihybrid cross (or test cross) Section: 2.2 Skill: Knowledge/Comprehension

7) A cross between a short pea plant and a tall pea plant results in a 1:1 genotypic and phenotypic ratio in the offspring. What are the genotypes of the parent plants?

Answer: Ss × ss (heterozygous × homozygous recessive) Section: 2.2 Skill: Application/Analysis 8) What is the probability of rolling one six-sided die and obtaining a 1 or a 2?

Answer: 1/6 + 1/6 = 2/6 = 1/3 Section: 2.4 Skill: Application/Analysis 9) What is the probability of rolling one six-sided die and obtaining any number but 6?

Answer: 1 - 1/6 = 5/6 Section: 2.4 Skill: Application/Analysis 10) What is the probability of rolling two six-sided dice and obtaining two 4s?

Answer: 1/6 × 1/6 = 1/36 Section: 2.4 Skill: Application/Analysis 11) What is the probability of rolling two six-sided dice and obtaining at least one 3?

Answer: (1/6) + (5/6)(1/6) = 6/36 + 5/36 = 11/36 Section: 2.4 Skill: Application/Analysis 12) What is the probability of rolling two six-sided dice and obtaining an odd number on at least one die?

Answer: 9/36 + 9/36 + 9/36 = 27/36 = 3/4 Prob. of rolling odd number the 1st roll only = 3/6 (odd) × 3/6 (even) = 9/36 Prob. of rolling odd number the 2nd roll only = 3/6 (even) × 3/6 (odd) = 9/36 Prob. of rolling odd number both rolls = 3/6 (odd) × 3/6 (odd) = 9/36 Prob. of any one of these three possible scenarios = addition rule Section: 2.4 Skill: Application/Analysis

For the following questions, determine which of the four rules of probability theory best fit the situation.

13) When calculating the probability of a given genotype in a trihybrid cross, you can generate a Punnett square. Which of the rules of probability can be used to calculate the joint probability of simultaneous inheritance of multiple alleles?

Answer: the product rule Section: 2.4 Skill: Synthesis/Evaluation 14) In a cross between individuals who are both heterozygous (carriers) for a recessive disease such as albinism, you would like to determine the risk of one or more children to inherit the recessive phenotype. Which of the rules of probability can be used to calculate the probability of a particular combination of events that each have two alternative outcomes?

Answer: binomial probability Section: 2.4 Skill: Synthesis/Evaluation 15) You have self-fertilized a plant with round seed that is heterozygous, and you want to determine what proportion of the offspring will be not only dominant, but true-breeding. Which of the rules of probability can be used to calculate the probability of obtaining a particular outcome when specific information about that outcome modifies the probability calculation?

Answer: conditional probability Section: 2.4 Skill: Synthesis/Evaluation 16) In a dihybrid cross, you want to calculate the probability that an F2 progeny of the cross will inherit both dominant phenotypes. Which of the rules of probability can be used to calculate the probability of obtaining that combination of alleles?

Answer: the product rule Section: 2.4 Skill: Synthesis/Evaluation 17) The statistical value obtained from the chi-square analysis refers to the probability that the deviations of between the observed numbers and the expected numbers are caused by what?

Answer: random chance Section: 2.5 Skill: Knowledge/Comprehension 18) A normal distribution curve contains all the possible experimental outcomes in graph form. The tall central segment of the curve represents the outcomes with the highest probability of occurrence. The average outcome, represented by the center of the data distribution, is known as what?

Answer: the mean (µ) Section: 2.5 Skill: Knowledge/Comprehension 19) The probability of experimental outcomes gets smaller toward the farthest left and right portions of the curve. The probability of a given observation compared to all possible outcomes is known as what?

Answer: the standard deviation (σ) Section: 2.5

Skill: Knowledge/Comprehension 20) Geneticists must be able to compare the outcomes they obtain in their experiments to the outcomes that might be expected to occur. Which test would they use to confirm that the difference between observed and expected can be attributed to chance?

Answer: chi-square test Section: 2.5 Skill: Knowledge/Comprehension 2.3 Fill-in-the-Blank Questions 1) One key to Mendel's success was choosing to observe ________ traits, which exhibit one of two possible phenotypes.

Answer: dichotomous Section: 2.1 Skill: Knowledge/Comprehension 2) The hereditary particles that are passed from one generation are called alleles in modern terminology.

This term had not been invented in Mendel's time; instead, he determined that two "________" (alleles) were present for each trait in a plant and together determined the phenotype of the trait.

Answer: elementen Section: 2.2 Skill: Knowledge/Comprehension 3) A ratio of 9:3:3:1 is expected among the F2 progeny of a dihybrid cross as a result of ________ of alleles at two loci.

Answer: independent assortment Section: 2.3 Skill: Knowledge/Comprehension 4) Binomial expansion is a complex genetic calculation requiring repetition and precision in the use of the product rule and the sum rule. A shortcut called ________ eliminates the need for these repetitive calculations and can be used for any number of expansions between 0 and the nth power to yield the size of each possible class and the total number of classes possible.

Answer: Pascal's triangle Section: 2.4 Skill: Knowledge/Comprehension 5) The P value is dependent on the number of ________, which is equal to the number of independent variables in an experiment.

Answer: degrees of freedom (df) Section: 2.5 Skill: Knowledge/Comprehension

2.4 Essay Questions 1) How did the study of physics with Professors Doppler and Ettinghausen influence Mendel's understanding of genetics?

Answer: Doppler, an experimental physicist famous for the Doppler effect, espoused a "particulate" view of physics and taught Mendel how to separate individual characteristics from one another in experiments. Professor Ettinghausen taught Mendel the mathematics of combinatorial analysis. Mendel would apply each of these lessons to his later research. Mendel's superior insight came principally from his familiarity with quantitative thinking and his understanding of the particulate nature of matter learned through the study of physics with Doppler. Central to Mendel's experimental success was counting the number of progeny with specific phenotypes. This logical and now routine component of data gathering was the key to Mendel's ability to formulate the hypotheses that explained his results.

Under Doppler and Ettinghausen, Mendel had learned to study individual properties of matter separately and to think in quantitative terms about combinations of outcomes.

Section: 2.1 Skill: Synthesis/Evaluation 2) Describe the traits that make Pisum sativum an ideal organism for genetic studies. Why did Mendel ultimately decide not to include exterior seed coat color (gray vs. white) as one of the traits he analyzed?

Answer: There are many varieties of peas with distinct heritable features in the form of dichotomous phenotypes that can be easily observed and quantified. In addition, mating of plants can be closely controlled. Since each pea plant has both sperm-producing (stamens) and egg-producing (carpels) organs, they can be self-crossed to generate true-breeding plants. After creating these true-breeding plants, Mendel could test for dominant or recessive inheritance patterns by cross-pollination (fertilization between different plants). Mendel initially selected an eighth trait producing either gray or white exterior seed coats. Early in his analysis, he saw that plants with purple flowers always had gray seed coats and that those with white flowers always had white seed coats. He speculated that flower color and seed-coat color were determined by the same genetic mechanism, and he was correct. (The pigment anthocyanin is produced by plants that have purple flower color and gray seed coats, but a mutation eliminates anthocyanin production in plants with white flowers and white seed coats.)

Section: 2.1 Skill: Synthesis/Evaluation 3) Describe the blending theory of heredity and how Mendel's results help to reject this theory.

Answer: The blending theory viewed the traits of progeny as a mixture of the characteristics possessed by the two parental forms. Under this theory, progeny were believed to display characteristics that were approximately intermediate between those of the parents. Mendel reasoned that if the blending theory were true, he would see evidence of it in each trait. If no blending were seen in individual traits, the blending theory would be disproved. F1 experimental results reject the blending theory of heredity because all F1 progeny have the same phenotype (i.e., the dominant phenotype) that is indistinguishable from the phenotype of one of the pure-breeding parents. This specifically contradicts the blending theory prediction that the F1 would display a mixture of the parental phenotypes. The persistence of the dominant phenotype and the reemergence of the recessive phenotype in the F2 also contradict the blending theory.

Section: 2.2 Skill: Synthesis/Evaluation

4) What are Mendel's first and second laws, and what do they state?

Answer: First Law: Law of Segregation The two alleles for each trait will separate from one another during gamete formation, and each allele will have an equal probability (1/2) of inclusion in a gamete. Random union of gametes at fertilization will unite one gamete from each parent to produce progeny in ratios that are determined by chance.

Second Law: Law of Independent Assortment During gamete formation, the segregation of alleles at one locus is independent of the segregation of alleles at another locus.

Section: 2.2 Skill: Synthesis/Evaluation 5) In Guinea pigs, short hair (S) is dominant over long hair (s), rough coat (R) is dominant over smooth coat (r), and black hair (B) is dominant over white hair (b). List the possible gametes that can be produced by each of the individuals below.

Answer: SSRRbb: SRb ssRrBB: sRB, srB SsRrbb: SRb, Srb, sRb, srb SsRrBb: SRB, SRb, SrB, Srb, sRB, sRb, srB, srb Section: 2.3 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 3 Cell Division and Chromosome Heredity 3.1 Multiple-Choice Questions 1) Mitosis is a process of cell division that results in A) two unique daughter cells.

B) four unique daughter cells.

C) two identical daughter cells.

D) three identical daughter cells.

E) four identical daughter cells.

Answer: C Section: 3.1 Skill: Knowledge/Comprehension 2) During which stage or phase of the cell cycle does the cell actively transcribe and translate all the protein products necessary for normal cellular structure and function?

A) M phase B) G1 C) G2 D) S E) G0 Answer: B Section: 3.1 Skill: Knowledge/Comprehension

3) During which stage or phase of the cell cycle does the cell replicate its chromosomes?

A) M phase B) G1 C) G2 D) S E) G0 Answer: D Section: 3.1 Skill: Knowledge/Comprehension 4) After which stage or phase of the cell cycle does the cell divide?

A) M phase B) G1 C) G2 D) S E) G0 Answer: A Section: 3.1 Skill: Knowledge/Comprehension

5) Certain kinds of cells (e.g., some cells in the eyes and bones) mature and differentiate into a state in which they have a specialized function but do not divide or progress through the cell cycle. These cells are "stuck" in which stage?

A) M phase B) G1 C) G2 D) S E) G0 Answer: E Section: 3.1 Skill: Application/Analysis 6) During cell division, chromosomal material is improperly divided between the two daughter cells such that one cell receives three copies of a chromosome, and the other cell receives only one. This is likely due to a defect in which process?

A) cytokinesis B) karyokinesis C) crossing over D) synapsis E) homologous recombination Answer: B Section: 3.1 Skill: Application/Analysis 7) Which organelle migrates during M phase to form the two opposite poles of the dividing cell and acts as the source of the spindle fiber microtubules?

A) centromere B) kinetochore C) centrosome D) chiasmata E) chromosome Answer: C Section: 3.1 Skill: Knowledge/Comprehension 8) Which structure is responsible for chromosome movement during cell division?

A) synaptonemal complex B) kinetochore microtubules C) metaphase plate D) nuclear envelope E) cell membrane Answer: B Section: 3.1 Skill: Knowledge/Comprehension

9) The separation of sister chromatids during anaphase I of mitosis is known as chromosome A) cytokinesis.

B) karyokinesis.

C) crossing over.

D) synapsis.

E) disjunction.

Answer: E Section: 3.1 Skill: Knowledge/Comprehension 10) How do protein kinases generally activate their target proteins?

A) The process of phosphorylation causes the target to change conformation.

B) The kinases bind to target proteins directly, forming a complex that activates the protein and lets them change conformations.

C) Kinases vary their concentrations through the cell cycle, so that proteins are activated only when kinase levels are high.

D) Kinases bind to target proteins and inactivate them, so kinase levels must drop for target protein activation.

E) The process of dephosphorylation causes the target to change conformations.

Answer: A Section: 3.1 Skill: Knowledge/Comprehension 11) Tumor suppressors are genes that A) promote advancement of the cell cycle.

B) phosphorylate proteins involved in the cell cycle.

C) block progression of the cell cycle.

D) prevent apoptosis in normal cells.

E) activate proto-oncogenes.

Answer: C Section: 3.1 Skill: Knowledge/Comprehension 12) Vinblastine is a commonly used chemotherapy drug that acts by interfering with the assembly of microtubules. How does it likely target cancer cells?

A) by suppressing cyclins B) by inhibiting protein kinases C) by inhibiting DNA synthesis D) by denaturing myosin and preventing the cleavage furrow from forming E) by disrupting mitotic spindle formation Answer: E Section: 3.1 Skill: Application/Analysis

13) If you mutate the cyclin-binding site on a Cdk such that Cdk cannot bind to cyclin, what effect would you expect to see?

A) Cdk will be unaffected.

B) The cell cycle will be up-regulated, and cells will proliferate.

C) Cdk will be inactive, so the target proteins will not be phosphorylated.

D) Cdk will be constitutively active, so the cell cycle will be unchecked.

E) Cdk will phosphorylate its target proteins regardless of the cell cycle.

Answer: C Section: 3.1 Skill: Synthesis/Evaluation 14) Mutations that inhibit Rb gene expression will have what effect on cell proliferation?

A) Cells will proliferate less, even if excess growth factors are present.

B) Cells will proliferate less because growth factors are absent.

C) Cells will proliferate more because growth factors are produced in excess.

D) Cells will proliferate more, regardless of growth factor concentration.

E) There will be no effect on cell proliferation; only apoptosis occurs.

Answer: D Section: 3.1 Skill: Application/Analysis 15) During meiosis I, when does homologous chromosome pairing and recombination occur?

A) prophase I B) pro-metaphase I C) metaphase I D) anaphase I E) telophase I Answer: A Section: 3.2 Skill: Knowledge/Comprehension 16) The binding of nonsister chromatids by a synaptonemal complex draws the homologs into close contact so that what process can occur?

A) cytokinesis B) karyokinesis C) independent assortment D) synapsis E) disjunction Answer: D Section: 3.2 Skill: Knowledge/Comprehension

17) Contact points between nonsister chromatids that mark the locations of DNA-strand exchange are called A) synaptonemal complex.

B) metaphase plate.

C) chiasmata.

D) kinetochore.

E) centrosome.

Answer: C Section: 3.2 Skill: Knowledge/Comprehension 18) What insight did Nettie Stevens's studies of the beetle Tenebrio molitor provide about the chromosome theory of inheritance?

A) Genes are located on chromosomes.

B) The number of autosomal chromosomes varied between gametes.

C) Crossing over occurs between homologous chromosomes.

D) Sex is determined by the presence of an X or Y chromosome in the male gamete.

E) Chromosomes assort independently into the haploid gametes.

Answer: D Section: 3.3 Skill: Knowledge/Comprehension 19) Morgan's analysis of a white-eyed male Drosophila mutation helped explain A) autosomal dominance.

B) random mutation.

C) crossing over.

D) independent assortment.

E) X-linked inheritance.

Answer: E Section: 3.3 Skill: Knowledge/Comprehension 20) Bridges showed that sex determination in Drosophila meant that A) females were XX or XXX, and males were XY or XXY. B) females were XX, and males were XY. C) females were XX, and males were XY or XO. D) females were XX or XXX, and males were XY or XO. E) females were XX or XO, and males were XY or XXY. Answer: D Section: 3.4 Skill: Knowledge/Comprehension

21) If a trait is X-linked recessive, who would express the trait?

A) homozygous dominant females and hemizygous recessive males B) heterozygous recessive females and hemizygous dominant males C) homozygous recessive females and hemizygous recessive males D) heterozygous dominant females and hemizygous dominant males E) the same proportions of females and males Answer: C Section: 3.5 Skill: Application/Analysis 22) If a trait is X-linked dominant, who would express the trait?

A) homozygous dominant females and hemizygous recessive males B) heterozygous recessive females and hemizygous dominant males C) homozygous recessive females and hemizygous recessive males D) heterozygous dominant females and hemizygous dominant males E) significantly more females than males Answer: D Section: 3.5 Skill: Application/Analysis 23) What dosage compensation mechanism is employed by female placental mammals?

A) synteny B) Y-inactivation C) X-inactivation D) X nondisjunction E) X chromosome crossing over Answer: C Section: 3.6 Skill: Knowledge/Comprehension 3.2 Short-Answer Questions 1) How many chromosomes are found in a human somatic cell nucleus?

Answer: 46 Section: 3.1 Skill: Knowledge/Comprehension 2) How many homologous pairs of chromosomes are found in a human somatic cell nucleus?

Answer: 23 Section: 3.1 Skill: Application/Analysis 3) How many sister chromatids are found in a human somatic cell nucleus during mitosis?

Answer: 92 Section: 3.1 Skill: Application/Analysis

4) What process is used to make mature germ-line cells, or gametes?

Answer: meiosis Section: 3.1 Skill: Knowledge/Comprehension 5) What is the short segment of the cell cycle during which cells divide?

Answer: the M phase Section: 3.1 Skill: Knowledge/Comprehension 6) If a cell begins to divide rapidly (as often seen in wound repair), what would you expect to happen to the G1 stage?

Answer: It would shorten.

Section: 3.1 Skill: Application/Analysis 7) At the end of the G1 stage, cells have the option of entering which two phases?

Answer: S phase or G0 Section: 3.1 Skill: Knowledge/Comprehension 8) The five substages of M phase are designed to accomplish two main goals equal partitioning of chromosomal material and cytoplasmic contents between the two cells. What are these two processes called?

Answer: karyokinesis and cytokinesis Section: 3.1 Skill: Knowledge/Comprehension 9) Microtubules are polar and have a plus and a minus end. Which end is anchored at the centrosome?

Answer: the minus end Section: 3.1 Skill: Knowledge/Comprehension 10) Why are chromosomes more easily visible under the microscope during metaphase as compared to prophase?

Answer: They condensed 10,000-fold.

Section: 3.1 Skill: Application/Analysis 11) Protein kinases help phosphorylate target proteins, but they are present in relatively constant concentrations throughout the cell cycle. Which proteins, which vary through the cell cycle, form a complex with the kinases to activate them only at specific times?

Answer: cyclins Section: 3.1 Skill: Knowledge/Comprehension

12) When cyclin D1 is expressed in normal levels, it stimulates the cell cycle. If overexpressed, it can promote cancer by allowing cells to overproliferate. A normal gene with the potential to cause cancer if mutated and/or overexpressed is known as what?

Answer: proto-oncogene Section: 3.1 Skill: Knowledge/Comprehension 13) Single-celled eukaryotes can survive and reproduce in both haploid and diploid forms. Yeast spend most of their life cycle and actively reproduce in which form?

Answer: haploid Section: 3.2 Skill: Knowledge/Comprehension 14) In meiosis I, the diploid number (2n) of chromosomes is reduced to the haploid number (n) because of the separation of what?

Answer: homologous chromosomes Section: 3.2 Skill: Knowledge/Comprehension 15) In meiosis II, sister chromatids separate and produce what?

Answer: four haploid gametes Section: 3.2 Skill: Knowledge/Comprehension 16) The gametes of the two sexes are often dramatically different in size and morphology. Both gametes have a haploid nucleus, so what do female gametes contain more of?

Answer: cytoplasm and organelles Section: 3.2 Skill: Application/Analysis 17) The number of recombination nodules along each homologous chromosome arm correlates closely with the average number of what?

Answer: crossover events Section: 3.2 Skill: Knowledge/Comprehension 18) This protein assures that sister chromatids of each chromosome remain firmly joined and can resist the pull of the kinetochore microtubules.

Answer: cohesin Section: 3.2 Skill: Knowledge/Comprehension 19) Females have two copies of the X chromosome, so they can be homozygous or heterozygous for any genes found on the X chromosome. Males have only one copy of the X chromosome; what term is used to describe this inheritance state?

Answer: hemizygous Section: 3.3 Skill: Knowledge/Comprehension

20) According to Bridges's proposal, exceptional phenotypes and unusual karyotypes were the result of rare mistakes in meiosis caused by the failure of X chromosomes to separate properly in either the first or second meiotic division in females. What term is used in referring to this phenomenon?

Answer: nondisjunction Section: 3.3 Skill: Knowledge/Comprehension 21) What is the mammalian transcription factor that elicits a cascade of gene transcription and developmental events that ultimately produce male internal and external structures?

Answer: SRY Section: 3.4 Skill: Knowledge/Comprehension 22) Any mechanism that compensates for differences in the number of copies of genes due to the different chromosome constitutions of males and females is called what?

Answer: dosage compensation Section: 3.6 Skill: Knowledge/Comprehension 3.3 Fill-in-the-Blank Questions 1) Of the trillions of cells in the human body, most are ________, the cells that form organs and tissues.

Answer: somatic cells Section: 3.1 Skill: Knowledge/Comprehension 2) The ________ varies among species (each species has its characteristic number of pairs) and is identified nonspecifically as 2n.

Answer: diploid number Section: 3.1 Skill: Knowledge/Comprehension 3) The ________ number, or n, is the number of chromosomes contained in the nuclei of gametes, the nonsomatic cells.

Answer: haploid Section: 3.1 Skill: Knowledge/Comprehension 4) In meiosis, interphase is followed by two successive rounds of cell division, meiosis I and II. Meiosis results in four haploid daughter cells because there is no ________ between these two meiotic cell divisions.

Answer: DNA replication Section: 3.2 Skill: Knowledge/Comprehension

5) Synapsis initiates formation of a protein bridge called the synaptonemal complex, a tri-layer protein structure that maintains synapsis by tightly binding ________ of homologous chromosomes to one another.

Answer: nonsister chromatids Section: 3.2 Skill: Knowledge/Comprehension 3.4 Essay Questions 1) What is the difference between cytokinesis and karyokinesis? Which process needs to be more precise for accurate cell division?

Answer: Karyokinesis is the equal partitioning of the chromosomal material in the nucleus of the parental cell between the nuclei of the two daughter cells. This process requires first that each of the chromosomes in the nucleus be fully and accurately duplicated and then that the duplicate copies of each chromosome be separated so that one copy goes to the nucleus of one daughter cell and the second copy goes to the other daughter nucleus. Karyokinesis is followed by cytokinesis, the partitioning of the cytoplasmic contents of the parental cell into the daughter cells.

Cytokinesis does not demand the same degree of equivalency required in karyokinesis. The cytoplasm of the parental cells contains an abundance of the proteins and organelles that the daughter cells require in order to function, so the division of this material need not be equal.

Section: 3.1 Skill: Synthesis/Evaluation 2) What are the three important differences between mitosis and meiosis I? Answer: In meiosis I, but not in mitosis, the following three processes occur: 1. Homologous chromosome pairing 2. Crossing over between homologous chromosomes 3. Segregation (separation) of the homologous chromosomes, which reduces chromosomes to the haploid number Section: 3.2 Skill: Synthesis/Evaluation 3) Describe how microtubules and cohesion work together to align the chromosomes along the metaphase plate during the process of sister chromatid cohesion.

Answer: Kinetochore microtubules are attached to the kinetochore at each centromere of sister chromatids. Because they are tethered to kinetochore microtubules from opposite centrosomes, the sister chromatids experience opposing forces that are critical to the positioning of chromosomes along an imaginary midline at the equator of the cell. This imaginary line is called the metaphase plate. The tension created by the pull of kinetochore microtubules is balanced by a companion process known as sister chromatid cohesion. Sister chromatid cohesion is produced by the protein cohesin that localizes between the sister chromatids and holds them together to resist the pull of kinetochore microtubules. As microtubules move chromosomes toward the midline of the cell, cohesin helps keep the sister chromatids together to ensure proper chromosome positioning and to prevent their premature separation.

Section: 3.2 Skill: Synthesis/Evaluation

4) How do cyclin-dependent kinases help regulate the cell cycle?

Answer: As cells move through the cell cycle, readiness to progress from one stage to the next is regularly assessed. Numerous cell cycle checkpoints are monitored for cell readiness by protein interactions. A common mechanism for this monitoring is carried out by protein complexes that join a protein kinase with a second protein known as a cyclin protein. Protein kinases catalyze protein phosphorylation–the addition of a phosphate group transferred from a nucleotide triphosphate such as ATP or GTP to a target protein. Phosphorylation changes the conformation of target proteins and can either activate or inactivate the target protein. Protein kinases are usually present continuously in cells at relatively steady concentrations. Cyclin proteins, however, are so named because their concentrations are cyclic and linked to the cell cycle stage. Cyclin protein production is stimulated by growth factor proteins that are produced by other cells. The protein kinase components of these complexes are activated only when they associate with a cyclin; thus, the protein kinases are called cyclin-dependent kinases, or Cdks. In their activated state, cyclin-Cdk complexes phosphorylate numerous target proteins and regulate cell cycle progression at various checkpoints.

Section: 3.1 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 4 Gene Interaction 4.1 Multiple-Choice Questions 1) Which mode of inheritance produces heterozygotes with phenotypes that differ from either homozygote but more closely resembles one homozygous phenotype than the other?

A) complete dominance B) incomplete dominance C) codominance D) epistasis E) incomplete penetrance Answer: B Section: 4.1 Skill: Knowledge/Comprehension 2) Which mode of inheritance results in the phenotype of a heterozygote being indistinguishable from that of an organism homozygous for the dominant allele?

A) complete dominance B) incomplete dominance C) codominance D) epistasis E) incomplete penetrance Answer: A Section: 4.1 Skill: Knowledge/Comprehension 3) Which mode of inheritance results in both alleles being detected equally in the heterozygous phenotype?

A) complete dominance B) incomplete dominance

C) codominance D) epistasis E) incomplete penetrance Answer: C Section: 4.1 Skill: Knowledge/Comprehension 4) A mutation results in an enzyme that is partially active compared to the wild-type allele. This type of "leaky" mutation is classified as A) amorphic.

B) hypomorphic.

C) hypermorphic.

D) neomorphic.

E) dominant negative.

Answer: B Section: 4.1 Skill: Knowledge/Comprehension 5) A mutation resulting in an inactive gene product is classified as A) amorphic.

B) hypomorphic.

C) hypermorphic.

D) neomorphic.

E) dominant negative.

Answer: A Section: 4.1 Skill: Knowledge/Comprehension 6) Two proteins interact to form a multimeric complex. When one of the proteins is mutated, there is a substantial loss of functional activity in the multimeric protein. This type of mutation is classified as A) amorphic.

B) hypomorphic.

C) hypermorphic.

D) neomorphic.

E) dominant negative.

Answer: E Section: 4.1 Skill: Knowledge/Comprehension 7) Many oncogenes result from mutations that cause a protein to be expressed in cells where it is normally not expressed or at inappropriate times during development. This type of mutation can be described as A) amorphic.

B) hypomorphic.

C) hypermorphic.

D) neomorphic.

E) dominant negative.

Answer: C Section: 4.1

Skill: Application/Analysis 8) A mutation results in a gene product with a novel function that is not normally found in wild-type organisms. This type of mutation is known as A) amorphic.

B) hypomorphic.

C) hypermorphic.

D) neomorphic.

E) dominant negative.

Answer: D Section: 3.6 Skill: Knowledge/Comprehension

9) Most combinations of different ABO alleles result in complete dominance of one allele. Which combination results in codominance?

A) IAi B) IAIB C) IBi D) ii E) either IAi or IBi Answer: B Section: 4.1 Skill: Application/Analysis 10) What is the genotype for individuals with blood type O? A) IAi B) IAIB C) IBi D) ii E) either IAi or IBi Answer: D Section: 4.1 Skill: Knowledge/Comprehension 11) What type of allele is often detected as a distortion in segregation ratios, where one class of expected progeny is missing?

A) dominant negative allele B) temperature-sensitive allele C) lethal allele D) partial dominance E) incomplete penetrance Answer: C Section: 4.1 Skill: Knowledge/Comprehension 12) Most people with the dominant mutant polydactyly allele have extra digits, but at least 25% have the normal number of digits. What is the genetic explanation for this observation?

A) dominant negative allele B) temperature-sensitive allele C) lethal allele D) partial dominance E) incomplete penetrance Answer: E Section: 4.2 Skill: Knowledge/Comprehension

13) Gene interactions in which an allele of one gene modifies or prevents expression of alleles of another gene is known as A) complete dominance.

B) incomplete dominance.

C) codominance.

D) epistasis.

E) incomplete penetrance.

Answer: D Section: 4.3 Skill: Knowledge/Comprehension 14) Bateson and Punnett crossed two white-flowered lines and saw all purple flowers in the F1 generation. They concluded this was an example of complementary gene interactions because a cross of the F1 plants yielded what ratio in the F2 generation?

A) 8 purple to 8 white B) 16 purple to 0 white C) 0 white to 16 purple D) 7 purple to 9 white E) 9 purple to 7 white Answer: E Section: 4.3 Skill: Application/Analysis 15) If a dominant allele of one gene completely suppresses the phenotypic expression of alleles of another gene, this is an example of A) recessive interaction.

B) dominant interaction.

C) recessive epistasis.

D) dominant epistasis.

E) dominant suppression.

Answer: E Section: 4.3 Skill: Knowledge/Comprehension 16) The 9:6:1 ratio seen in the dihybrid cross of summer squash indicates what genetic relationship between the two genes controlling fruit shape?

A) recessive interaction B) dominant interaction C) recessive epistasis D) dominant epistasis E) dominant suppression Answer: B Section: 4.3 Skill: Knowledge/Comprehension

Refer to the the table to answer the following questions.

17) In the biosynthetic pathway for conversion from homoserine to methionine, you identify a Neurospora crassa double mutant Met1/Met2. This mutant will grow only if which supplement(s) are added to the minimal media?

A) homoserine only B) cysteine only C) homocysteine only D) cystathionine and homocysteine E) methionine only Answer: E Section: 4.3 Skill: Synthesis/Evaluation 18) Which step is catalyzed by the enzyme responsible for the Met 3 mutant?

A) homoserine → cysteine B) cysteine → cystathionine C) cystathionine → homocysteine D) homocysteine → methionine E) methionine → homoserine Answer: B Section: 4.3 Skill: Synthesis/Evaluation 19) Independent assortment predicts a 9:3:3:1 ratio with four different phenotypes in the F2 progeny. If the alleles are epistatic, what would you predict?

A) more than four phenotypes B) fewer than four phenotypes C) heterozygotes with a novel phenotype between the dominant and recessive homozygotes D) 1/3 of the progeny with the dominant phenotypes and 2/3 recessive E) no change in the 9:3:3:1 ratio Answer: B Section: 4.3 Skill: Application/Analysis

20) Two pure-breeding mutant plants produce white flowers. When they are crossed, all of the progeny have wild-type purple flowers. What does this genetic complementation tell you?

A) The genes are part of distinct biosynthetic pathways.

B) The two lines exhibit different mutations in the same gene.

C) More than one gene is involved in determining the phenotype.

D) The allele is pleiotropic.

E) The allele exhibits incomplete dominance.

Answer: C Section: 4.3 Skill: Application/Analysis 4.2 Short-Answer Questions 1) A metabolic reaction requires 40 units of enzymatic activity to proceed. If a dominant allele "D" can generate 40 units of enzyme and a mutant allele "d" generates 20 units of enzyme, what can be said of the dominant wild-type allele?

Answer: D is haplosufficient.

Section: 4.1 Skill: Application/Analysis 2) A metabolic reaction requires 10 units of enzymatic activity to proceed. If a dominant allele "D" can generate 8 units of enzyme and a recessive allele "d" can generate 2 units of enzyme, what can be said of the dominant wild-type allele?

Answer: D is haploinsufficient.

Section: 4.1 Skill: Application/Analysis 3) Are loss-of function mutations more likely to be dominant or recessive?

Answer: recessive Section: 4.1 Skill: Knowledge/Comprehension 4) What are the three categories of loss-of-function mutations?

Answer: amorphic (null), hypomorphic (leaky), dominant negative Section: 4.1 Skill: Knowledge/Comprehension 5) What are the two categories of gain-of-function mutations?

Answer: neomorphic and hypermorphic Section: 4.1 Skill: Knowledge/Comprehension 6) You cross a white flower with a red flower, and the offspring are all pink. This is an example of what type of inheritance?

Answer: incomplete dominance Section: 4.1 Skill: Application/Analysis

7) You cross a white flower with a red flower, and the offspring are white with red spots. This is an example of what type of inheritance?

Answer: codominance Section: 4.1 Skill: Application/Analysis 8) The Sbe1 gene controlling pea shape is an example of codominance at the molecular level; the inheritance pattern is seen by detection of how many DNA or RNA bands in heterozygous plants?

Answer: two Section: 4.1 Skill: Knowledge/Comprehension 9) The four different human blood types are caused by how many alleles?

Answer: three: IA, IB, and i Section: 4.1 Skill: Knowledge/Comprehension 10) Individuals with blood type A have which blood antigen(s)?

Answer: A antigen Section: 4.1 Skill: Knowledge/Comprehension 11) Individuals with blood type AB have which blood antibody(ies)?

Answer: none; neither anti-A nor anti-B Section: 4.1 Skill: Knowledge/Comprehension 12) An individual is a "universal acceptor," meaning he or she will not have an adverse reaction to a transfusion with any type of blood, has which blood type?

Answer: AB Section: 4.1 Skill: Application/Analysis 13) Which antigen, expressed on the surface of all red blood cells, is modified by A- or B-transferases?

Answer: H antigen Section: 4.1 Skill: Knowledge/Comprehension 14) The allele responsible for the Siamese coat-color pattern produces an unstable tyrosinase enzyme.

This type of gene product is an example of what type of allele?

Answer: temperature-sensitive Section: 4.1 Skill: Knowledge/Comprehension

15) King George III of England and other members of the royal family were afflicted with a series of strange, seemingly unrelated symptoms including abdominal pain, rapid pulse, convulsions, and insanity. It has been determined that he likely suffered from porphyria, caused by a mutation in a single allele. What is the genetic term describing the alteration of multiple, distinct traits of an organism by a mutation in a single gene?

Answer: pleiotropy Section: 4.2 Skill: Application/Analysis 16) Which type of pathway is responsible for melanin production the biosynthetic pathway, signal transduction pathway, or developmental pathway?

Answer: biosynthetic pathway Section: 4.3 Skill: Application/Analysis 17) In familial hypercholesterolemia, individuals who are homozygous for the mutation lack LDL cholesterol receptors in the liver, causing high levels of serum cholesterol. This is a defect in which type of gene pathway the biosynthetic, signal transduction, or developmental pathway?

Answer: signal transduction pathway Section: 4.3 Skill: Application/Analysis 18) To better understand which genes are involved in developmental pathways, geneticists use experimental analyses of mutant phenotypes. What is this analytic approach called?

Answer: genetic dissection Section: 4.3 Skill: Knowledge/Comprehension 19) Numerous genes contribute to the wild-type red eye color in Drosophila. Two genes, brown and vermilion, produce eye pigments. What does the white gene do?

Answer: transports pigments to eye cells Section: 4.3 Skill: Knowledge/Comprehension 20) Mendel studied tall and short pure-breeding lines of pea plants. Inherited genetic variation would dictate that one line would produce tall plants and the other short plants, but the genes and phenotypes are influenced by what external factor?

Answer: environmental influence Section: 4.3 Skill: Knowledge/Comprehension

4.3 Fill-in-the-Blank Questions 1) ________ mutations decrease or eliminate gene activity.

Answer: Loss-of-function Section: 4.1 Skill: Knowledge/Comprehension 2) ________ mutations cause overexpression or result in new functions.

Answer: Gain-of-function Section: 4.1 Skill: Knowledge/Comprehension 3) IA and IB alleles exhibit ________ over the i allele.

Answer: complete dominance Section: 4.1 Skill: Knowledge/Comprehension 4) If an organism with a particular genotype fails to produce the corresponding phenotype, the organism is said to be ________ for the trait.

Answer: nonpenetrant Section: 4.1 Skill: Knowledge/Comprehension 5) In Labrador retrievers, coat color is controlled by gene interaction in which homozygosity for a recessive allele can mask the phenotypic expression of a second gene. This genetic interaction is known as ________ and has a characteristic ________ ratio.

Answer: recessive epistasis; 9:3:4 Section: 4.3 Skill: Knowledge/Comprehension 4.4 Essay Questions 1) Proper cross-matching of blood type is essential for safe blood transfusion. What is the general rule for safe blood transfusion, and what are the consequences if a patient receives the incorrect blood type by accident? If a patient's blood type is unknown, what is the safest course of action?

Answer: The general rule for safe blood transfusion is that the recipient blood must not contain an antibody that reacts with an antigen in the donated blood. When such a reaction occurs, blood clots produced by clumping blood cells form at the site of transfusion. The clots block circulation, deprive tissues of oxygen, and can potentially cause life-threatening complications. Thus, if a patient's blood type is not known, they are given type O-negative blood because it contains no antigens for the patient's own blood to react with.

Section: 4.1 Skill: Synthesis/Evaluation

2) You have crossed two Mexican hairless dogs, and the offspring are 1/3 hairy and 2/3 hairless. Given this phenotypic ratio, draw the Punnett square for this cross. What are the genotypes of the P1 and F1 dogs in this cross? List the predicted genotype as well as the phenotype for each of the offspring. Which genotypes are hairless, and which are hairy? Can you design a genetic cross that would yield a truebreeding hairless line (where all offspring are hairless)?

Answer: The 1:2 ratio in the offspring provides strong evidence that this trait involves a lethal allele.

Thus the Punnett square for the cross would be

You cannot design a true-breeding cross, because the hh genotype is lethal. Thus a maximum of 2/3 of all of the offspring from any cross are predicted to be hairless.

Section: 4.1 Skill: Synthesis/Evaluation 3) Describe the difference between incomplete penetrance and variable expressivity. It is often difficult to pinpoint the cause of incomplete penetrance or variable expressivity. What possible interactions may be responsible?

Answer: In incomplete penetrance, a genotype is not expressed by every organism in which it is present. In variable expressivity, the organisms that share a genotype express the corresponding phenotype to different degrees. Both are explained by genetic or nongenetic interactions that modify or prevent the consistent expression of a genotype.

Three kinds of interactions may be responsible: (1) other genes that act in ways that modify the expression of the mutant allele, (2) environmental or developmental (i.e., nongenetic) factors that interact with the mutant allele to modify its expression, and (3) some combination of other genes and environmental factors interacting to modify expression of the mutation. In inbred laboratory strains of model genetic organisms, variation in genetic factors can be eliminated experimentally to allow separation of gene-gene and gene-environment variability, something that cannot be done in organisms such as humans.

Section: 4.2 Skill: Synthesis/Evaluation

4) Explain how the ch (Himalayan) allele and tyrosinase control the Siamese coat-color pattern. What is the underlying reason that certain parts of a Siamese cat (e.g., tail and ears) are darker than the cat's trunk?

Answer: The tyrosinase enzyme produced by the hypomorphic ch (Himalayan) allele is unstable and is inactivated at a temperature very near the normal body temperature of most mammals. Thus ch is a temperature-sensitive allele. The parts of cats that are farthest away from the core of the body (the paws, ears, tail, and tip of the nose) at most times tend to be slightly cooler than the trunk. At these cooler extremities, the temperature-sensitive tyrosinase produced by the ch allele remains active, producing pigment in the hairs there. However, in the warmer central portion of the body, the slightly higher temperature is enough to cause the tyrosinase produced by the ch allele to denature, or unravel. This inactivates the enzyme and leads to an absence of pigment in the central portion of the body. Animals that are chch or chc have the Himalayan phenotype. The final allele in the series, c, is a null allele that does not produce functional tyrosinase.

Section: 4.3 Skill: Synthesis/Evaluation 5) Phenylketonuria (PKU) is caused by the absence of the enzyme phenylalanine hydroxylase, which catalyzes the first step of the pathway that breaks down the amino acid phenylalanine, a common component of dietary protein. Explain how environmental intervention is commonly practiced to prevent the development of this human autosomal recessive condition.

Answer: PKU damages neurons because the body is unable to break down dietary phenylalanine, ultimately causing irreversible mental retardation and neuron death. There are two main ways in which the environment has been changed to affect the development of PKU. First, newborns are routinely screened for PKU so they can be placed on a phenylalanine-free diet and will avoid PKU complications.

Second, individuals living with PKU are warned of foods containing phenylalanine by strict labeling of foods (e.g., foods and beverages containing the artificial sweetener aspartame). Thus, while they will always have the mutated allele, they are able to avoid developing the disease through strict control of diet.

Section: 4.3 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 5 Genetic Linkage and Mapping in Eukaryotes 5.1 Multiple-Choice Questions 1) Syntenic genes can assort independently when A) they are very close together on a chromosome.

B) they are located on different chromosomes.

C) crossing over occurs rarely between the genes.

D) they are far apart on a chromosome and crossing over occurs frequently between the genes.

E) they are far apart on a chromosome and crossing over occurs very rarely between the genes.

Answer: D Section: 5.1 Skill: Knowledge/Comprehension 2) The alleles of linked genes tend to

A) segregate together during gamete production.

B) assort independently.

C) be mutated more often than unlinked genes.

D) experience a higher rate of crossing over.

E) assort independently and show a higher rate of crossing over.

Answer: A Section: 5.1 Skill: Knowledge/Comprehension 3) Genetic linkage leads to the production of a significantly greater than expected number of gametes containing chromosomes with A) allele combinations that are different from parental combinations.

B) parental combinations of alleles.

C) mutant alleles.

D) dominant alleles.

E) recessive alleles.

Answer: B Section: 5.1 Skill: Knowledge/Comprehension 4) The syntenic genes A and Z are linked. A cross between two parents, AAZZ and aazz, produces F1 progeny with the AaZz genotype. What are the possible arrangements of alleles on the F1 progeny's chromosomes?

A) AZ/az B) Az/aZ C) Aa/Zz D) AZ/Az or az/aZ E) A/Z or a/z Answer: A Section: 5.1 Skill: Knowledge/Comprehension

5) In a dihybrid cross exhibiting complete genetic linkage, what would you expect?

A) one parental allele combination occurring more frequently than another B) two equally frequent gametes containing only parental allele combinations C) only recombinant gametes D) one parental allele combination occurring more frequently than another and no recombinant gametes E) two equally frequent gametes containing only parental allele combinations and no recombinant gametes Answer: E Section: 5.1 Skill: Knowledge/Comprehension 6) In a two-point test-cross analysis, a dihybrid F1 fly is crossed to a A) pure-breeding mate with a dominant phenotype (AABB).

B) pure-breeding mate with a recessive phenotype (aabb).

C) heterozygote mate (AaBb).

D) mate that is dominant for one gene and recessive for the other (AAbb or aaBB).

E) second dihybrid F1 fly (sibling cross).

Answer: B Section: 5.1 Skill: Knowledge/Comprehension 7) If you are given a recombination frequency of 34% between genes X and Y and 27% between X and Z, can you predict the order of the three genes?

A) Yes; the order is X-Z-Y. B) Yes; the order is X-Y-Z. C) Yes; the order is Z-X-Y. D) No; based on this data alone, the order could be Z-Y-X or X-Y-Z. E) No; based on this data alone, the order could be X-Z-Y or Z-X-Y. Answer: E Section: 5.2 Skill: Application/Analysis 8) What type of test would you use to determine whether observed data constitute evidence of genetic linkage or are simply a case of chance variation from expected values?

A) test cross B) two-point test-cross C) three-point test-cross D) chi-square analysis E) recombination frequency (r) calculation Answer: D Section: 5.2 Skill: Knowledge/Comprehension

9) Incomplete genetic linkage of three genes in a trihybrid produces eight genetically different gamete genotypes. How many genotypes are produced in a four-gene cross with incomplete genetic linkage?

A) 4 B) 8 C) 16 D) 20 E) 24 Answer: C Section: 5.2 Skill: Application/Analysis 10) Assuming three genes are linked, how many recombinant genotypes would you expect and at what frequency?

A) two recombinant genotypes, more frequent than expected by chance B) four recombinant genotypes, more frequent than expected by chance C) six recombinant genotypes, more frequent than expected by chance D) four recombinant genotypes, less frequent than expected by chance E) six recombinant genotypes, less frequent than expected by chance Answer: E Section: 5.2 Skill: Knowledge/Comprehension 11) For a given cross, the expected number of double recombinants is 20 and the observed number of double recombinants is 15. What is the coefficient of coincidence (c)?

A) 0.25 B) 1.33 C) 20 D) 0.75 E) 15 Answer: D Section: 5.3 Skill: Application/Analysis 12) For a given cross, the expected number of double recombinants is 20 and the observed number of double recombinants is 15. What is the interference calculation (I)?

A) 0.25 B) 1.33 C) 20 D) 0.75 E) 15 Answer: A Section: 5.3 Skill: Application/Analysis

13) Where does crossing over occur?

A) at the centromeres B) at the telomeres C) at nodules that occur randomly at "hotspots" along the synaptonemal complex D) at nodules that occur in specific, evenly spaced locations along a chromosome E) at both the centromeres and telomeres Answer: C Section: 5.4 Skill: Knowledge/Comprehension 14) When does recombination occur?

A) before DNA replication B) at the two-strand stage, when each member of a homologous chromosome pair is composed of two sister chromatids, and before DNA replication C) at the four-strand stage, when each member of a homologous chromosome pair is composed of two sister chromatids, and before DNA replication D) at the two-strand stage, when the sister chromatids have been separated, and after DNA replication E) at the four-strand stage, when each member of a homologous chromosome pair is composed of two sister chromatids, and after DNA replication Answer: E Section: 5.4 Skill: Knowledge/Comprehension 15) Which type of crossover event produces no recombinant chromosomes?

A) single-strand crossover B) two-strand double crossover C) three-strand double crossover D) four-strand double crossover E) None of the above; all crossover events produce recombinant chromosomes.

Answer: B Section: 5.4 Skill: Knowledge/Comprehension 16) Which type of crossover event produces two parental and two recombinant chromosomes in gametes?

A) single-strand crossover B) two-strand double crossover C) three-strand double crossover D) four-strand double crossover E) None of the above; all crossover events produce only recombinant chromosomes.

Answer: C Section: 5.4 Skill: Knowledge/Comprehension

17) Which type of crossover event produces all four recombinant chromosomes?

A) single-strand crossover B) two-strand double crossover C) three-strand double crossover D) four-strand double crossover E) None of the above; all crossover events produce a combination of parental and recombinant chromosomes.

Answer: D Section: 5.4 Skill: Knowledge/Comprehension 18) For a given haplotype, the frequencies of alleles for gene Y are Y = 0.65 and Y' = 0.35, and the frequencies at gene Z are Z = 0.25 and Z' = 0.75. What is the predicted frequency of the Y'Z' haplotype?

A) 0.16 B) 0.26 C) 0.74 D) 0.84 E) 1.1 Answer: B Section: 5.4 Skill: Synthesis/Evaluation 19) Intragenic recombination can be detected A) based on recombination between homologs carrying different mutant alleles.

B) by the recovery of 100% double-mutant chromosomes.

C) by the recovery of both wild-type and double-mutant chromosomes.

D) based on recombination between homologs with different mutant alleles and recovery of 100% double-mutant chromosomes.

E) based on recombination between homologs with different mutant alleles and recovery of both wildtype and double-mutant chromosomes.

Answer: E Section: 5.4 Skill: Knowledge/Comprehension 20) The Huntington disease (HD) allele was discovered A) by mapping it to blood enzyme genes, based on Lod scores of -2.

B) through pedigree analysis alone.

C) by linkage to a DNA genetic marker, D4S10 with a Lod score of 8.53.

D) through mapping to known homologs in other organisms such as mice.

E) through direct sequencing of mutations identified by SNPs. Answer: C Section: 5.8 Skill: Knowledge/Comprehension

5.2 Short-Answer Questions 1) What is observed when syntenic genes are so close to one another that they are unable to assort independently?

Answer: genetic linkage Section: 5.1 Skill: Application/Analysis 2) Chromosomes that do not reshuffle the alleles of linked genes are identified as what?

Answer: parental chromosomes Section: 5.1 Skill: Application/Analysis 3) Recombination analysis allows for the mapping of genes because the frequency of crossing over is proportional to what?

Answer: distance between genes Section: 5.1 Skill: Knowledge/Comprehension 4) What organism exhibits complete genetic linkage, meaning there is no recombination between homologous chromosomes?

Answer: male Drosophila or Diptera Section: 5.1 Skill: Knowledge/Comprehension 5) Incomplete genetic linkage is far more common than complete linkage. Recombination between the homologs produces what type(s) of gametes?

Answer: parental and nonparental Section: 5.1 Skill: Knowledge/Comprehension 6) Two genes, A and X, exhibit incomplete linkage. The frequency of each parental gamete (AX and ax) is 45%. What is the approximate frequency of the Ax gamete?

Answer: 5% Section: 5.1 Skill: Application/Analysis 7) There are three genes located in the gene order A-B-C on a chromosome. Would you expect the recombination frequency to be higher between A and B or A and C? Answer: higher from A to C (farther apart) Section: 5.1 Skill: Application/Analysis 8) Bateson and Punnett saw early evidence of gene linkage. What terms did they use to describe genes that remained together frequently and the loss of parental alleles in favor of new, nonparental phenotypes in the F2 progeny?

Answer: coupling and repulsion Section: 5.1 Skill: Knowledge/Comprehension

9) Morgan realized that Bateson and Punnett had detected genetic linkage but couldn't explain their results because they performed the wrong cross. What type of cross did Morgan use for the analysis of genetic linkage of autosomal genes in Drosophila?

Answer: two-point test-cross analysis Section: 5.1 Skill: Knowledge/Comprehension 10) In a two-point test-cross analysis, what are the "two points" being tested?

Answer: the two genes Section: 5.1 Skill: Knowledge/Comprehension 11) What unit of physical distance between genes on a chromosome provides a convenient way to link the recombination frequencies for linked genes with their positions and order along a chromosome?

Answer: map unit (m.u.) or centimorgan (cM) Section: 5.2 Skill: Knowledge/Comprehension 12) What type of test would you use to determine the relative order of more than two genes?

Answer: three-point test-cross Section: 5.3 Skill: Knowledge/Comprehension 13) In most tests of genetic linkage, the number of double crossovers is less than the number expected due to what effect, which limits the number of crossovers that can occur in a short length of chromosome?

Answer: interference (I) Section: 5.3 Skill: Knowledge/Comprehension

14) Refer to this map to answer the following questions:

a) What is the distance between gene E and gene F? b) Assuming I = 0, what is the probability of no crossovers between gene D and gene E? c) Assuming I = 0, what is the probability of no crossovers between gene E and gene F? d) Considering both gene pairs, what is the proportion of nonrecombinant gametes? e) What is the predicted frequency of each parental gamete (DEF/def)? f) What is the recombination frequency between gene D and gene E? g) What is the recombination frequency between gene E and gene F? h) What is the frequency of two single recombinant gametes between genes D and E? i) What is the frequency of two single recombinant gametes between genes E and F? j) What is the frequency of each of the double-recombinant gametes, DeF and dEf? k) Fill in the blanks below to prove that the sum of the frequencies of the eight predicted gamete genotypes is equal to 1.0: _______ + _______ + _______ + _______ + _______ + _______ + _______ + _______ = 1 DEF def Def dEF DEf deF DeF dEf Answer: a) 12 cM b) 92% c) 88% d) (0.92)(0.88) = 0.8096 e) (0.5)(0.8096) = 0.4048 f) 8% g) 12% h) (0.08)(0.88)(0.5) = 0.0352 i) (0.12)(0.92)(0.5) = 0.0552 j) (0.08)(0.12)(0.5) = 0.0048 k) 0.4048 + 0.048 + 0.0352 + 0.0352 + 0.0552 + 0.0552 + 0.0048 + 0.0048 = 1 Section: 5.3 Skill: Synthesis/Evaluation 15) When constructing a genetic map of Zea mays, Creighton and McClintock used genetic markers as well as structural differences in the homologous copies of chromosome 9 that can be seen under the microscope. What are these structural differences called?

Answer: cytological markers Section: 5.4 Skill: Knowledge/Comprehension

16) What is the highest possible frequency of recombination between linked genes that can be generated by any type of crossover event?

Answer: 50% Section: 5.4 Skill: Knowledge/Comprehension 17) Does recombination frequency between linked genes increase as the distance between genes gets larger or smaller?

Answer: larger Section: 5.4 Skill: Knowledge/Comprehension 18) Recombination frequency differs between males and females. Which sex has a higher rate of recombination, heterogametic (males) or homogametic (females)?

Answer: homogametic (females) Section: 5.4 Skill: Knowledge/Comprehension 19) If there is a maximum Lod score of 4.2 at = 0.31, what can you say about the linkage and the distance between two genes?

Answer: linked and 31 cM apart Section: 5.5 Skill: Application/Analysis 20) If too little time has passed for crossing over to randomize haplotypes or if natural selection favors certain haplotypes, what type of linkage would you expect to see?

Answer: linkage disequilibrium Section: 5.6 Skill: Knowledge/Comprehension 21) Ascospores in yeast are not arranged in a particular order. Thus, the four haploid ascospores contained within a yeast ascus are known as what?

Answer: unordered tetrad Section: 5.7 Skill: Knowledge/Comprehension 5.3 Fill-in-the-Blank Questions 1) The alleles of syntenic genes can be reshuffled by crossing over between homologous chromosomes to produce ________.

Answer: recombinant chromosomes Section: 5.1 Skill: Knowledge/Comprehension 2) The tendency of alleles of linked genes to segregate together during meiosis can be broken by ________.

Answer: crossing over Section: 5.1 Skill: Knowledge/Comprehension

3) Genetic linkage can be spotted by comparing the observed frequencies of ________ with the frequencies expected (assuming independent assortment).

Answer: gamete genotypes/progeny phenotypes Section: 5.1 Skill: Knowledge/Comprehension 4) In early genetic analysis, 5% recombination is equal to ________ map unit(s) (m.u.) or centimorgan(s) (cM) of distance between linked genes.

Answer: 5 Section: 5.2 Skill: Application/Analysis 5) The specific array of alleles making up a set of linked genes on a single chromosome is called a ________.

Answer: haplotype Section: 5.6 Skill: Knowledge/Comprehension 5.4 Essay Questions 1) What is the relationship between linked genes and syntenic genes? Are syntenic genes always linked, or are linked genes always syntenic? Describe what is meant by each term.

Answer: Linked genes are always inherited together with no independent assortment. Syntenic genes are genes found on the same chromosome. Linked genes are always found on the same chromosome, but they are unable to sort independently because there is no crossing over observed between these two genes. Syntenic genes can be unlinked, and their alleles will assort independently, if they are far enough apart on the chromosome for crossing over to generate independent assortment of the alleles.

Section: 5.1 Skill: Synthesis/Evaluation 2) The mapping of the human genome in the mid-1980s was integral for identifying new human genetic markers because of the identification of what types of polymorphic DNA markers?

Answer: SNPs and restriction fragments Section: 5.5 Skill: Synthesis/Evaluation Gene-mapping methods that have been successfully used in model organisms are difficult to apply to human gene mapping, due to the fewer number of offspring and the lack of controlled matings.

3) Describe why allelic phase is a major obstacle in mapping human genes.

Answer: Allelic phase is the arrangement of different alleles of linked genes on homologous parental chromosomes. Allelic phase is required for genetic linkage analysis because you need to know which alleles are unique to "parental" and "recombinant" gametes.

Section: 5.5 Skill: Synthesis/Evaluation

4) How is Newton Morton's statistical method helpful for calculating the overall probability of genetic linkage when allelic phase is unknown?

Answer: Morton's method determines whether genetic linkage exists between genes for which allelic phase is unknown by comparing the likelihood of obtaining the genotypes and phenotypes observed in a pedigree if two genes are linked versus the likelihood of getting the same pedigree structure if the genes assort independently. This approach assesses the probability of genetic linkage between two genes at a time and compares the probability of genetic linkage to the probability of independent assortment of the genes.

Section: 5.5 Skill: Synthesis/Evaluation 5) What is the Lod score?

Answer: The ratio of these two likelihoods gives the "odds" of genetic linkage, and the Logarithm of the odds ratio generates the Lod score, a statistical value representing the probability of genetic linkage between two genes. A Lod score is a statistic that can attain significance favoring genetic linkage if the probability of genetic linkage is sufficiently greater than the probability of independent assortment, or significance against genetic linkage if the probability of independent assortment is sufficiently greater than the linkage probability. Lod scores can be interpreted for individual families, or they can be added together for as many families as are analyzed.

Section: 5.5 Skill: Synthesis/Evaluation 6) What would a Lod score of 3.2 tell you about genetic linkage? What information does the Θ value tell you?

Answer: Scores of greater than 3.0 provide evidence of genetic linkage. Θ values that correspond to a significant Lod score indicate an approximate distance (in cM) between linked genes.

Section: 5.5 Skill: Synthesis/Evaluation 7) Describe the three types of tetrads.

Answer: Parental ditypes (PD) contain four haploid cells with the same alleles seen in the parental class.

Nonparental ditypes (NPD) contain four recombinant haploid cells because the two parental alleles have recombined to form a nonparental combination of alleles.

Tetratype tetrads (TT) contain four different kinds of haploid cells: two parental spores and two recombinant spores. When two genes are not linked, tetratypes arise when a crossover occurs between one of the two genes and its centromere.

Section: 5.7 Skill: Synthesis/Evaluation

Tetrad analysis can be used to identify genetic linkage in haploid eukaryotes. Parental and recombinant gametes contained in an ascus can be analyzed to map genes.

8) What ratios of these three tetrads would you expect to see if two genes are linked or unlinked? Why is one type of tetrad more or less common?

Answer: If two genes are linked, you will see that PD > TT > NPD. PD tetrads are most common; they are produced when no crossover occurs between genes and when two-strand double crossover takes place. TT tetrads are less frequent than PD; they occur when single crossovers or three-strand double crossovers take place. NPD tetrads are least frequent; they form only when four-strand double crossover occurs.

If PD = NPD, then the two genes are unlinked. This ratio indicates that either the genes are located on different chromosomes or they are syntenic but very far apart.

Section: 5.7 Skill: Synthesis/Evaluation 9) The products of meiosis yield the following results: PD = 1200 TT = 750 NPD = 50 What is the recombination frequency?

Answer: RF = [(NPD + 1/2TT)/total # tetrads] × 100 = [(50 + 1/2(750)/2000] × 100 = 21.25 m.u.

Section: 5.7 Skill: Synthesis/Evaluation 10) Crossing over occurs frequently during meiosis. Can crossing over occur during mitosis? Explain why mitotic crossing over may or may not be possible.

Answer: Mitotic crossover is a rare event. Mitotic crossing over occurs only in diploid (somatic) cells; thus, to study mitotic crossing over in a haploid organism, diploid cells have to be created artificially.

Mitotic crossing over most likely occurs when homologous chromosomal segments are accidentally paired in somatic cells since it requires that after DNA replication, but before homologous chromosomes randomly align along the metaphase plate, the homologous chromosomes come together and interact like chromosomes during prophase I in meiosis. Opportunities for this unusual kind of mitotic interaction are rare but variable by species. Although rare, mitotic crossing over is important in some organisms. Some fungi that do not have a sexual cycle use mitotic crossing over to introduce genetic variation. In humans, mitotic crossing over is thought to be important in allowing recessive cancercausing mutations such as RB. Section: 5.8 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 6 Genetic Analysis and Mapping in Bacteria and Bacteriophage 6.1 Multiple-Choice Questions 1) Why does conjugation with an Hfr donor result in a much higher rate of gene transfer than conjugation with an F+ donor?

A) An F+ donor makes fewer pili.

B) An F+ donor's T strand is slow to reach the pilus.

C) The Hfr donor transfers genes from the chromosome and not the plasmid.

D) The Hfr donor segment must undergo recombination in the recipient.

E) The F+ donor segment must undergo recombination in the recipient.

Answer: C Section: 6.1 Skill: Synthesis/Evaluation 2) In phi X 174 phage, whose genome is a single-stranded, circular, DNA known as the - strand, replication occurs by the rolling circle method. In which order must the events of replication occur?

A) nicking of the - strand, binding of relaxase protein to the 5′ end, DNA synthesis of a + strand, separation of + from - strands, circularization of both strands B) synthesis of a + strand, nicking at the origin, binding of protein to the 5′ end, DNA synthesis, displacement of the + strand when replication passes the origin, circularization of the + strand C) synthesis of a + strand, nicking of both strands to separate them, recircularization of both strands D) binding of a protein at the origin of the - strand, synthesis of a + strand, release of the + strand Answer: B Section: 6.1 Skill: Synthesis/Evaluation 3) A large region of the F plasmid (~33 kb) is called the transfer region (tra) and contains about 40 genes necessary for conjugation. The gene traA codes for the protein pilin that is polymerized to form a pilus, using a number of other tra genes. The pilus is a hollow cylinder whose units disassemble once contact is made in order to "reel in" the recipient cell. What other functions of the tra genes are required for transfer?

A) DNA unwinding, and restriction of contact with F+ cells B) DNA replication enzymes C) resistance to phage infection D) integration of F genes into the recipient's chromosome E) integration of F genes into the recipient's plasmid Answer: A Section: 6.1 Skill: Application/Analysis

4) When F+ conjugation occurs successfully, only one copy of the F plasmid single-stranded DNA is transferred. What else must minimally occur?

A) circularization of the recipient's new DNA B) circularization plus replication of the new recipient DNA C) stopping the rolling circle replication after one copy length, circularization, and replication D) continuation of rolling circle replication in the donor cell and replication in the recipient cell E) rolling circle replication in the recipient cell Answer: C Section: 6.1 Skill: Application/Analysis 5) To integrate an F plasmid's genes into a host chromosome, there must be DNA recombination at homologous regions on the two circular strands. What is the minimum number of such crossover events needed for integration?

A) 1 B) 2 C) 3 D) 4 E) 8 Answer: A Section: 6.1 Skill: Application/Analysis 6) When F+ conjugation occurs successfully, only one copy of the F plasmid single-stranded DNA is transferred. What else must minimally occur?

A) circularization of the recipient's new DNA B) circularization plus replication of the new recipient DNA C) stopping the rolling circle replication after one copy length, circularization, and replication D) continuation of rolling circle replication in the donor cell and replication in the recipient cell E) rolling circle replication in the recipient cell Answer: C Section: 6.1 Skill: Application/Analysis Refer to this text to answer the following questions.

To map genes of a bacterial strain, conjugation must be interrupted at given times. Suppose you have Hfr cells of genotype a+b+c+d+e+strR and F= cells of genotype a-b-c-d-e-strS and you combine these two cultures in liquid medium in four blenders at time 0. After intervals of 3, 6, 9, and 12 minutes, you turn on successive blenders.

7) What role does the blender play in the experiment?

A) serves as a culture container only B) acts as a mechanism to bring cells into contact C) speeds up molecular movement of DNA D) shears the sex pili of conjugants E) hastens conjugation events Answer: C Section: 6.2

Skill: Application/Analysis 8) The resulting cultures were then plated on medium containing streptomycin. Why? A) to eliminate non-conjugating Hfr cells B) to eliminate all non-conjugated cells C) to eliminate non-conjugated F- cells D) to selectively eliminate cells that have taken in F+ genes Answer: C Section: 6.2 Skill: Application/Analysis 9) This graph shows the results of plating the cultures of each blender. What is the correct order of the genes on this chromosome?

A) a b c d e B) a c e b d C) b e c a (d is unknown) D) d a c e b E) b e c a d Answer: B Section: 6.2 Skill: Application/Analysis 10) Which of the following allows the compilation of time-of-entry maps?

A) the use of many replicates of each experiment B) the use of appropriate controls repeatedly C) the fact that gene entry can proceed in either direction D) the observation that Hfr genes recombine so efficiently E) the recognition of phenotype interaction Answer: C Section: 6.3 Skill: Application/Analysis

11) In five Hfr strains, each of which was used to build a time-of-entry map, the genes entered the recipient cells as follows: Strain 1: S L A C T F Strain 2: N P F T C A Strain 3: T F P N U Y Strain 4: S H Y U N P Strain 5: U N P F T C Which of the following represents a correct compilation of these results?

A) S L A C T F P N H C U B) N P F T S L A C H U T C) T C A L S P N U Y H D) U N P F T C A L S T F E) S L A C T F P N U Y H Answer: E Section: 6.2 Skill: Application/Analysis 12) An F′ donor includes F DNA plus a segment of bacterial chromosome DNA. If conjugation is interrupted before the entire F′ chromosome transfers, what could be the expected consequences?

A) integration of some but not all F′ genes into the recipient's chromosome B) a smaller F′ plasmid than had been present in the donor C) failure of the F′ strand to circularize in the recipient D) change of the recipient from F- to F+ E) formation of a recipient that is a stable partial diploid Answer: C Section: 6.3 Skill: Application/Analysis 13) In a partially diploid F′ exconjugant, there are two alleles for lac (1 and 2) and two alleles for his (3 and 4). The phenotype of this cell is identical to a lac 1 his 4 cell. What is the best interpretation?

A) Lac 1 and 2 and his 3 and 4 share only lac 1 and his 4 phenotypes.

B) Lac 1 complements lac 2 and his 3 complements his 4.

C) Lac 1 and his 4 are dominant.

D) Lac 1 and his 4 are wild types.

E) Lac 1 shows its phenotype only if his 4 is also present.

Answer: C Section: 6.3 Skill: Application/Analysis

14) A competent Hemophilus strain is individually transformed by three genes (a, b, and c) at a rate of 0.002% each. However, cotransformation frequencies for each pair are as follows: a and b 0.38 % a and c 0.72% b and c 0.19% What is the correct conclusion from this data?

A) Genes b and c are farthest apart.

B) Gene order is c, b, a.

C) Cotransformation of b and c is so frequent, they must be one gene.

D) Gene a is closer to b than to c.

E) Genes a and c are farthest apart.

Answer: A Section: 6.4 Skill: Application/Analysis 15) In his analysis of rII in phage, Benzer assayed complementation. Suppose you try your hand at this with several mutants and get the following results (+ = complementation): 42 and 62 + 42 and 83 62 and 63 62 and 74 + 42 and 74 42 and 75 + 62 and 75 Mutation 42 has previously been shown to be in a different gene than mutation 62. What can you conclude?

A) Mutations 42 and 75 are in the same gene.

B) Mutations 62 and 74 are in the same gene.

C) Mutations 62, 63, and 74 are in the same gene.

D) Mutations 42 and 74 are in the same gene.

E) Mutation 63 is in a gene with neither 62 nor 42.

Answer: D Section: 6.6 Skill: Application/Analysis 16) In generalized transduction, a phage introduces a segment of donor DNA into the recipient cell. This is followed by recombination of the donor fragment with the recipient chromosome. Which of the following must occur?

A) circularization of the donor fragment before recombination B) a pair (or even number) of crossovers between the donor segment and the recipient C) degradation of one of the two strands of phage genome D) replication of the donor segment before recombination E) a single crossover between the donor segment and the recipient chromosome Answer: B Section: 6.5 Skill: Application/Analysis

17) In an imagined transduction experiment with alleles a+, b+, and c+, a+ alone was transduced into an auxotroph 1400 times, b+ alone 1200 times, and c+ alone 970 times. Genes a+ and b+ were both found 10 times, b+ and c+ 2 times, and a+ and c+ one time. What can you conclude?

A) Cotransduction is too infrequent to be used to map genes.

B) Genes a and c cotransduce only accidentally.

C) Genes a and b are more closely linked than a and c or b and c.

D) Genes a and c are found together only if double crossovers have occurred.

E) Transduction of gene c is very rare.

Answer: C Section: 6.5 Skill: Application/Analysis Refer to this text to answer the following questions.

Bacterial gene transfer is also used in the lab to introduce genes into organisms of different species, genera, phyla, or even kingdoms or domains. A major example is the use of a Ti plasmid of the soil bacterium Agrobacterium tumefaciens to transfer genes into dicotyledonous plants. The bacterium infects the plant root in the soil and transfers some of the Ti plasmid genes (T DNA) into the plant nucleus. Recombination may occur, and T DNA expression is controlled by plant compounds generated in response to wounding and infection.

18) The T DNA is generated when there is a nick that creates a primer for replication, followed by transfer of a single-stranded DNA piece that converts to a double-stranded piece in the plant nucleus.

Which of these processes is Ti plasmid transfer most like?

A) conjugation B) generalized transduction C) specialized transduction D) transformation Answer: A Section: 6.1 Skill: Application/Analysis 19) This method can be used to introduce a gene for an antigenic protein into an edible plant, such as a potato, to produce an edible vaccine (e.g., for cholera). What has to be made a part of the Ti plasmid?

A) a gene for cholera resistance B) a gene for a subunit of cholera toxin, including its regulatory elements C) a gene for that part of the human immune system that neutralizes cholera D) a molecule of anti-cholera antibody E) a polypeptide that can attach the T DNA to plant DNA Answer: B Section: 6.3 Skill: Application/Analysis

20) Assuming you have appropriately engineered the Ti plasmid to carry the gene in question, how could you best and most safely test whether your potatoes are now working as a vaccine?

A) Feed the engineered potato to a population at high risk for cholera.

B) Cook the potatoes to denature the toxin and feed them to mice susceptible to cholera.

C) Feed pieces of raw potato to susceptible mice and test them for antibodies.

D) Emulsify the potatoes in a liquid that can be injected into experimental animals.

Answer: C Section: 6.3 Skill: Synthesis/Evaluation 21) Benzer's analysis of phage genomes included deletion mapping. Since recombination could not occur in the area of a deletion, an infection by two phage strains, one of which has a deletion and the other a point mutation, which results in no wild-type recombinants being produced indicates ________?

A) that the deletion includes the position of the point mutation in the same gene B) that the deletion includes the position of the point mutation in a different gene C) that the point mutation is dominant D) that the point mutation is recessive E) the distance between the two types of mutation Answer: A Section: 6.6 Skill: Application/Analysis 6.2 Short-Answer Questions 1) Name the two cycles in which bacteriophages act when they infect bacteria.

Answer: lysis and lysogeny Section: 6.5 Skill: Knowledge/Comprehension 2) What is the difference between F+ and F- bacteria?

Answer: F+ bacteria have an F plasmid.

Section: 6.1 Skill: Knowledge/Comprehension 3) Describe two features of a plasmid.

Answer: two of the following: circular, double-stranded DNA, autonomous, or includes genes Section: 6.1 Skill: Knowledge/Comprehension 4) Which form of bacterial recombination requires no vector?

Answer: transformation Section: 6.4 Skill: Knowledge/Comprehension 5) Name the three usual phases of bacterial growth in a liquid medium.

Answer: lag (phase), log (phase), stationary (phase) Section: 6.1 Skill: Knowledge/Comprehension

6) If you count 73 plaques on a bacterial plate when you had added 0.1 ml of a 10-4 dilution of phage, what was the initial concentration of the (undiluted) phage?

Answer: 73 × 104 × 10 = 7.3 × 106 pfu/ml Section: 6.5 Skill: Application/Analysis 7) In a bacterial cross of an Hfr donor a+b+ with a recipient F- a-b-, what would be a possible recombinant?

Answer: a+b- or a-b+ Section: 6.2 Skill: Application/Analysis 8) In a transduction experiment, what would constitute evidence of linkage?

Answer: cotransduction (of genes) Section: 6.5 Skill: Knowledge/Comprehension 9) All three types of bacterial recombination share one characteristic. What is it?

Answer: one-way transfer of gene(s), or transfer from donor to recipient Section: 6.1 Skill: Knowledge/Comprehension 10) One kind of plasmid in bacteria is called an R plasmid. What kinds of genes are typical of such plasmids?

Answer: antibiotic-resistant genes Section: 6.1 Skill: Knowledge/Comprehension 11) What is the role of the relaxosome in conjugation?

Answer: binds to F origin, or cleaves T strand, or binds to 5′ end to begin transfer Section: 6.1 Skill: Knowledge/Comprehension 12) In conjugation, as the 5′ end of the T strand begins to move across the pilus, what form of replication occurs in the donor cell?

Answer: rolling circle (replication) Section: 6.1 Skill: Knowledge/Comprehension 13) In Hfr transfer, what happens to linear DNA that does not recombine with the host chromosome?

Answer: (enzymatic) degradation Section: 6.1 Skill: Knowledge/Comprehension 14) If conjugation occurs with an F′ donor, how will the recipient be characterized genetically?

Answer: partially diploid Section: 6.3 Skill: Knowledge/Comprehension

15) What characterizes a bacterial cell that can undergo transformation?

Answer: competence, or ability to take up double-stranded DNA Section: 6.4 Skill: Knowledge/Comprehension 16) What distinguishes generalized versus specialized transduction?

Answer: generalized uses lytic phage; specialized uses lysogenic phage Section: 6.5 Skill: Knowledge/Comprehension 17) Jacob and colleagues developed a bacterial linkage map based on time. What kind of recombination was used?

Answer: conjugation Section: 6.2 Skill: Knowledge/Comprehension 18) In an E. coli conjugation experiment, the donor cell is found to be thr+, leu+, his+, and the F- cell is thr-, leu-, and his-. When the mating is interrupted after 10 minutes and the recipients are plated, they are found to be leu+ and his+ but thr-. What can you conclude?

Answer: Leu and his are close together.

Section: 6.2 Skill: Application/Analysis 19) In a conjugation experiment, the Hfr donor is thr+, leu+, his+, str. The recipient is thr-, leu-, and his- strS. Why would you use medium containing streptomycin to analyze your results?

Answer: to identify strR recipients Section: 6.2 Skill: Application/Analysis 20) Arabinose sugar is used by ara+ bacteria but cannot be used by ara- mutants. When both ara3 and ara5 mutants are introduced into the same recipient via a vector, and the recipients still cannot use arabinose, what can you conclude?

Answer: The mutations are in separate genes.

Section: 6.6 Skill: Application/Analysis 21) Seymour Benzer's fine structure studies of genomes used mutations found in specific regions of what kind of genome?

Answer: bacteriophage (phage) Section: 6.6 Skill: Knowledge/Comprehension 22) What is meant by a mutation that Benzer referred to as nonrevertible?

Answer: due to a deletion Section: 6.6 Skill: Knowledge/Comprehension

6.3 Fill-in-the-Blank Questions 1) Some bacteriophages can be either lytic or lysogenic. These are called ________ phages.

Answer: temperate Section: 6.5 Skill: Knowledge/Comprehension 2) Prototrophs are bacteria that can grow on ________ medium.

Answer: minimal Section: 6.1 Skill: Knowledge/Comprehension 3) Once a phage genome has become part of the host chromosome, it is called a ________.

Answer: prophage Section: 6.5 Skill: Knowledge/Comprehension 6.4 Essay Questions 1) When a soil bacterium infects a plant root and introduces its Ti plasmid, how does this affect both the infectious bacterium and the plant?

Answer: A soil bacterium, Agrobacterium tumefaciens, contains a tumor-inducing (Ti) plasmid that is ~200 kb and contains the T-DNA and the genes required to transfer that DNA to the plant cell.

Bacterial plasmids are circular, so DNA must be linearized in order to be able to incorporate into the plant's linear chromosomes. When A. tumefaciens infects the plant, the bacterium must cut the T-DNA out of the circular plasmid. DNA is nicked at the right border sequence, yielding a region with singlestranded DNA. Single-stranded binding proteins attach to this single stranded DNA. DNA synthesis will displace the single-stranded region and a second nick at the left border region will release the single stranded T-DNA fragment. Through the process of single strand DNA transfer (similar to bacterial conjugation), this fragment can be transferred to the plant cell. The DNA is targeted to the plant cell nucleus, where it can be incorporated into the host genome. Through homologous recombination, the DNA becomes integrated into the host genome, where the gene products will be transcribed and translated as any other plant gene by the plant's own transcription and translation machinery.

Section: 6.1 Skill: Synthesis/Evaluation

2) Conjugation, transduction, and transformation all occur in nature. Which do you expect would more likely account for the acquisition of the genes for resistance to an antibiotic? Why? Which do you think would best account for the acquisition of genes from one bacterial species by another of a different genus? Why? Answer: Through the process of lateral gene transfer, genes located on a circular strand of DNA called an R-plasmid contain antibiotic-resistant genes. Through conjugation, an antibiotic-resistant bacterium can transfer the R-plasmid to a non-resistant bacterium. Because this R-plasmid provides a selective growth advantage to some bacteria, they may "collect" many copies of various antibiotic resistance genes from other bacteria.

Transfer of genes between bacterial species likely involves transformation, the process by which extracellular fragments of DNA released when a donor bacterial cell lyses can be absorbed across the cell membrane of a competent recipient cell as transforming DNA. Specialized transduction involves bacteriophages, which can transfer host DNA from one host to the next through aberrant excision of a lysogenic prophage. However, it is more likely that gene transfer between species occurred through transformation rather than transduction.

Section: 6.5 Skill: Application/Analysis 3) Which, if any, characteristics of bacterial gene transfer is/are useful in analyzing eukaryotic gene transfer?

Answer: Lateral gene transfer is the process by which genes are transferred between species, as opposed to vertical gene transfer, in which genes are inherited from ancestors. This phenomenon is common in prokaryotes, and can occur by conjugation, transformation, and transduction. There is evidence of natural lateral gene transfer in eukaryotes, but it is far more relevant in terms of artificial gene transfer and genetic engineering. Genes from one organism can be successfully expressed in a different organism, e.g. expression of GFP, luciferase, and various reporter genes. If you want to introduce a gene of interest into a eukaryotic cell, you will most often use transformation. Cells are incubated with the DNA of interest, and then subjected to artificial selection to select for cells that have taken up the DNA from the environment. This same process occurs naturally in bacteria and in single-celled eukaryotes, but can be a valuable asset for genetic engineering.

More recently, specially engineered viruses have also been used to introduce DNA into a eukaryotic cell as a strategy for creating transgenic cell lines or for gene therapy. This process is similar to the mechanism by which bacteriophages transfer DNA between bacteria through the process of transduction.

Thus, studying the mechanisms of lateral gene transfer is both helpful and relevant to understanding the processes by which eukaryotes acquire new genes naturally or through intentional genetic engineering.

Section: 6.1 Skill: Synthesis/Evaluation

4) How are selective media used in studying bacterial recombination?

Answer: Selective media permits certain bacteria to grow while suppressing the growth of other bacteria. During bacterial recombination, you can accurately assess transfer of genes using selective media in two ways.

First, you can use selective media to identify bacteria that have acquired a resistance gene and are recombinant. For example, E. coli can be transformed with a plasmid containing your gene of interest and a selectable marker, often an antibiotic resistance gene. You can add antibiotic to a bacterial culture and only the bacteria that have taken up the plasmid will be resistant. This will allow for identification of sequences which have been transferred between bacteria.

Second, you can use selective media to identify bacteria which have a mutation in a metabolic pathway by looking for mutants that are unable to complete normal metabolic processes. For instance, let's say you are studying a metabolic pathway and wish to identify bacteria that have mutations in this pathway.

You can grow the bacteria in media with and without the final product of the pathway. If the pathway is mutated and one of the enzymes is absent, the bacteria must be supplemented with certain nutrients to be able to grow. This will allow you to identify transformants, or mutants of interest.

Section: 6.1 Skill: Application/Analysis 5) Explain when, why, and how recombination in bacteria is like crossing over in diploid cells.

Answer: Both bacterial recombination and crossing over during meiosis require several factors. First, both processes require nicking of DNA. In bacteria, this is when the DNA is nicked in preparation for the transfer of single-stranded DNA via conjugation. In eukaryotes, one strand of the DNA is nicked during the formation of chiasmata. This allows for strand invasion and crossing over between homologous chromosomes.

Second, both processes use similar enzymes to mediate recombination. DNA helicases are involved in relieving torsional strain in the circular and linear chromosomes. Ligases are necessary for repairing the phosphodiester bonds within the DNA backbone that were created by the single-stranded breaks. Singlestranded binding proteins (SSBs) assist in stabilizing the otherwise unstable ssDNA and allow it to undergo processes such as replication, recombination, and transcription.

Third, both processes require pairing of homologous sequences. In bacteria, when the chromosome DNA from the donor cell is transferred to a recipient bacterium, the homologous parts of the donor and recipient DNA molecules can undergo recombination that leads to a change in the genotype of the recipient cell. In eukaryotes, proper pairing of homologous sequences is necessary for even exchange of chromosomes during crossing over. If homologous sequences are unevenly pairing, this can result in a region of chromosome being deleted or duplicated in the resulting gametes.

Section: 6.4 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 7 DNA Structure and Replication 7.1 Multiple-Choice Questions 1) In 1928, Griffith found that injecting which strains of Pneumococcus caused the mouse to die?

A) living type RII B) heat-killed type SIII C) heat-killed type RII D) heat-killed type SIII and living type RII E) heat-killed type RII and living type RII Answer: D Section: 7.1 Skill: Knowledge/Comprehension 2) The process by which Pneumococcus transfers DNA between living type RII and heat-killed type SIII cells is known as A) transformation.

B) replication.

C) transduction.

D) ligation.

E) conjugation.

Answer: A Section: 7.1 Skill: Knowledge/Comprehension 3) Avery, Macleod, and McCarty expanded on Griffith's experiment to prove that DNA is the hereditary molecule required for transformation. What treatment of the heat-killed SIII bacteria extract resulted in the mouse living?

A) destroying type SIII lipids and polysaccharides B) destroying type SIII DNA with DNase C) destroying type SIII RNA with RNase D) destroying type SIII proteins with protease E) the control group, in which all components are intact Answer: B Section: 7.1 Skill: Knowledge/Comprehension 4) In the Hershey-Chase experiment, what was labeled by growing bacteriophage in 32P-containing medium?

A) ghost phage particles B) proteins C) lipids D) DNA E) RNA Answer: D Section: 7.1 Skill: Knowledge/Comprehension 5) Which of the following are classified as pyrimidines?

A) adenine and thymine B) guanine and cytosine C) adenine and guanine D) thymine and cytosine E) adenine and uracil

Answer: D Section: 7.2 Skill: Knowledge/Comprehension 6) What type of bond is formed between the hydroxyl group of one nucleotide and the phosphate group of an adjacent nucleotide, forming the sugar-phosphate backbone of DNA? A) glycosidic bond B) phosphodiester bond C) ester linkage D) hydrogen bond E) ionic bond Answer: B Section: 7.2 Skill: Knowledge/Comprehension 7) What is the difference between a nucleotide and a nucleoside?

A) Nucleotides are found in DNA, while nucleosides are found in RNA. B) Nucleotides are involved in eukaryotic DNA replication, while nucleosides are used in bacterial DNA replication.

C) Nucleosides contain only deoxyribose sugars.

D) A nucleoside with a phosphate ester linked to the sugar is a nucleotide.

E) Nucleosides are purines, while nucleotides are pyrimidines.

Answer: D Section: 7.2 Skill: Knowledge/Comprehension 8) What types of bonds are formed between complementary DNA bases?

A) glycosidic bonds B) phosphodiester bonds C) covalent bonds D) hydrogen bonds E) ionic bonds Answer: D Section: 7.2 Skill: Knowledge/Comprehension

9) If complementary DNA strands were arranged in a parallel manner, what would you expect to see?

A) Complementary nucleotides would be attracted to each other, forming ionic bonds that would make the helix stable but not uniform in width.

B) Charges of complementary nucleotides would repel and no hydrogen bonds would form.

C) Complementary nucleotides would line up properly; but fewer hydrogen bonds would form, so the strands could be more easily pulled apart.

D) Purines would pair with both purines and pyrimidines due to the closer alignment of the parallel strands.

E) There would be no discernable difference between DNA strands aligned in a parallel versus antiparallel manner.

Answer: B Section: 7.2 Skill: Application/Analysis 10) What is one difference between DNA replication in bacteria versus eukaryotes?

A) Eukaryotic chromosomes are replicated bidirectionally, while bacterial chromosomes are replicated in one direction.

B) Eukaryotic chromosomes have many origins of replication and replicate bidirectionally, while bacteria have only one origin of replication and replicate unidirectionally.

C) Bacterial chromosomes are replicated bidirectionally, while eukaryotic chromosomes are replicated in one direction.

D) Eukaryotic chromosomes have many origins of replication, while bacteria have only one origin of replication.

E) The process is identical in bacterial and eukaryotic DNA replication.

Answer: D Section: 7.3 Skill: Application/Analysis 11) During DNA replication in E. coli, what is the role of DnaA? A) DnaA is a region of DNA recognized by the replisome where replication is initiated.

B) DnaA bends DNA, hydrolyzing the hydrogen bonds in the 13-mer region of oriC and creating an open complex.

C) DnaA is a helicase that uses ATP to hydrolyze hydrogen bonds and separate DNA strands.

D) DnaA carries the helicase protein to the open region of DNA where hydrolysis and strand separation will occur.

E) DnaA prevents reannealing of the separated strands so replication can proceed.

Answer: B Section: 7.4 Skill: Application/Analysis

12) If SSB is not present during DNA replication, what would you expect to see?

A) The replisome complex would not assemble on the oriC region.

B) The DNA cannot bend, so hydrogen bonds in the 13-mer region of oriC remain intact.

C) Helicase activity is inhibited, so DNA strands cannot be separated.

D) SSB carries the helicase protein to the open region of DNA, so hydrolysis and strand separation will not occur.

E) SSB prevents reannealing of the separated strands, so strands would quickly reanneal and DNA replication cannot proceed.

Answer: E Section: 7.4 Skill: Application/Analysis 13) DNA helicase inhibitors are well studied as potential drug targets. What would you expect to see if DNA helicase activity is inhibited?

A) The replisome complex would not assemble on the oriC region.

B) The DNA cannot bend, so hydrogen bonds in the 13-mer region of oriC remain intact.

C) Helicase catalyzes ATP hydrolysis and DNA strands separation, so the helix cannot be unwound and strands will not separate.

D) Helicase carries the SSB protein to the open region of DNA, so hydrolysis and strand separation will not occur.

E) Helicase prevents reannealing of the separated strands, so strands would quickly reanneal and DNA replication cannot proceed.

Answer: C Section: 7.4 Skill: Application/Analysis 14) What is required for DNA polymerase to initiate DNA strand synthesis?

A) a short DNA primer synthesized by the enzyme primase B) a short RNA primer synthesized by the enzyme primase C) ATP and a short DNA primer synthesized by the enzyme topoisomerase D) ATP and a short RNA primer synthesized by the enzyme topoisomerase E) DNA polymerase initiates DNA strand synthesis without requiring any additional enzymes.

Answer: B Section: 7.4 Skill: Knowledge/Comprehension 15) What enzyme(s) is/are responsible for removal of RNA primers and joining of Okazaki fragments?

A) DNA polymerase I B) DNA polymerase III C) DNA ligase D) DNA polymerase I and DNA ligase E) DNA polymerase III and DNA ligase Answer: D Section: 7.4 Skill: Knowledge/Comprehension

16) DNA polymerase III is functional, but it seems to exhibit extremely low processivity. This is likely a defect in what structure?

A) the τ proteins B) the pol III holoenzyme C) the clamp loader D) the topoisomerase enzyme E) the sliding clamp Answer: E Section: 7.4 Skill: Application/Analysis 17) The extraordinary accuracy of the DNA polymerase III enzyme lies in its ability to "proofread" newly synthesized DNA, a function of the enzyme's A) 3′-to-5′ exonuclease activity.

B) 5′-to-3′ exonuclease activity.

C) 3′-to-5′ polymerase activity.

D) 5′-to-3′ polymerase activity.

E) 3′-to-5′ helicase activity.

Answer: A Section: 7.4 Skill: Knowledge/Comprehension 18) Which of these components would you use to distinguish whether a sample has been prepared for a Sanger sequencing reaction or for a polymerase chain reaction?

A) DNA template B) DNA polymerase C) dNTPs D) ddNTPs E) DNA primer Answer: D Section: 7.5 Skill: Synthesis/Evaluation 19) Which functional groups have been altered in a ddNTP compared to a dNTP? A) The ddNTPs have a 2′ OH and a 3′ H, while dNTPs have a 2′ H and a 3′ OH. B) The ddNTPs have a 2′ H and a 3′ H, while dNTPs have a 2′ H and a 3′ OH. C) The ddNTPs have a 2′ OH and a 3′ OH, while dNTPs have a 2′ H and a 3′ H. D) The ddNTPs have a 2′ H and a 3′ H, while dNTPs have a 2′ OH and a 3′ OH. E) The ddNTPs have a 2′ H and a 3′ OH, while dNTPs have a 2′ H and a 3′ H. Answer: B Section: 7.5 Skill: Knowledge/Comprehension

20) Which of the following temperature cycles would you expect to see in a standard polymerase chain reaction (from denaturation to annealing to extension)?

A) 95° → 72° → 55° B) 45° → 72° → 95° C) 95° → 55° → 72° D) 95° → 72° → 95° E) 75° → 95° → 45° Answer: C Section: 7.5 Skill: Knowledge/Comprehension 7.2 Short-Answer Questions 1) Based on Griffith's results, what would you expect if you injected both heat-killed type RII and living type SIII? Answer: The mouse lives.

Section: 7.1 Skill: Application/Analysis 2) What are the three parts of a DNA nucleotide?

Answer: deoxyribose, base, phosphate Section: 7.1 Skill: Application/Analysis 3) What is the molecular diameter of the DNA helix (in angstroms)?

Answer: 20 Ǻ Section: 7.2 Skill: Knowledge/Comprehension 4) The regions where DNA-binding proteins can make direct contact with exposed nucleotides are known as what?

Answer: major and minor grooves Section: 7.2 Skill: Knowledge/Comprehension 5) The major and minor grooves of DNA are features of the helix that can be attributed to this process, which involves tight packing of DNA bases in a duplex.

Answer: base stacking Section: 7.2 Skill: Knowledge/Comprehension 6) DNA polymerase catalyzes DNA strand elongation by removing what from the dNTPs? Answer: two phosphates (pyrophosphate group) Section: 7.2 Skill: Knowledge/Comprehension

7) What type of experimental technique did Huberman and Riggs use to produce the first evidence of bidirectional replication in bacteria?

Answer: pulse-chase labeling Section: 7.3 Skill: Application/Analysis 8) What is the purpose of the DNA sequences found at the origins of replication?

Answer: binding replication enzymes Section: 7.4 Skill: Knowledge/Comprehension 9) What well-characterized sequence in E. coli is 245 bp, AT-rich, and divided into three 13-mers and four 9-mers?

Answer: oriC Section: 7.4 Skill: Knowledge/Comprehension 10) In E. coli, which protein uses ATP to hydrolyze hydrogen bonds holding complementary strands together?

Answer: DnaB Section: 7.4 Skill: Knowledge/Comprehension 11) During DNA replication in E. coli, how does the cell prevent the DNA strands from their natural tendency to seek maximum stability and reanneal?

Answer: SSB prevents reannealing.

Section: 7.4 Skill: Application/Analysis 12) What is the overall direction of DNA strand elongation?

Answer: 5′ to 3′ Section: 7.4 Skill: Knowledge/Comprehension 13) What structure in E. coli delivers primase and accessory proteins to the oriC site to synthesize the primer needed for DNA replication?

Answer: primosome Section: 7.4 Skill: Knowledge/Comprehension 14) During DNA replication, the continuously elongated daughter strand is referred to as what?

Answer: leading strand Section: 7.4 Skill: Knowledge/Comprehension 15) What did Okazaki's discovery of the short segments of DNA reveal about DNA replication on the lagging strand?

Answer: It is discontinuous.

Section: 7.4

Skill: Application/Analysis 16) Eukaryotic polymerases interact with which protein that acts as the sliding clamp in eukaryotic DNA replication?

Answer: PCNA Section: 7.4 Skill: Knowledge/Comprehension 17) Which enzyme, a combination of several proteins and a molecule of tRNA, acts as the template for the telomere repeat DNA sequence?

Answer: telomerase Section: 7.4 Skill: Knowledge/Comprehension 18) How might you choose a region of DNA for a PCR primer so as to increase the temperature necessary for primer annealing (to minimize nonspecific PCR products)?

Answer: higher GC concentrations Section: 7.5 Skill: Application/Analysis 19) What characteristic of the polymerase isolated from Thermus aquaticus makes it unique and highly useful for maintaining the enzyme efficiency during PCR? Answer: heat stability Section: 7.5 Skill: Knowledge/Comprehension 20) What are the three steps of each PCR cycle?

Answer: denaturation, annealing, extension Section: 7.5 Skill: Knowledge/Comprehension 7.3 Fill-in-the-Blank Questions 1) The large, complex aggregation of proteins and enzymes that assembles at the replication fork is known as the ________.

Answer: replisome Section: 7.4 Skill: Knowledge/Comprehension 2) Comparison of conserved sequences among related species usually leads to the identification of ________, which illustrate the nucleotides most often found at each position of DNA in the conserved region.

Answer: consensus sequences Section: 7.4 Skill: Knowledge/Comprehension 3) DNA polymerase requires a primer sequence to provide a ________ functional group to which the new DNA nucleotide can be added.

Answer: 3′ OH

Section: 7.4 Skill: Knowledge/Comprehension 4) ________ is the term for multiprotein complexes in which a core enzyme is associated with additional protein components leading to its function.

Answer: Holoenzyme Section: 7.4 Skill: Knowledge/Comprehension 5) ________ must be added to a PCR to direct the polymerase where to begin synthesis.

Answer: PCR primers Section: 7.5 Skill: Knowledge/Comprehension 7.4 Essay Questions 1) Describe the DNA helix structure proposed by Watson and Crick. What is the diameter of the helix and the distance between bases? What types of bonds are involved in stabilizing the DNA backbone, and what types of bonds are found between complementary base pairs? Why would a sequence containing high concentrations of G and C require more energy to break the bonds?

Answer: The DNA model contains two DNA polymer strands in a right-handed twist around a common axis. The strands are orientated in an antiparallel direction from 5′ to 3′ and 3′ to 5′. The backbone consists of alternating sugar-phosphate groups stabilized by phosphodiester bonds, and it is located on the outside of the helix. Complementary base pairs are stacked in the center of the helix and are stabilized by hydrogen bonds. There are three hydrogen bonds between G and C, but only two between A and T. Thus, GC-rich regions require higher temperatures to break the bonds. The base pairs, which are perpendicular to the axis, are stacked with 10 base pairs per turn and a helical repeat of 34 Ǻ.

Section: 7.2 Skill: Synthesis/Evaluation 2) Briefly describe the Meselson and Stahl experiment that indicated that DNA replication is semiconservative. How would their results have differed if DNA replication was actually conservative in nature?

Answer: Meselson and Stahl grew bacteria in the presence of 15N and then replaced the 15N media with the "lighter" 14N media. DNA was extracted at various intervals to collect several successive generations. Using high-speed CsCl density-gradient ultracentrifugation, they observed that after the first generation, the "heavy" DNA (15N/15N) was not present, and the DNA was intermediate between light and heavy (14N/15N). After the second generation, there were two bands with equal amounts of intermediate and light bands (14N/15N and 14N/14N). These results validated the semiconservative model, in which one strand of the parent DNA is present in each daughter strand of replicated DNA. If the conservative model were correct, they would have expected to observe DNA molecules with two distinct densities after generation 1 (15N/15N and 14N/14N). Their first-generation results excluded this model.

Section: 7.2 Skill: Synthesis/Evaluation

3) Describe the function of the oriC. What do you predict will happen to the cell if 100 bp between the 13-mer and 9-mer sequence is deleted as indicated in the figure?

Answer: A consensus sequence is the portion of a conserved region that carries out the essential function that is the focus of natural selection. The 13-mer and 9-mer consensus sequences that are part of oriC have been maintained by natural selection because they have essential functional roles in replication initiation, as we explain in the following section. Beyond the presence of the consensus sequences themselves, natural selection also acts to maintain specific spacing between the consensus sequences. Spacing is important because DNA-binding proteins must assemble at sites where consensus sequences are located. Different proteins are attracted to different consensus sequences, and each protein must have the physical space to bind to DNA and to interact with the other proteins bound to the consensus sequence region. The spacing between consensus sequence elements is also important to allow room on the DNA molecule for protein binding and interaction between bound proteins. If you delete this region, replication will not proceed because there is not enough space for replisome assembly.

Section: 7.4 Skill: Synthesis/Evaluation 4) Many antibacterial agents and chemotherapy agents act as topoisomerase inhibitors. Why would inhibition of topoisomerase cause cell death in bacterial and eukaryotic cells?

Answer: During DNA replication, as DNA strands unwind it creates torsional stress that accumulates as the unwound region gets larger. Without free ends, covalently closed circular DNA readily accumulates superhelical twists (supercoiled DNA). The accumulating stress could break the DNA template at random locations, potentially leading to a breakdown of DNA replication and, ultimately, to cell death.

Topoisomerases prevent this lethal event by catalyzing a controlled cleavage and rejoining of DNA, thus enabling over-twisted strands to unwind. Topoisomerase inhibitors block the ligation step of the cell cycle, generating single- and double-stranded breaks that harm the integrity of the genome and ultimately result in apoptosis of the cancer cells or bacteria.

Section: 7.4 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 8 Molecular Biology and Transcription and RNA Processing

8.1 Multiple-Choice Questions 1) Which type of mRNA is translated within a cell?

A) tRNA B) mRNA C) rRNA D) siRNA E) all of the above Answer: B Section: 8.1 Skill: Knowledge/Comprehension 2) Which type of research technique was used to track newly synthesized RNA within a eukaryotic cell?

A) in situ hybridization B) band shift assay C) pulse-chase D) Southern blotting E) DNA footprint protection assay Answer: C Section: 8.1 Skill: Knowledge/Comprehension 3) If bacteria are infected with a bacteriophage and briefly exposed to radioactive uracil, what results would you expect to see immediately after exposure?

A) radioactive mRNA in the cytoplasm B) radioactive mRNA in the nucleus C) radioactive DNA in the cytoplasm D) radioactive DNA in the nucleus E) radioactive proteins in the nuclear membrane Answer: A Section: 8.1 Skill: Knowledge/Comprehension 4) You wish to create a mutation in which you prevent access of RNA polymerase to the gene. Which region of a gene would you target?

A) coding region B) start codon C) promoter sequence D) terminator sequence E) stop codon Answer: C Section: 8.2 Skill: Application/Analysis

5) A gene has acquired a mutation in which the protein product has an "extra" 50 amino acids at the end.

Which region of the gene was likely mutated?

A) coding region B) start codon C) promoter sequence D) terminator sequence E) stop codon Answer: E Section: 8.2 Skill: Application/Analysis 6) How does the antibiotic rifampicin prevent transcription of RNA? A) Rifampicin blocks binding of RNA polymerase to the promoter.

B) Rifampicin prevents chain elongation by preventing formation of the phosphodiester bond.

C) Rifampicin prevents the chain termination, so transcripts are too long and rendered nonfunctional.

D) Rifampicin blocks ribosomal RNAs from forming functional ribosomes.

E) Rifampicin blocks translation by preventing the mRNA from being processed.

Answer: B Section: 8.2 Skill: Knowledge/Comprehension 7) Which region(s) of a gene are not found within the mRNA transcript?

A) promoter region B) termination region C) stop codon D) promoter and stop codon E) promoter and termination regions Answer: E Section: 8.2 Skill: Knowledge/Comprehension 8) What is the consensus sequence for the Pribnow box from these sequences?

A) TTGATA B) TATGAT C) TATTAT D) ACCA E) TAGTAT Answer: B Section: 8.2 Skill: Application/Analysis

9) What is the -35 consensus sequence for the following sequences?

A) TTGATA B) TATGAT C) TATTAT D) ACAA E) TAGTAT Answer: A Section: 8.2 Skill: Application/Analysis 10) In a given bacterium, transcription of housekeeping genes is normal, but genes involved in nitrogen metabolism, stress, and chemotaxis are disrupted. Which sigma subunit is intact?

A) σ28 B) σ32 C) σ35 D) σ54 E) σ70 Answer: E Section: 8.2 Skill: Application/Analysis 11) If you want to prevent rho-dependent transcription, which sequences would you target to mutate the rho utilization (rut) site?

A) coding region B) start codon C) promoter sequence D) terminator sequence E) stop codon Answer: D Section: 8.3 Skill: Application/Analysis

12) You want to design a drug that prevents transcription of mRNAs but does not affect transcription of other RNAs. What enzyme would you target?

A) RNA polymerase I B) RNA polymerase II C) RNA polymerase III D) ribozyme E) methyl transferase Answer: B Section: 8.3 Skill: Application/Analysis 13) You suspect a region of DNA contains a promoter and wish to confirm that a sequence of interest binds proteins. Which experiment would you use?

A) in situ hybridization B) band shift assay C) pulse-chase assay D) Southern blotting E) northern blotting Answer: B Section: 8.3 Skill: Application/Analysis 14) Which assay allows you to identify the exact location of the protein-binding sequence within a promoter?

A) western blotting B) in situ hybridization C) pulse-chase assay D) Southern blotting E) DNA footprint protection assay Answer: E Section: 8.3 Skill: Application/Analysis

15) Following a DNA footprint protection assay, which banding pattern would you expect to see if you identify a potential promoter region?

A)

B)

C)

D)

E)

Answer: C

Section: 8.3 Skill: Application/Analysis 16) Nucleoli contain many tandem repeat copies of which genes, transcribed by RNA polymerase I? A) mRNA B) tRNA C) rRNA D) siRNA E) all of the above Answer: C Section: 8.3 Skill: Knowledge/Comprehension 17) Which enzyme is required to initiate 5′ capping of eukaryotic mRNA transcripts by removing the terminal phosphate group?

A) ribozyme B) guanylyl transferase C) methyl transferase D) adenylyl cyclase E) phosphodiesterase Answer: B Section: 8.4 Skill: Knowledge/Comprehension 18) A cell has a defect in polyadenylation. Which RNA transcripts would not be affected by this defect because they are not polyadenylated?

A) histone proteins B) transcription factors C) transmembrane proteins D) DNA binding proteins E) SR proteins Answer: A Section: 8.4 Skill: Application/Analysis 19) A cell exhibits specific defects in the alternative intron splicing pathway. Which proteins are likely mutated in this cell?

A) histone proteins B) transcription factors C) transmembrane proteins D) DNA binding proteins E) SR proteins Answer: E Section: 8.4 Skill: Application/Analysis

20) In humans, the 30S pre-RNA transcript yields three rRNA segments following transcription and RNA cleavage. What RNA transcripts are generated by the 30S pre-RNA transcript in E. coli?

A) two rRNAs (15S each) B) three rRNAs (5S, 16S, and 23S) C) two tRNA transcripts D) three rRNAs (5.8S, 18S, and 28S) and two tRNA transcripts E) three rRNAs (5S, 16S, and 23S) and two tRNA transcripts Answer: E Section: 8.4 Skill: Knowledge/Comprehension 8.2 Short-Answer Questions 1) What are the two chemical structures found in RNA but not DNA? Answer: ribose sugar, uracil base Section: 8.1 Skill: Knowledge/Comprehension 2) Which three types of RNAs are found in both prokaryotes and eukaryotes?

Answer: mRNA, tRNA, and rRNA Section: 8.1 Skill: Application/Analysis 3) Name three functional RNAs unique to eukaryotes.

Answer: snRNA, siRNA, and miRNA Section: 8.1 Skill: Knowledge/Comprehension 4) miRNA is capable of regulating protein production through which process?

Answer: RNA interference Section: 8.1 Skill: Knowledge/Comprehension 5) Which type of RNA interacts with nuclear proteins to form a ribonucleoprotein complex responsible for intron removal?

Answer: snRNA Section: 8.1 Skill: Knowledge/Comprehension 6) Catalytically active RNAs that can activate processes such as self-splicing are called what?

Answer: ribozymes Section: 8.2 Skill: Knowledge/Comprehension 7) You wish to prevent transcription of all three types of RNA in bacteria. How many different types of RNA polymerase would you need to inhibit?

Answer: one Section: 8.2 Skill: Application/Analysis

8) What are the mechanisms for transcription termination in bacteria?

Answer: intrinsic and rho-dependent Section: 8.2 Skill: Knowledge/Comprehension 9) Transcription of inverted repeats produces an mRNA with complementary segments that fold to form what type of secondary structure?

Answer: stem-loop (hairpin) Section: 8.2 Skill: Knowledge/Comprehension 10) What molecule is required for activation of rho protein?

Answer: ATP Section: 8.2 Skill: Knowledge/Comprehension 11) Eukaryotic promoters are diverse, which means that eukaryotes have how many different RNA polymerases?

Answer: 3 Section: 8.3 Skill: Knowledge/Comprehension 12) A plant cell exhibits defects in transcription of transfer RNA genes. Which polymerase is likely to be mutated in this cell?

Answer: RNA polymerase III Section: 8.3 Skill: Knowledge/Comprehension 13) Prokaryotes have a -10 and -35 consensus sequence in their promoter. What are the three eukaryotic promoter sequences elements (or "boxes")?

Answer: GC-rich, CAAT, and TATA boxes Section: 8.3 Skill: Knowledge/Comprehension 14) What proteins aid in the recognition of the promoter sequence and binding of RNA polymerase II in eukaryotes?

Answer: transcription factors (TFs) Section: 8.3 Skill: Knowledge/Comprehension 15) What proteins are bound to enhancers, forming a protein "bridge" that bends the DNA over the promoter to initiate transcription?

Answer: activator proteins Section: 8.3 Skill: Knowledge/Comprehension

16) What proteins bind to silencer sequences, forming a protein "bridge" that bends the DNA over the promoter and prevents transcription?

Answer: repressor proteins Section: 8.3 Skill: Knowledge/Comprehension 17) Spliceosome components are recruited to 3′ and 5′ splice sites by SR proteins bound to what DNA sequences?

Answer: exonic splicing enhancers Section: 8.4 Skill: Knowledge/Comprehension 18) The rat α-tropomyosin (α-Tm) gene produces nine different mature mRNAs proteins from a single gene using which three "alternative" mechanisms?

Answer: splicing, promoters, and polyadenylation Section: 8.4 Skill: Application/Analysis 19) Processing of the 30S pre-RNA transcript in humans produces what three rRNAs after enzymatic cleavage?

Answer: 5.8S, 18S, 28S Section: 8.4 Skill: Knowledge/Comprehension 20) One type of RNA editing involves inserting uracils into edited mRNA with the assistance of which type of specialized RNA? Answer: guide RNA (gRNA) Section: 8.4 Skill: Knowledge/Comprehension 8.3 Fill-in-the-Blank Questions 1) RNA polymerase is called a ________, meaning it is an intact complex with full enzymatic capacity.

Answer: holoenzyme Section: 8.2 Skill: Knowledge/Comprehension 2) In intrinsic termination, inverted repeat DNA sequences followed immediately by a string of ________ produce an mRNA stem-loop followed by a string of ________.

Answer: adenines; uracils Section: 8.2 Skill: Knowledge/Comprehension 3) The 5S rRNA and tRNA genes have internal promoters located downstream of the transcription start site, called ________.

Answer: internal control regions Section: 8.3 Skill: Knowledge/Comprehension

4) Polyadenylation begins with the binding of ________ near a six-nucleotide mRNA sequence, AAUAAA, downstream of the stop codon.

Answer: cleavage and polyadenylation specificity factor (CPSF) Section: 8.4 Skill: Knowledge/Comprehension 5) Intron self-splicing occurs when two ________ reactions excise the intron and allow exons to ligate.

Answer: transesterification Section: 8.4 Skill: Knowledge/Comprehension 8.4 Essay Questions 1) In the following prokaryotic DNA sequence, label the highlighted regions.

Using the prokaryotic DNA sequence above, describe the relationship between the highlighted regions and how mutations in each region may affect gene expression.

Answer: The -10 (Pribnow) and -35 consensus sequences are important because of their nucleotide content, their location relative to one another, and their location relative to the start of transcription.

Mutation in a consensus sequence is likely to alter how efficiently a protein binds to the promoter, and to decrease the amount of gene transcription. In contrast, mutations between consensus sequences are unlikely to alter gene transcription, because the sequences in these intervening regions do not bind tightly to RNA polymerase. The nucleotides between -10 and -35 are important as spacers between the consensus elements, but their specific sequences are not critical.

Section: 8.2 Skill: Synthesis/Evaluation 2) Describe how band shift assays and DNA footprint protection assays can be used to identify promoter regions in DNA. Answer: A band shift assay uses two identical samples of DNA fragments that contain suspected promoter consensus sequences. The control sample has no transcriptional proteins added, while the experimental sample includes transcriptional proteins.

DNA footprint protection also begins with two identical samples of DNA fragments containing suspected promoter consensus sequences. All fragments are end-labeled with 32P, and once again the experimental DNA is mixed with transcriptional proteins, but the control sample is not. In this experiment, both samples are exposed to DNase I that randomly cuts DNA that is not protected by protein. The samples are run in separate lanes of an electrophoresis gel, and each end-labeled fragment produced is identified by autoradiography.

Section: 8.3 Skill: Synthesis/Evaluation

3) Describe the mechanism(s) used by bacteria and eukaryotes for transcription termination.

Answer: Bacterial transcription termination can be intrinsic or rho-dependent, and the type of termination is dependent on the presence of certain DNA sequences. In intrinsic termination, inverted repeat DNA sequences followed immediately by a string of adenines produce an mRNA stem-loop followed by a poly-U string. In rho-dependent termination, a rho protein binds to mRNA and catalyzes the release of the transcript from RNA polymerase.

Each of the eukaryotic RNA polymerases uses a different mechanism to terminate transcription.

Transcription by RNA polymerase III is terminated in a mechanism that resembles E. coli intrinsic termination. The RNA pol III transcribes a terminator sequence that creates a string of uracils in the transcript. The RNA pol III terminator sequence does not contain an inverted repeat, however, so there is no stem-loop structure near the 3′ end of RNA. Transcription by RNA pol I is terminated at a 17-bp consensus sequence that binds transcriptionterminating factor I (TTFI). A large rRNA precursor transcript is cleaved about 18 nucleotides upstream of the TTFI binding site, so the consensus sequence does not appear in the mature transcript.

Termination of transcription by RNA pol II is not completely clear, but it is known that the 3′ end of mRNA is not generated by transcriptional termination. The 3′ end of the pre-mRNA is created by enzymatic action that removes a segment from the 3′ end of the transcript and replaces it with a string of adenine nucleotides, the poly-A tail. This step of pre-mRNA processing is thought to be associated with subsequent termination of transcription.

Section: 8.3 Skill: Synthesis/Evaluation 4) Describe the three main difference bacterial and eukaryotic transcripts.

Answer: • Eukaryotic transcripts are more stable than bacterial transcripts.

• Eukaryotes exhibit a separation, in time and in location, between transcription and translation.

Bacteria lack a nucleus, which leads to coupling of transcription and translation.

• Eukaryotic genes have introns, which also leads to alternative splicing of transcripts. Bacteria do not have introns, and thus do not have intron splicing mechanisms.

Section: 8.4 Skill: Synthesis/Evaluation

5) What are the three types of posttranscriptional processing, and what are the consequences of preventing each of these modifications in terms of gene expression?

Answer: (a) 5′ capping, the addition of a modified nucleotide to the 5′ end of mRNA. The 5′ cap has several functions, including (1) protecting mRNA from rapid degradation, (2) facilitating mRNA transport across the nuclear membrane, (3) facilitating subsequent intron splicing, and (4) enhancing translation efficiency by orienting the ribosome on mRNA. (b) 3′ polyadenylation, cleavage at the 3′ end of mRNA and addition of a tail of multiple adenines to form the poly-A tail. The 3′ poly-A tail has several functions, including (1) facilitating transport of mature mRNA across the nuclear membrane, (2) protecting mRNA from degradation, and (3) enhancing translation by enabling ribosomal recognition of messenger RNA. (c) Intron splicing, RNA splicing to remove introns and ligate exons. Intron splicing requires exquisite precision to remove all intron nucleotides accurately without intruding on the exons, and without leaving behind additional nucleotides, so that the mRNA sequence encoded by the ligated exons will completely and faithfully direct synthesis of the correct polypeptide.

Disrupting any one of the transcriptional processes would ultimately disrupt normal gene expression by resulting in mRNA degradation, reduced translation, or improper splicing and, thus, production of the wrong gene product.

Section: 8.4 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 9 The Molecular Biology of Translation 9.1 Multiple-Choice Questions 1) What method can be used to separate ribosomal protein subunits?

A) polymerase chain reaction B) gel electrophoresis C) two-dimensional gel electrophoresis D) CsCl gradient ultracentrifugation E) cryo-EM Answer: C Section: 9.1 Skill: Knowledge/Comprehension 2) The Shine-Dalgarno sequence A) is a purine-rich consensus sequence found in the 5′ UTR of the mRNA. B) is a pyrimidine-rich consensus sequence found in the 3′ UTR of the mRNA. C) is a purine-rich consensus sequence found in the 16S rRNA subunit.

D) is a region of the tRNA molecule involved in formation of charged tRNAs. E) is a consensus sequence involved in the termination of translation.

Answer: A Section: 9.2 Skill: Knowledge/Comprehension 3) During translation initiation in prokaryotes, the amino acid on the initiator tRNA is

A) methionine (Met).

B) N-formylmethionine (fMet).

C) acetylated.

D) IF-1.

E) added using ATP as the energy source.

Answer: B Section: 9.2 Skill: Knowledge/Comprehension 4) Identification of ribosomal proteins involved two-dimensional gel electrophoresis, which separates the proteins on the basis of A) mass.

B) folded shape.

C) charge.

D) mass and charge.

E) mass, charge, and folded shape.

Answer: D Section: 9.2 Skill: Application/Analysis

5) How does the eukaryotic initiation complex locate the true start codon?

A) The true start codon is the first ATG encountered downstream of the Shine-Dalgarno sequence.

B) The initiation complex moves the small ribosomal subunit through the 5′ UTR, scanning for the start AUG. C) The preinitiation complex moves the ribosome through the 3′ UTR, scanning for the Kozak sequence.

D) The true start codon is the first ATG encountered downstream of the Kozak sequence.

E) The true start codon is the formyl-ATG, which will encode for fMet in the protein.

Answer: B Section: 9.2 Skill: Application/Analysis 6) Which step in translation initiation is unique to eukaryotes?

A) formation of the preinitiation complex B) formation of the initiation complex C) scanning D) ribosome assembly E) translocation of the ribosome in the 3′ direction Answer: C Section: 9.3 Skill: Knowledge/Comprehension 7) Elongation factors translocate the ribosome in the 3′ direction by a distance of A) one codon.

B) one nucleotide.

C) two nucleotides.

D) two codons.

E) three codons.

Answer: A Section: 9.3 Skill: Knowledge/Comprehension 8) What would you expect to find bound to the stop codon at the A site?

A) a charged tRNA with the anticodon TAG B) a charged tRNA with the anticodon ATC C) an uncharged tRNA D) a release factor E) Nothing binds to a stop codon, which is why the peptide is released.

Answer: D Section: 9.2 Skill: Knowledge/Comprehension 9) A polycistronic mRNA contains multiple what?

A) promoters B) Shine-Dalgarno sequences C) Kozak sequences D) mRNAs E) polypeptide-producing sequences Answer: E Section: 9.3

Skill: Knowledge/Comprehension 10) Which of the following are synonymous codons?

A) UUC and UUG B) UCC and UAC C) CGA and AGA D) AGG and AGC E) AAG and AAC Answer: C Section: 9.4 Skill: Application/Analysis 11) What result would you expect if you mutate one of the four arms of a tRNA? A) The tRNA will fit into the A site but will not release the peptide at the P site.

B) The tRNA will not be recognized by tRNA synthetase and cannot be charged.

C) The tRNA will be charged with the wrong amino acid.

D) The tRNA will not be able to undergo traditional complementary base pairing.

E) There will be no effect on function, so long as the anticodon region is intact.

Answer: B Section: 9.4 Skill: Application/Analysis 12) Heinz Fraenkel-Conrat induced point mutations in the coat proteins of TMV and looked for mutations in the resulting proteins. By examining single-base changes in the mRNA, Conrat proved that A) the genetic code is overlapping.

B) the genetic code is nonoverlapping.

C) the genetic code is redundant.

D) there are 20 amino acids.

E) the genetic code is triplet.

Answer: B Section: 9.5 Skill: Application/Analysis 13) A mutagen has introduced a frame-shift mutation by adding one nucleotide base. Which of the following would be classified as a reversion mutation for this particular mutant?

A) deleting 2 bases B) adding 1 base C) deleting 1 base or adding 1 base D) deleting 1 base or adding 3 bases E) deleting 1 base or adding 2 bases Answer: E Section: 9.5 Skill: Application/Analysis

14) You have identified a bacterial protein that has retained the starting fMet in its protein sequence.

Which of the following is likely true of this protein?

A) It is likely nonfunctional, since bacteria use posttranslational cleavage of fMet to make functional proteins.

B) It will show improper protein sorting and will likely remain in the ER. C) It will be not be able to be chemically modified, so it will be sent to the Golgi for secretion.

D) It will be a functional bacterial protein, since all functional proteins must begin with fMet. E) It will likely form a disulfide bond with a second peptide chain, forming a protein complex.

Answer: A Section: 9.5 Skill: Application/Analysis 15) Experiments by Chapeville and colleagues proved that the genetic code derives its specificity through what interactions?

A) hydrogen bonds and disulfide bonds that maintain the shape of the tRNAs B) complementary base pairing between amino acids and tRNA C) complementary base pairing between tRNA and mRNA D) ionic bonds between the small and large subunits of the ribosome E) hydrogen bonding between the ribosomal subunits and the mRNA Answer: C Section: 9.5 Skill: Knowledge/Comprehension 16) In the unlikely event that a tRNA has been charged with the wrong amino acid, what high-fidelity enzyme is likely to blame?

A) peptidyl transferase B) DNA polymerase I C) DNA polymerase III D) aminoacyl synthetase E) aminoacyl peptidase Answer: D Section: 9.5 Skill: Application/Analysis 17) Which of these choices represents one possible corresponding mRNA sequence that can be transcribed from the following DNA template?

5′ - CTGTATCCTAGCACCCAAATCGCATTAGGAC - 3′ A) 5′ - ATG CGA TTT GGG TGC TAG - 3′ B) 5′ - AUG CGA UUU GGG UGC UAG - 3′ C) 5′ - AUG CGA UUU GGG UGC - 3′ D) 5′ - CTA GCA CCC AAA TCG CAT TAG - 3' E) 3′ - GGA CAU AGG UAC GUG GGU UUA GCG UAA UCC UG - 5′ Answer: B Section: 9.6 Skill: Application/Analysis

18) Given the following mRNA sequence, what is the amino acid sequence for the corresponding polypeptide?

5′ - AUG CGA UUU GGG UGC UAG - 3′ A) 5′-Met-Arg-Phe-Gly-Stop-3′ B) C—Met-Arg-Leu-Glu—N C) N—Met-Arg-Phe-Gly-Cy—C D) N—Met-Asp-Phe-Gly-Trp—C E) N—Arg-Phe-Gly-Stop—C Answer: C Section: 9.6 Skill: Application/Analysis 19) Given the following mRNA sequence, which of the following mutations would affect the protein sequence?

5′ - AUG CAG UUA GCG UGC UAG - 3′ A) 5′ - AUG CAA UUA GCG UGC UAG - 3′ B) 5′ - AUG CAG UUA GCA UGC UAG - 3′ C) 5′ - AUG CAG UUG GCG UGC UAG - 3′ D) 5′ - AUG CAA UUA GCG UGU UAG - 3′ E) 5′ - AUG CAC UUA GCA UGC UAG - 3′ Answer: E Section: 9.6 Skill: Application/Analysis 20) Which single-base mutation would insert a premature stop codon into the following protein sequence?

N—Met-Gln-Leu-Arg-Cys—C A) 5′ - AUG CAG AUA GCG UGC UAG - 3′ B) 5′ - AUG AAG UUA GCG UGC UAG - 3′ C) 5′ - AUG CAG UAA GCG UGC UAG - 3′ D) 5′ - AUG CAG UUA UUG UGC UAG - 3′ E) 5′ - AUG CAG UUA GCG UGC AAG - 3′ Answer: C Section: 9.6 Skill: Application/Analysis

9.2 Short-Answer Questions 1) Ribosomal subunits are measured in which units, which describe the speed of sedimentation of a substance during centrifugation?

Answer: Svedberg units (S) Section: 9.1 Skill: Application/Analysis 2) What molecule provides the energy to form charged tRNAs? Answer: GTP Section: 9.2 Skill: Knowledge/Comprehension 3) The preinitiation complex forms when the start codon is recognized through binding of 16srRNA and what region of mRNA? Answer: Shine-Dalgarno sequence Section: 9.2 Skill: Knowledge/Comprehension 4) What are the three phases of translation?

Answer: initiation, elongation, termination Section: 9.2 Skill: Knowledge/Comprehension 5) If the pH gradient in a two-dimensional gel is lost, which step during two-dimensional electrophoresis would be affected?

Answer: isoelectric focusing (first dimension) Section: 9.2 Skill: Application/Analysis 6) What reaction provides the energy for the eukaryotic ribosome to search for the start ATG during the process of scanning?

Answer: ATP hydrolysis Section: 9.2 Skill: Application/Analysis 7) During elongation, where does the charged tRNA is recruited to which location on the ribosome?

Answer: A site Section: 9.2 Skill: Knowledge/Comprehension 8) If you are designing an antibiotic that inhibits peptide bond formation, what protein would you target?

Answer: peptidyl transferase Section: 9.2 Skill: Application/Analysis

9) Binding of what protein initiates translation-termination events that result in polypeptide release and dissociation of ribosomal subunits?

Answer: release factor (RF) Section: 9.3 Skill: Knowledge/Comprehension 10) In a polyribosome, which end of the mRNA would have the shortest polypeptides?

Answer: 5′ end Section: 9.3 Skill: Application/Analysis 11) Thanks to flexible base pairing, the wobble nucleotides in anticodons can be any of the RNA nucleotides or which modified nucleotide?

Answer: inosine (I) Section: 9.4 Skill: Knowledge/Comprehension 12) Eighteen of the amino acids have two or more synonymous codons. Which two amino acids are the exceptions?

Answer: Met and Trp Section: 9.4 Skill: Application/Analysis 13) How many different aminoacyl-tRNA synthetases can be found in a given cell?

Answer: 20 Section: 9.4 Skill: Application/Analysis 14) Khorana used synthetic mRNAs to determine genetic code possibilities. To do so, he translated synthetic mRNA in vitro in the presence of individual 14C-labeled amino acids. What polypeptide did he identify using the repeating dinucleotide Poly-UG? Answer: Cys-Val-Cys-Val Section: 9.5 Skill: Application/Analysis 15) Sidney Brenner designed an experiment to determine if the triplet code was overlapping or nonoverlapping. If the code were overlapping, how many complete codons would the following sequence encode before encountering a stop codon?

5′ - AUGCGAUUAUAGUGC - 3′ Answer: 9 (AUG UGC GCG CGA GAU AUU UUA UAU AUA UAG) Section: 9.5 Skill: Application/Analysis 16) Sidney Brenner proved that the triplet code is, in fact, nonoverlapping. Assuming a nonoverlapping code, how many complete codons would the following sequence encode before encountering a stop codon? 5′ - AUGCGAUUAUAGUGC - 3′ Answer: 3 (AUG CGA UUA UAG) Section: 9.5 Skill: Application/Analysis

17) Many antibiotics cause mispairing between codons and anticodons. What fatal effect would this mispairing have on the cell?

Answer: defective protein synthesis Section: 9.6 Skill: Application/Analysis 18) What process is targeted by the antibiotic streptomycin?

Answer: translation initiation Section: 9.6 Skill: Knowledge/Comprehension 19) In the Golgi, what type of protein modification will signal the polypeptide destination by determining the receptor to which the peptide will bind?

Answer: glycosylation Section: 9.6 Skill: Application/Analysis 20) In the ER, misfolded proteins are identified and bound by what molecules?

Answer: chaperones Section: 9.6 Skill: Knowledge/Comprehension 9.3 Fill-in-the-Blank Questions 1) ________ help control ribosome formation and binding of the initiator tRNA. Answer: Initiator factor proteins Section: 9.1 Skill: Knowledge/Comprehension 2) In eukaryotes, the initiation factor proteins eIF1A and eIF3 join with ________ to form the preinitiation complex.

Answer: charged tRNAMet Section: 9.2 Skill: Knowledge/Comprehension 3) Bacteria group their genes such that they share a single promoter and the mRNA transcript synthesizes several different polypeptides. Collectively, these are referred to as ________ mRNAs, which are part of the operon system.

Answer: polycistronic Section: 9.3 Skill: Knowledge/Comprehension 4) Using mathematical reasoning, a triplet genetic code gives a possible ________ different codons, while a doublet genetic code would yield ________ different codons. Thus, the triplet code accounts for 20 known amino acids and indicates redundancy in the genetic code.

Answer: 64; 16 (4n, where n = # of nucleotides per codon) Section: 9.4 Skill: Knowledge/Comprehension

5) A signal sequence at the N-terminus of newly synthesized proteins will direct the protein to the ________ for protein packaging.

Answer: endoplasmic reticulum (ER) Section: 9.6 Skill: Knowledge/Comprehension 9.4 Essay Questions 1) Describe the process of two-dimensional gel electrophoresis. What are the two dimensions referring to? How is this technique different from conventional gel electrophoresis, and when would it be used?

Answer: In 2-D gel electrophoresis, proteins are separated in the first dimension by a version of gel electrophoresis known as isoelectric focusing. In this procedure, proteins are separated exclusively by their charge. A protein's pH environment affects its charge, and every protein has a pH–called the isoelectric point–at which it has neutral charge and cannot move in an electrical field. In isoelectric focusing, proteins migrate through a pH gradient to their isoelectric point, where they stop.

Once isoelectric focusing is complete, protein separation takes place in the second dimension that uses sodium dodecyl sulfate (SDS) gel electrophoresis. SDS is a strong anionic detergent that denatures proteins by disrupting the interactions that keep them folded. Denatured proteins migrate through the gel at a rate determined by their mass, that is, by the number of amino acids they contain. In the SDS gel dimension of two-dimensional gel electrophoresis, each protein has a unique starting point corresponding to its isoelectric point. Proteins with large mass (more amino acids) migrate a short distance in the second dimension, whereas proteins with small mass (fewer amino acids) migrate a greater distance.

One-dimensional gel electrophoresis uses a buffered solution to maintain constant pH throughout the gel, while isoelectric focusing gels contain a pH gradient. Therefore, two-dimensional gel electrophoresis separates proteins based on charge and mass.

Section: 9.2 Skill: Application/Analysis

2) Starting translation at the authentic (correct) start codon is essential for translation of the correct polypeptide. Errant translation starting at the wrong codon, or even at the wrong nucleotide of the start codon, may produce an abnormal polypeptide and result in a nonfunctional protein. Compare and contrast the mechanisms used by bacteria and eukaryotes to identify the authentic start codon during translation initiation.

Answer: In bacteria (specifically E. coli), six components are involved in translation initiation: (1) mRNA, (2) small ribosomal subunit, (3) the large ribosomal subunit, (4) the initiator tRNA, (5) three essential initiation factor proteins, and (6) GTP. The preinitiation complex forms when the authentic start codon sequence is identified by base pairing that occurs between the 16S rRNA in the 30S ribosome and a short mRNA sequence located a few nucleotides upstream of the start codon in the 5′ UTR of mRNA. The complex knows where to assemble on the mRNA because of the Shine-Dalgarno sequence, a purine-rich sequence of about six nucleotides located three to nine nucleotides upstream of the start codon. A complementary pyrimidine-rich segment containing the sequence UCCUCC is found near the 3′ end of 16S rRNA, and it pairs with the ShineDalgarno sequence to position the mRNA on the 30S subunit such that it recognizes the correct start codon.

In eukaryotes, the preinitiation complex is recruited to the 5′-cap region of mRNA, located at the end of the 5′ UTR. Once the eukaryotic initiation complex is formed, it embarks on a process called scanning, in which it uses ATP hydrolysis to move the small ribosomal subunit through the 5′ UTR in search of the start codon. About 90% of eukaryotic mRNAs use the first AUG encountered by the initiation complex as the start codon. The initiation complex is able to accurately locate the authentic start codon because the codon is embedded in a consensus sequence, the Kozak sequence, which reads 5' ACCAUGG - 3'.

Section: 9.2 Skill: Application/Analysis 3) Prokaryotes and eukaryotes show differences in posttranscriptional modifications of mRNA. What are these differences, and why do eukaryotes show more modifications than prokaryotes?

Answer: In bacteria, the coupling of transcription and translation allows ribosomes to engage in translation of the 5′ region of mRNAs whose 3′ end is still under construction by RNA polymerase.

Translation of the mRNA transcripts begins before transcription is complete, since there are no membrane-bound organelles to provide physical separation of the two processes. In eukaryotes, however, transcription and translation are uncoupled. Transcription takes place in the nucleus, where pre-mRNA is processed to form mature mRNA. Translation occurs in the cytoplasm after release of mature mRNA. Because the processes are physically (and temporally) separated, the mRNA receives a 5′ cap and 3′ tail to make the transcript stable and to help the ribosome locate the transcript when it exits the nucleus.

Section: 9.3 Skill: Application/Analysis

4) Polypeptides must be sorted after translation. Does protein sorting occur in both prokaryotes and eukaryotes? Describe the process of protein sorting and explain how the signal sequence is involved.

What would you expect to see if the signal sequence has been mutated or deleted?

Answer: Protein sorting is a process that uses signal sequences of amino acids at the N-terminal end of a polypeptide to sort proteins and direct them to their cellular destinations.

Protein sorting is needed in bacterial cells because of the many proteins specifically destined for the cell membrane. In eukaryotes, however, protein sorting is far more complex than in bacteria; proteins are dispatched to particular cellular organelles, such as the chloroplast, mitochondrion, lysosome, and nucleus, and certain proteins are secreted from the cells. If the signal sequence is mutated, proteins may not be secreted or processed properly and will begin to accumulate within the cell. Ultimately, this will result in cell death and impaired protein secretion.

Section: 9.6 Skill: Application/Analysis 5) Describe what is meant by a "conformational disease" and give an example. Why do you think these diseases exhibit late age of onset?

Answer: Diseases resulting from the death of cells due to accumulation of misfolded proteins are classified as conformational diseases. Several human protein conformational diseases are known, and they are often neurodegenerative disorders and dementias. For example, Alzheimer disease is produced by the accumulation of misfolded β-amyloid protein that leads to the destruction of certain brain cells. A familial form of Parkinson disease produced by the accumulation of misfolded α-synuclein protein also results in the destruction of certain brain cells. Similarly, Huntington disease is produced by the destruction of neurons due to the accumulation of misfolded huntingtin protein. Each of these diseases has a late age of onset that is due to gradual cell death caused by the presence of abnormal aggregates of misfolded protein. These diseases are neuronal because that is where the genes encoding the proteins are expressed.

Section: 9.6 Skill: Application/Analysis 6) Compare and contrast the mechanism of action of two commonly used antibiotics. Include details about which steps of translation are inhibited, and explain how the antibiotic accomplishes this inhibition.

Answer: Antibiotics target different aspects of microbe biology to kill microorganisms. Familiar antibiotics such as tetracycline, streptomycin, and chloramphenicol target different stages of microbial translation, as do less familiar antibiotics such as erythromycin, puromycin, and cycloheximide. Each antibiotic contains a different active compound that takes advantage of unique features of bacterial translation to disrupt the production of bacterial proteins while not interfering with the translation of proteins in our cells. That is how the antibiotic selectively kills the bacteria and does not harm our cells.

Answers may vary, but some potential answers are listed below.

Section: 9.6 Skill: Application/Analysis Genetics: An Integrated Approach (Sanders) Chapter 10 The Integration of Genetic Approaches: Understanding Sickle Cell Disease 10.1 Multiple-Choice Questions 1) Hemoglobin, the protein responsible for binding and transporting oxygen, forms a tetramer. What is the protein composition of the most common form of hemoglobin, hemoglobin A (HbA)?

A) α1β1 B) α1β2 C) α2β1 D) α2β2 E) α2βS Answer: D Section: 10.1 Skill: Knowledge/Comprehension 2) Sickle cell is a codominant disease, meaning that heterozygotes exhibit an intermediate blood cell phenotype (a mixture of normal and sickle cells) due to the blend of hemoglobin tetramers in each red blood cell. Which of the following tetramer structures would you expect to find in a carrier of SCD?

A) α2βA2 B) α2βSβA C) α2βS2 D) α2βA2 and α2βSβA E) α2βA2 and α2βSβA and α2βS2 Answer: E Section: 10.1 Skill: Application/Analysis 3) The mutation that causes SCD causes what kind of change in the structure of hemoglobin?

A) decreases heme-binding ability B) fewer chains of amino acids (trimeric rather than tetrameric structure) C) a single amino acid substitution that alters the structure of the tetramer D) addition of a single amino acid that causes a longer protein subunit E) loss of a single amino acid that causes a frameshift and premature stop codon Answer: C Section: 10.2 Skill: Knowledge/Comprehension

4) Pauling used gel electrophoresis to confirm that homozygous individuals produce a single form of βglobin protein that differs depending on the homozygous genotype as well as that heterozygotes produce proteins of both types in approximately equal concentration. What characteristic(s) of the bands proved this?

A) the positions of the protein bands B) the densitometry peaks of each band C) the shape of each band D) both the positions and the densitometry peaks of each band E) both the shape and the position of each band Answer: D Section: 10.2 Skill: Synthesis/Evaluation 5) Ingram examined hemoglobin structural variation by breaking down the hemoglobin protein into many fragments by chemical treatment and subjecting the fragments to electrophoresis to separate them in a gel; and then in a second dimension, perpendicular to the first, he separated them by chromatography. This two-step approach is commonly referred to as A) hybridization.

B) electrophoretic mobility.

C) densitometry.

D) restriction fragment analysis.

E) peptide fingerprint analysis.

Answer: E Section: 10.2 Skill: Knowledge/Comprehension 6) Which of the following techniques describes either the rate of a molecule's migration or its final position in a gel?

A) hybridization B) electrophoretic mobility C) densitometry D) restriction fragment analysis E) peptide fingerprint analysis Answer: B Section: 10.2 Skill: Knowledge/Comprehension 7) Which of the following techniques quantifies the amount of protein present in a gel lane by measuring how much light is blocked from passing through the gel by the presence of a band of protein?

A) hybridization B) electrophoretic mobility C) densitometry D) restriction fragment analysis E) peptide fingerprint analysis Answer: C Section: 10.2 Skill: Knowledge/Comprehension

8) Molecular charge is an important characteristic influencing protein migration on a gel. What accounts for the variation in the molecular charge of proteins?

A) the intronic sequences of the protein B) the amino acid composition of the protein C) the negative charge of the sugar-phosphate backbone D) the amount of protein loaded into the wells of the gel E) the presence of secondary structures in the protein Answer: B Section: 10.2 Skill: Knowledge/Comprehension 9) You set up an EcoRI restriction digest for a sample of genomic DNA from an individual with a hemoglobin mutation. To your surprise, you notice a different pattern of bands in the affected individual you have identified a new SNP! What can you conclude from the pattern of bands from the gel below?

A) The SNP resulted in the loss of an EcoRI site.

B) The SNP resulted in the creation of a new EcoRI site.

C) The SNP resulted in the loss of an exon.

D) The SNP results in the creation of a new HindIII site.

E) There was no change in the genomic DNA sequence.

Answer: B Section: 10.2 Skill: Application/Analysis 10) The βS and βA alleles can be distinguished on the basis of A) variations in the length of their DNA restriction fragments.

B) variations in the length of the amino acid sequence.

C) different electrophoretic mobilities caused by a single amino acid substitution.

D) variations in the length of both the DNA restriction fragments and the amino acid sequence.

E) variations in the DNA restriction fragments and different electrophoretic mobilities of the protein products due to a single amino acid substitution.

Answer: E Section: 10.1 Skill: Knowledge/Comprehension

11) Thalassemia is a recessive blood disease caused by reduced synthesis of one of the globin genes.

While SCD is a mutation in the structure of hemoglobin, thalassemias can also be caused by mutations in the regulatory elements, causing too little globin to be produced and resulting in anemia. You've designed a DNA probe that you want to use to screen a family for a rare mutation in the β-globin gene that causes thalassemia. First you will use the probe for northern blotting. What information will the northern blot tell you?

A) whether the mRNA for β-globin is being translated into functional proteins in a given cell population B) whether the amount of the β-globin gene being transcribed (mRNA levels) is within normal range C) the size of the mRNA being transcribed D) the presence of a particular mutation by restriction fragment analysis E) whether the amount of the β-globin gene transcript is within normal range, and the size of the mRNA gene product Answer: E Section: 10.2 Skill: Application/Analysis 12) The same DNA probe used to measure mRNA levels by northern blotting can also be used on a Southern blot with a restriction digest of genomic DNA. What new information can be gained from Southern blotting (that wasn't already obtained by the northern blot analysis)?

A) the level of protein that is being translated from the mRNA transcript B) the presence of a unique restriction fragment pattern indicating a mutation in a known restriction site C) the size of the genomic DNA fragments, including both introns and exons D) the size of the genomic DNA fragments and the level of protein being translated E) the size of the genomic DNA fragments and the presence of a unique restriction fragment pattern indicating a mutation in a known restriction site Answer: E Section: 10.2 Skill: Application/Analysis 13) How is the DNA visualized on a Southern blot?

A) ethidium bromide B) radioactively labeled probes C) chemically labeled probes D) both ethidium bromide and chemically labeled probes E) either radioactively labeled probes or chemically labeled probes Answer: E Section: 10.2 Skill: Knowledge/Comprehension

14) You are interested in designing a molecular probe to detect the region of βS SNP, which includes a DdeI (CTNAG) restriction site. Which of the following molecular probe sequences would anneal to the correct position?

A) ACTCCTGAGGAGAAG B) TGAGGACTCCTCTTC C) GTGCACCTGACTCCT D) CACGTGGACTGAGGA E) CACGUGGACUGAGGA Answer: A Section: 10.2 Skill: Synthesis/Evaluation 15) How is gel electrophoresis used to demonstrate the molecular basis of SCD? A) by revealing that the protein products of the βA and βS alleles have distinctive electrophoretic band patterns B) by revealing variations in the length of a DNA restriction fragment that distinguishes the βA and βS alleles C) by revealing variations in the exon splicing of the mRNA products of the βA and βS alleles D) by revealing variations in the electrophoretic mobility of proteins and the restriction fragment length polymorphisms of DNA E) by revealing variations in the electrophoretic mobility of proteins, the restriction fragment length polymorphisms of DNA, and the splice variants of the mRNA Answer: D Section: 10.2 Skill: Application/Analysis

16) At a biological level, βAβS heterozygotes suffer fewer cases of malaria than βAβA homozygotes, and when they do get malaria, their disease is less severe. What is the biological explanation for the heterozygous advantage?

A) The βAβA homozygotes have a shortened average life span of their red blood cells, and the shorter RBC life spans interrupt the developmental cycle of Plasmodium larvae by causing the parasites to mature and reproduce rapidly.

B) The βAβS heterozygotes have a shortened average life span of their red blood cells, and the shorter RBC life spans interrupt the developmental cycle of Plasmodium larvae by preventing many of the immature parasites from reaching maturity.

C) The sickle-cell shape of the RBCs in βAβS heterozygotes causes the Plasmodium larvae to become stuck in the RBC aggregates and prevents reproduction.

D) The anemia associated with SCD makes the βAβS heterozygotes less attractive to mosquitoes, so they experience fewer mosquito bites and less opportunity for infection by Plasmodium larvae.

E) The βAβS heterozygotes are bitten by mosquitoes less frequently, and when they are infected, the sickle-cell shape of the RBCs in βAβS heterozygotes causes the Plasmodium larvae to become stuck in the RBC aggregates and prevents reproduction.

Answer: B Section: 10.3 Skill: Knowledge/Comprehension

17) Given what you know of the inheritance pattern of SCD, is the following pedigree representative of a typical SCD pedigree?

A) Yes; this is a dominant pedigree, and SCD exhibits dominant inheritance.

B) Yes; this is a recessive pedigree, and SCD exhibits recessive inheritance.

C) No; this is a dominant pedigree, and SCD exhibits recessive inheritance.

D) No; this is a recessive pedigree, and SCD exhibits dominant inheritance.

E) Another generation is needed before you can determine mode of inheritance.

Answer: C Section: 10.3 Skill: Synthesis/Evaluation 18) Given what you know of the inheritance pattern of SCD, what can be said of individual II-5?

A) He is a carrier because his mother was affected and his son was affected.

B) He is unaffected, and the mutant allele is passed from II-4 to their son.

C) His children with II-4 will all have a 50% chance of being affected (βSβS).

D) He is a carrier, and his children with II-4 will have a 50% chance of being affected (βSβS) and a 50% chance of being a carrier (βAβS).

E) He is a carrier, and his children with II-4 will have a 50% chance of being affected (βSβS) and a 50% chance of being unaffected (βAβA).

Answer: D Section: 10.3 Skill: Synthesis/Evaluation 19) From a standpoint of natural selection, we see that the heterozygous advantage for βAβS individuals is balanced by the disadvantage to those with SCD or the wild-type alleles (and no malaria protection).

The action of these conflicting trends produces an overall increase in the mutant allele until it reaches a state where the gain and loss of βS alleles is equal. What is the term for this stable condition that allows mutant alleles to remain in a particular population?

A) equilibrium frequency B) heterozygote fitness C) restriction fragment length polymorphism D) single nucleotide polymorphism E) balanced polymorphism Answer: A Section: 10.3 Skill: Knowledge/Comprehension

20) What term is used to describe the situation when the loss of an allele because of selection against one of its phenotypes is balanced by natural selection in favor of the allele for another phenotype?

A) equilibrium frequency B) heterozygote fitness C) restriction fragment length polymorphism D) single nucleotide polymorphism E) balanced polymorphism Answer: E Section: 10.3 Skill: Knowledge/Comprehension 10.2 Short-Answer Questions 1) As part of the maturation process, red blood cells lose their nucleus. What is the consequence for cell division in these cells?

Answer: They cannot divide.

Section: 10.1 Skill: Application/Analysis 2) A couple has four children; one is affected by SCD, and two are carriers of the disease. What can be said of the genotypes of the parents?

Answer: Both are heterozygotes.

Section: 10.1 Skill: Application/Analysis 3) The rate of molecule movement in electrophoretic gels depends on three parameters of molecular structure. Each of these parameters individually is important in determining how a particular molecule migrates, but they can also interact with one another to produce a characteristic migration rate for each molecule. What are the three parameters?

Answer: mass, charge, shape Section: 10.2 Skill: Knowledge/Comprehension 4) Given the following gel, which DNA fragment has a lower electrophoretic mobility?

Answer: Band 1 Section: 10.2 Skill: Application/Analysis

5) Given the following gel, which DNA fragment has a lower molecular mass?

Answer: Band 2 Section: 10.2 Skill: Application/Analysis 6) Given the following gel, which protein has a more negative charge?

Answer: Band 2 Section: 10.2 Skill: Application/Analysis 7) Given the following gel, which protein is more likely to have a globular shape?

Answer: Band 2 Section: 10.2 Skill: Application/Analysis

8) You have digested a molecule of DNA and want to identify a specific fragment of interest. The DNA is subjected to gel electrophoresis, but you get two bands that are very close in size. What could you use to determine which band is the correct one?

Answer: hybridization of a molecular probe or Southern blotting Section: 10.2 Skill: Synthesis/Evaluation 9) The different globin genes in each cluster are sequentially expressed in tightly controlled patterns during the various stages of prenatal and adult development. What gene structure helps regulate globin gene transcription by controlling both the timing and the starting point for transcribing DNA into mRNA? Answer: promoters Section: 10.1 Skill: Knowledge/Comprehension 10) Sodium dodecyl sulfate (SDS) is a detergent commonly added to protein samples before gel electrophoresis. The purpose of the negatively charged SDS is to neutralize the charges and denature the protein to remove secondary structures. Thus, the proteins can be separated on a gel based solely on what?

Answer: size or molecular weight Section: 10.2 Skill: Knowledge/Comprehension 11) Before transferring DNA from a gel to the membrane in Southern blotting, the DNA must be denatured (usually by soaking the gel in NaOH). Why is this step necessary?

Answer: to make the DNA single stranded Section: 10.2 Skill: Application/Analysis 12) Hemoglobin is made up of four proteins from two different gene clusters, the α-globin gene cluster and the β-globin gene cluster. The two clusters are related, having evolved from a common ancestral gene through gene duplication, based on strong similarities in which three gene characteristics?

Answer: sequence, structure, and function Section: 10.1 Skill: Knowledge/Comprehension 13) Which of the following images represents the expected results of a DNA gel for three individuals with different genotypes homozygous dominant, heterozygote, and homozygous recessive?

Answer: the image on the left Section: 10.2 Skill: Application/Analysis

14) What information allows you to predict the pattern of DNA bands expected in a Southern blot of an individual with the wild-type allele?

Answer: location of restriction sites or restriction mapping Section: 10.2 Skill: Knowledge/Comprehension 15) Based only on this Southern blot and with no knowledge of which of the β-globin SNPs you are looking at, which lane(s) might represent(s) a heterozygous individual?

Answer: lane 2 Section: 10.2 Skill: Application/Analysis 16) Given the following Southern blot results, which lane represents a heterozygous individual?

Answer: lane 2 Section: 10.2 Skill: Application/Analysis

17) Given the following plasmid, what are the sizes of the fragments you would expect from an XhoI digest?

Answer: 7 kb, 2 kb, 1 kb Section: 10.2 Skill: Application/Analysis 18) Given the following plasmid, what are the sizes of the fragments you would expect from a NotI digest?

Answer: 7.5 kb, 2.5 kb Section: 10.2 Skill: Application/Analysis

19) A SNP results in a loss of the XhoI site indicated, so you design a molecular probe that spans the lost XhoI site. Which lane represents the expected Southern blot of an individual who is a carrier (heterozygous) for the mutant allele?

Answer: lane 3 Section: 10.2 Skill: Synthesis/Evaluation 10.3 Fill-in-the-Blank Questions 1) The bone marrow must continuously generate new red blood cells because the life span of a typical red blood cell is only ________. The life span is even shorter in individuals with sickle-cell trait, which accounts for the protective effects against malaria.

Answer: 100-120 days Section: 10.1 Skill: Knowledge/Comprehension 2) Globin genes contain ________ segments that ultimately encode the amino acids of the protein and interspersed ________ sequences that do not contribute to encoding the protein product of the gene.

Answer: exon; intron Section: 10.1 Skill: Knowledge/Comprehension 3) In contrast to the nucleic acid probes used to detect DNA or RNA target sequences, molecular probes used to detect target proteins in western blots are ________.

Answer: antibodies Section: 10.2

Skill: Knowledge/Comprehension 4) The intercalating agent ________ allows researchers to visualize DNA and RNA on a gel when the gel is exposed to UV light.

Answer: ethidium bromide (EtBr) Section: 10.2 Skill: Knowledge/Comprehension 5) Burkitt's lymphoma is a cancer of the lymphatic system. The endemic variant, common to equatorial Africa, commonly affects children who are also affected by chronic malaria. A recent study found that individuals with AA hemoglobin were twice as likely to develop Burkitt's lymphoma as those with the AS genotype. We saw a similar benefit of certain genotypes in cases of sickle cell disease. Recall that individuals who are βAβS, βAβC, or βAβE have a survival and reproductive advantage over individuals who are homozygous βAβA (and therefore succumb more easily to malaria) and over those who are homozygous for the mutant alleles (and therefore suffer from hereditary anemia). Resistance to malaria and Burkitt's lymphoma are both an example of ________ advantage.

Answer: heterozygous Section: 10.3 Skill: Application/Analysis 10.4 Essay Questions 1) Being a carrier of the sickle cell trait, although generally not a serious problem, can pose significant risks for athletes, particularly those in intense training regimens. The death of Florida State University linebacker Devaughn Darling, a carrier of the sickle cell trait, in a spring conditioning drill refocused attention on sickle cell abnormalities, including both sickle cell disease and sickle cell trait (carriers).

High altitude, aggressive exercise, and high temperatures can all lead to acute rhabdomyolysis in individuals with sickle cell trait. Why do you think an individual who is a carrier of the sickle cell trait may be more likely to experience the symptoms of SCD during rigorous exercise?

Answer: Individuals who are carriers might never experience any symptoms of SCD, because they have enough normal RBCs to carry oxygen required for normal activities. However, if they are subjected to extreme conditions, they may have adverse reactions because they still have some sickle-shaped cells in their bloodstream. The hemoglobin contained in red blood cells of people with SCD is less stable than wild-type (normal) hemoglobin when oxygen concentration is low (due to high altitudes, high temperatures, or intense exercise). This instability can cause hemoglobin to collapse into linear, crystallike molecules at low oxygen levels. Due to its high concentration in red blood cells, the collapse of hemoglobin into linear structures deforms red cells into the sickle shapes first seen by Ernest Irons.

These sickle cells can clog the bloodstream, leading to intense cramping and pain often confused with heat or exercise-related exhaustion. However, it is clearly a more dangerous and potentially fatal problem.

Section: 10.1 Skill: Synthesis/Evaluation

2) Describe the hypothesis that James Neel proposed to describe the inheritance pattern of SCD. What experimental methods did he use to prove his hypothesis?

Answer: In 1949, James Neel identified the inheritance pattern of SCD in a study that examined 42 parents of children with SCD. None of these parents had SCD. Neel was testing the genetic hypothesis that SCD is produced by a recessive allele and that the parents of children with SCD are heterozygous carriers. He reasoned that if his hypothesis were correct, all the parents of SCD patients would show evidence of a low level of sickle-shaped red blood cells. In Neel's tests, the blood of all 42 parents of SCD patients displayed a small number of sickle-shaped red blood cells, confirming the hypothesis that the inheritance of SCD is explained by the presence of two alleles in heterozygous carrier parents. The validation of Neel's hypothesis confirms that in SCD families, two alleles, βA and βS, can occur at the βglobin gene, and that there are three genotypes, βAβA (homozygous wild type), βAβS (heterozygous carrier), and βSβS (sickle cell disease). The hypothesis predicts that when two heterozygous carriers produce a child, the likelihood of the child having SCD is 1 in 4.

Section: 10.1 Skill: Application/Analysis 3) We saw that Southern blotting of DNA from the βA allele produces two DNA bands corresponding to fragment lengths of 1150 bp and 200 bp based on the location of two DdeI sites. In contrast, Southern blot analysis of βS-allele DNA produces a single DNA restriction fragment, measuring 1350 bp in length, since one of the DdeI sites is altered by SNP variation. Why is Southern blotting necessary to identify the correct bands if you know the sizes, why can't you just identify your band on a gel? Why is the location of the probe (spanning the predicted lost DdeI site) important?

Answer: In a Southern blot, you are digesting genomic DNA, and the restriction digest may yield hundreds or thousands of bands. If you were to visualize the genomic digest on a gel using ethidium bromide, the fragments would look like a smear of DNA because you will have fragments of every possible length. It would be impossible to isolate any one band with any degree of accuracy. Instead of trying to visualize these bands on a gel, the digested DNA is denatured into single-stranded DNA and transferred to a membrane. The probe is added, and it will anneal only to its specific complementary DNA sequence.

By running a DNA molecular weight marker, or "DNA ladder," you can approximate the size of band you are interested in. However, by designing a probe that spans the DdeI site in question, you can assure that your probe will anneal to two fragments in the wild-type allele (because the molecular probe has been cut in half at the same restriction site). If there is no DdeI site, you will have only one band. Thus, this strategy allows you to look at the total number of bands seen on the autoradiograph, not just their size differences.

Section: 10.2 Skill: Application/Analysis

4) Northern blotting is not useful in detecting the SNPs associated with the βA and βS alleles. Explain why.

Answer: The DNA sequences of the βA and βS alleles are identical except for the SNP that distinguishes the sequences of one from the other. Upon transcription, each allele produces an mRNA molecule containing 664 nucleotides. The single-nucleotide substitution that differentiates the two alleles does not alter the length of the mRNA transcript. Considering that the molecular attribute producing electrophoretic mobility differences among mRNAs is the total length of the molecule, it is not surprising that in this instance there is no difference in the electrophoretic mobilities of the mRNA transcripts of these two alleles, because the lengths of their mRNAs are identical. A northern blot analysis performed on mRNA from individuals with the three β-globin genotypes detects the same single mRNA band for each genotype. Consequently, northern analysis is not useful in detecting variation in this case.

Section: 10.2 Skill: Synthesis/Evaluation 5) We can detect the SNP in the βS allele because it destroys a DdeI restriction site, leading to DNA restriction fragment variation revealed by Southern blot analysis.

Either two or three restriction sequences for the restriction endonuclease DdeI can occur near the βglobin gene, depending on the allele. One DdeI restriction sequence is eliminated from the βS allele by the SNP in the mutant allele because of a mutation in the double-stranded target sequence 5′ - CTNAG 3′.

Suppose the wild-type allele sequence is CCTGAG, while a newly identified SNP is CCTTAG. Would this same restriction fragment analysis technique be appropriate for detecting this particular SNP? What other technique(s) could you use to identify this particular mutation?

Answer: The DdeI target sequence is 5′ - CTNAG - 3′, where N indicates that any of the four nucleotides (A, T, C, or G) can occur in the middle of the 5-bp sequence as long as the variable nucleotide is flanked by CT and AG dinucleotide combinations. Thus, you would be unable to detect the difference between CCTGAG and CCTTAG by restriction fragment analysis. Several other techniques may help detect this particular mutation: (1) You could design a molecular probe specific for the mutant sequence, although it may be difficult to detect a one-base-pair change in sequence. The probe would have to be very short, and the annealing conditions would need to be fairly stringent so that the probe would not anneal based solely on the homologous regions around the mutation.

(2) You could also look for changes at the protein level. This mutation occurred in an exon region, so you might expect to see a change in the amino acid sequence (depending on the translation frame). For instance, if GAG is one codon and TAG is the SNP's codon, you would see a Glu → Stop mutation and could look for the presence of a truncated protein in the mutant allele.

(3) Alternatively, it may be easiest to design a short primer upstream of this mutation and sequence the region directly using genomic DNA as a template for PCR (a technique that is described in detail in later chapters).

Section: 10.2 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 11 Chromosome Structure 11.1 Multiple-Choice Questions

1) Where are bacterial chromosomes located within the cell?

A) nucleus B) nucleoid C) nucleolus D) cytoplasm E) mitochondria Answer: B Section: 11.1 Skill: Knowledge/Comprehension 2) You have isolated DNA from actively replicating cells. Where would you expect to find this DNA on a gel?

A) DNA would be highly supercoiled and would be found in the band seen in the positive pole (bottom) of Lane 1.

B) DNA would be predominantly relaxed and would be found in the band seen in the positive pole (bottom) of Lane 1.

C) DNA would be highly supercoiled and would be found in the band seen in the negative pole (top) of Lane 1.

D) DNA would be predominantly relaxed and would be found in the band seen in the negative pole (top) of Lane 1.

E) DNA would be both relaxed and supercoiled, so it would appear in a pattern similar to Lane 2.

Answer: A Section: 11.1 Skill: Application/Analysis 3) You have isolated DNA for PCR and accidentally added topoisomerase instead of polymerase to your sample. What would you expect if you ran this DNA sample on a gel?

A) The DNA would show increased supercoiling and have a lower electrophoretic mobility.

B) The DNA would show decreased supercoiling and have a higher electrophoretic mobility.

C) The DNA would show increased supercoiling and have a higher electrophoretic mobility.

D) The DNA would show decreased supercoiling and have a lower electrophoretic mobility.

E) The DNA would show decreased supercoiling, but electrophoretic mobility will be unaffected.

Answer: D Section: 11.1 Skill: Application/Analysis 4) A bacterium inherits a mutation that renders its topoisomerase enzyme nonfunctional. What phenotype would you expect to see in this bacterium?

A) The cell would accumulate broken DNA because overwinding cannot be corrected, so the cell will be unable to replicate and will die.

B) The cell would accumulate supercoiled DNA, but otherwise it is perfectly normal.

C) The cell would accumulate relaxed DNA, which would make DNA replication and cell division highly disorganized and lead to cell death.

D) The cell would accumulate relaxed DNA, but otherwise it is perfectly normal.

E) The cell would be unaffected because bacteria have many different topoisomerases that are involved in replication.

Answer: A Section: 11.1 Skill: Application/Analysis 5) How might a thermophilic bacterium use chromosome structure to protect against heat-induced DNA degradation?

A) It expresses excess amounts of SMC proteins that allow the DNA to more tightly condense during periods of high heat.

B) It can express a specialized topoisomerase that induces negative supercoiling, which provides additional chromosomal stability.

C) It expresses a specialized topoisomerase that induces positive supercoiling, which provides additional chromosomal stability.

D) It expresses enzymes that inhibit topoisomerase during high heat, which prevents DNA from unwinding and becoming susceptible to heat degradation.

E) It expresses helicase, an enzyme that nicks the DNA helix and allows it to assume a more stable conformation and thus to resist heat-induced degradation.

Answer: C Section: 11.1 Skill: Knowledge/Comprehension

6) What is meant by the "beads on a string" model of chromatin?

A) The beads are the histones, and the string is the transcriptionally active DNA loops.

B) The beads are the nucleosomes, and the string is the linker DNA. C) The beads are molecules of DNA polymerase that attach to the DNA string.

D) The beads are ribosomes, and the string is the mRNA that has been transcribed from active chromatin.

E) The beads are the heterochromatic regions that are tightly compacted, and the strings are euchromatic regions that are being actively transcribed.

Answer: B Section: 11.2 Skill: Knowledge/Comprehension 7) Under normal cellular conditions, what is the diameter of the chromatin as it forms the solenoid structure?

A) 3 nm B) 10 nm C) 30 nm D) 34 nm E) 100 nm Answer: C Section: 11.2 Skill: Application/Analysis 8) A cell can form 10-nm chromatin fibers, but not 30-nm fibers. Which molecule has likely been removed or mutated in this cell?

A) H1 B) H2A C) H3 D) topoisomerase E) SMC proteins Answer: A Section: 11.2 Skill: Application/Analysis 9) Which histone protein is not part of the nucleosome structure?

A) H1 B) H2A C) H2B D) H3 E) H4 Answer: A Section: 11.2 Skill: Knowledge/Comprehension

10) Active transcription occurs in regions of chromatin loops that are located A) near the MARS. B) a large distance away from the MARS. C) within the linker DNA. D) within the telomeres.

E) within the euchromatin and constitutive heterochromatin.

Answer: B Section: 11.2 Skill: Application/Analysis 11) Typically, methylation of nucleosome N-terminal tails leads to A) removal of the protein components of the chromatin from the DNA. B) relaxed packaging of the chromatin and increased transcription.

C) tighter packaging of the chromatin and reduced transcription.

D) increased amounts of euchromatin relative to heterochromatin.

E) activation of topoisomerase.

Answer: C Section: 11.2 Skill: Knowledge/Comprehension 12) Typically, acetylation of histone tails leads to A) removal of the protein components of the chromatin from the DNA. B) relaxed packaging of the chromatin and increased transcription.

C) tighter packaging of the chromatin and reduced transcription.

D) increased amounts of euchromatin relative to heterochromatin.

E) activation of topoisomerase.

Answer: B Section: 11.2 Skill: Application/Analysis 13) Which type of chromosome has no p arms?

A) metacentric B) submetacentric C) acrocentric D) subacrocentric E) telocentric Answer: E Section: 11.3 Skill: Knowledge/Comprehension 14) You have identified a chromosome that has both p and q arms, but notice that the p arms are composed of highly repetitive DNA. What type of chromosome is this?

A) metacentric B) submetacentric C) acrocentric D) subacrocentric E) telocentric Answer: C Section: 11.3

Skill: Application/Analysis 15) How many pairs of chromosomes are found in a typical human somatic cell's karyotype?

A) 0 B) 22 C) 23 D) 44 E) 46 Answer: C Section: 11.3 Skill: Application/Analysis 16) You are trying to take a biopsy and prepare a karyotype from an ovary cell, but you accidentally selected an ovum instead. How many pairs of chromosomes are found in the karyotype of a typical human ovum?

A) 0 B) 22 C) 23 D) 44 E) 46 Answer: A Section: 11.3 Skill: Application/Analysis 17) You are trying to prepare a DNA karyotype, but this time you forgot to add trypsin to your chromosome preparation. What effect would you expect to see?

A) Dyes would stain the chromosomes darker because more proteins are present to take up the dye.

B) Dyes would not stain the chromosomes as expected, because proteins would not be removed from the chromatin and so would block the dye from accessing the DNA. C) Chromosomes do not condense properly, so it would be very difficult to isolate individual chromosomes.

D) Chromosomes would be degraded quickly because the trypsin is necessary to promote DNA stability by adding more scaffold proteins to the chromatin.

E) The chromosomes would be more difficult to pair up because the trypsin acts as a dye, creating a banding pattern that distinguishes between chromosomes.

Answer: B Section: 11.3 Skill: Application/Analysis 18) What technique would you use to detect a target sequence in an intact chromosome using a labeled molecular probe?

A) PCR B) in situ hybridization C) western blotting D) Sanger sequencing E) karyotyping Answer: B Section: 11.3 Skill: Knowledge/Comprehension

19) What is the function of interchromosomal domains in gene transcription?

A) These domains separate chromosome territories and are areas of active transcription.

B) These domains act as regions for formation of chiasmata during crossing over.

C) These domains act as gene-rich regions positioned toward the nuclear membrane to maximize gene transcription.

D) These domains contain no chromatin and act as channels for the movement of proteins, RNAs, and enzymes among the chromosomal territories.

E) These domains act as the regions of the chromosome recognized during meiosis when homologous chromosomes are matched up and aligned at the metaphase I plate.

Answer: D Section: 11.4 Skill: Synthesis/Evaluation 20) The flies with variegated eye color had what abnormal chromosome structure?

A) X-chromosome inversion that moves the w+ gene to a different orientation on the chromosome in some cells following x-ray exposure B) X-chromosome translocation that moves the w+ gene to a region of heterochromatin on another chromosome in some cells following x-ray exposure C) a mutation in the w+ gene following x-ray exposure that causes the gene to be overexpressed in a tissue-specific manner D) X-chromosome deletion that deleted the w+ gene in some cells following x-ray exposure E) a mutation in a somatic chromosome that causes the w+ gene to be translocated to the Xchromosomes following x-ray exposure Answer: A Section: 11.4 Skill: Knowledge/Comprehension 11.2 Short-Answer Questions 1) Which proteins participate in DNA bending and contribute to folding and condensation of the chromosome?

Answer: small nucleoid-associated proteins Section: 11.1 Skill: Knowledge/Comprehension 2) What is the purpose of DNA supercoiling?

Answer: nucleoid chromosomal compaction Section: 11.1 Skill: Knowledge/Comprehension 3) What type of DNA supercoiling is the most common?

Answer: negative Section: 11.1 Skill: Knowledge/Comprehension 4) What type of chromosome structure is commonly found in hyperthermophiles?

Answer: positive supercoiling Section: 11.1

Skill: Knowledge/Comprehension 5) Nucleosomes are comprised of how many molecules of histones?

Answer: 8 Section: 11.2 Skill: Knowledge/Comprehension 6) What molecular interaction allows histones to interact with DNA? Answer: positive charges of histones Section: 11.2 Skill: Knowledge/Comprehension 7) What is the diameter of a nucleosome?

Answer: 10 nm Section: 11.2 Skill: Knowledge/Comprehension 8) What is the name of the structure formed when the 10-nm fibers of chromatin form a cylindrical filament of coiled nucleosomes?

Answer: solenoid structure Section: 11.2 Skill: Knowledge/Comprehension 9) Nonhistone proteins make up the protein framework that gives the chromosomes their shape. What is this structure called?

Answer: chromosome scaffold Section: 11.2 Skill: Knowledge/Comprehension 10) Chromosome condensation compacts the chromosomes so they can be efficiently separated at which phase of mitosis?

Answer: anaphase Section: 11.2 Skill: Application/Analysis 11) Which proteins would you target if you wanted to disrupt the chromosome superstructure for chromatin condensation that ultimately produces the characteristic shape of the metaphase chromosome?

Answer: scaffold proteins Section: 11.2 Skill: Application/Analysis 12) If you want to prevent chromatin loops from being anchored to the chromosome scaffold, which regions of the DNA would you target?

Answer: MARS (matrix attachment regions) Section: 11.2 Skill: Application/Analysis 13) Nucleosome displacement during transcription initiation is referred to as what?

Answer: chromatin remodeling

Section: 11.2 Skill: Knowledge/Comprehension 14) To preserve epigenetic memory in the daughter cells, the cell retains these "old" proteins during DNA replication.

Answer: histones Section: 11.2 Skill: Knowledge/Comprehension 15) If you block histone deacetylase, what effect would you expect to see on transcriptional activity?

Answer: continued/prolonged transcription Section: 11.2 Skill: Application/Analysis 16) Karyotypes are a method for displaying chromosomes by grouping them into homologous pairs based on what two distinguishing factors?

Answer: size and banding pattern Section: 11.3 Skill: Knowledge/Comprehension 17) You are attempting to isolate cells for your karyotype, and you find one cell in each of the phases (prophase, metaphase, anaphase, and telophase). Which one is best suited for a karyotype?

Answer: metaphase Section: 11.3 Skill: Application/Analysis 18) If heterochromatin spreading does not reach the new location of the w+ gene, and it remains in the euchromatic region, the gene produces what phenotype?

Answer: red eyes Section: 11.4 Skill: Application/Analysis 19) If heterochromatic spreading encompasses the new location of the w+ gene, what is the phenotype of the cells?

Answer: white eyes Section: 11.4 Skill: Application/Analysis 20) What gene is critical for establishment of X-inactivation in mammals?

Answer: Xist Section: 11.4 Skill: Knowledge/Comprehension

11.3 Fill-in-the-Blank Questions 1) The total amount of hereditary information in an entire set of chromosomes is collectively referred to as an organism's ________.

Answer: genome Section: 11.1 Skill: Knowledge/Comprehension 2) ________ are extrachromosomal DNA molecules that are not part of the bacterial genome.

Answer: Plasmids Section: 11.1 Skill: Knowledge/Comprehension 3) The DNA-protein complex that comprises eukaryotic chromosomes is called ________.

Answer: chromatin Section: 11.2 Skill: Knowledge/Comprehension 4) Each nucleus in a human somatic cell contains ________ chromosomes.

Answer: 46 Section: 11.2 Skill: Knowledge/Comprehension 5) The ability of DNA-binding proteins to initiate transcription can be hindered by the presence of ________ on the promoter regions of the DNA. Answer: nucleosomes Section: 11.2 Skill: Knowledge/Comprehension 11.4 Essay Questions 1) Quinolones are a group of broad-spectrum antibiotics that specifically target topoisomerase. Explain how this contributes to cell death in bacteria. Would you expect these drugs to kill hyperthermophilic prokaryotes too? Why or why not?

Answer: Supercoiled DNA exists naturally in circular chromosomes of bacteria, but it can become "overwound" as it does during DNA replication. Overwinding puts unsustainable torsional stress on DNA, and if the stress is not relieved, DNA will break at random. Supercoiled DNA is partially unwound by topoisomerases that relax supercoiled DNA in a controlled manner. If you inhibit topoisomerase, the cell cannot correct overwinding and the chromosome breaks, causing DNA damage and cell death. You might expect that these drugs would not necessarily kill the hyperthermophilic bacteria, because most organisms exhibit negative supercoiling and have those types of topoisomerases.

The hyperthermophiles exhibit positive supercoiling, which is induced by specialized topoisomerases that are not the same ones involved in negative supercoiling. Thus, depending on how specific the quinolones mode of action may be, negative but not positive supercoiling can be inhibited.

Section: 11.1 Skill: Synthesis/Evaluation

2) Eukaryotic chromosomes are a combination of proteins and DNA, which allows the cell to perform which four essential functions?

Answer: 1. compacts the DNA so it can fit in the nucleus 2. stabilizes the DNA to protect it from damage 3. promotes chromosome condensation required for cell division 4. regulates DNA replication and gene transcription Section: 11.2 Skill: Synthesis/Evaluation 3) What are the common types of epigenetic modifications? Describe how epigenetic modifications can assist in predicting whether a gene is active or inactive in a cell.

Answer: Epigenetic modifications consist of the addition and removal of chemical groups such as methyl groups (methylation and demethylation), acetyl groups (acetylation and deacetylation), and phosphate groups (phosphorylation and dephosphorylation) near the amino-terminal (N-terminal) ends of histone proteins. These epigenetic marks alter how tightly DNA and nucleosomes bind together. The tighter the association, the less likely it is that transcription will occur; conversely, looser association between nucleosomes and DNA increases transcription. Methylation of nucleosome N-terminal tails often leads to tighter packaging of chromatin that reduces transcription, whereas acetylation of histone tails loosens chromatin and leads to more transcription. Phosphorylation is detected in both more compact and less compact chromatin. Epigenetic modifications alter the strength of association between nucleosomes and DNA and can lead to the displacement of nucleosomes (i.e., to remodeling of chromatin) to make regulatory DNA sequences more accessible to DNA-binding proteins.

Section: 11.4 Skill: Synthesis/Evaluation 4) Histone deacetylase (HDAC) inhibitors are commonly used as mood stabilizers or in the treatment of neurodegenerative diseases. What effect would a HDAC inhibitor have on the target cells?

Answer: A histone deacetylase removes acetyl groups from the histone tails. This process results in tighter chromosome compaction and thus less transcription of the genes in that region of the chromosome. HDAC inhibitors block this enzyme, resulting in hyperacetylation of histones and an increase in gene expression in the target cells, which can increase transcription (and translation) of gene products that may be decreased in the diseased tissues.

Section: 11.4 Skill: Synthesis/Evaluation 5) Tortoiseshell cats have a coat-color gene on their X chromosome. What process in eukaryotes explains why males can have solid orange or black fur, and only females can have the tortoiseshell pattern of fur?

Answer: The gene encoding coat color is X-linked, and male cats have a single copy of the X chromosome. They can therefore express either the O or o allele, resulting in orange or black coat, respectively. Female cats, on the other hand, have two copies of the X chromosomes. Thus, they can be heterozygous and express either orange or black. The tortoiseshell pattern of black and orange fur is due to X-chromosome inactivation. In black patches, the X chromosome with the dominant O allele has been inactivated. In orange patches, the X chromosome with the recessive o allele is active so the fur is black.

Since X-chromosome inactivation is random, the coat will appear mottled and have patches of both black and orange.

Section: 11.4 Skill: Synthesis/Evaluation

6) Given what you know about coat color in cats, list the possible genotypes for these individuals: • orange male • orange female • black male • black female • tortoiseshell Answer: O-, OO, o-, oo, Oo (females only) Section: 7.4 Skill: Application/Analysis Genetics: An Integrated Approach (Sanders) Chapter 12 Gene Mutation, DNA Repair, and Homologous Recombination 12.1 Multiple-Choice Questions 1) Studies of gene mutation frequencies have shown that A) mutations are rare, and genomes are generally stable.

B) mutation frequencies differ among organisms and also between genes, suggesting certain genes are more susceptible to mutation.

C) mutation frequencies are consistent between organisms, and each region of DNA is equally susceptible to random mutations.

D) Both A and B are correct.

E) Both A and C are correct.

Answer: D Section: 12.1 Skill: Knowledge/Comprehension 2) The adaptation mutation hypothesis predicts that A) mutations occur at random in organisms.

B) separate bacterial cultures may contain different frequencies of a particular gene mutation.

C) environmental change cannot trigger mutations in a specific gene.

D) different bacterial cultures exposed to the same stress will have similar frequencies of a particular mutation.

E) bacterial cultures exposed to multiple types of environmental stress will consistently resist mutation.

Answer: D Section: 12.1 Skill: Knowledge/Comprehension 3) Which types of mutation are possible thanks to the redundant nature of the genetic code?

A) silent B) missense C) nonsense D) frameshift E) splice site Answer: A Section: 12.2 Skill: Knowledge/Comprehension

4) Given the DNA sequence 5′-TAC AAA ATA CAG CGG-3′, which of these sequences represents a nonsense mutation?

A) 5′-TAG AAA ATA CAG CGG-3′ B) 5′-TAC AAA TAC AGC GGG-3′ C) 5′-TAC AAG ATA CAG CGG-3′ D) 5′-TAC AAA ATA CAC CGG-3′ E) 5′-TAC AAA ATA CAG AGG-3′ Answer: A Section: 12.2 Skill: Application/Analysis 5) Given the DNA sequence 5′-TAC AAA ATA CAG CGG-3′, which of these sequences represents a missense mutation?

A) 5′-TAG AAA ATA CAG CGG-3′ B) 5′-TAC AAA TAC AGC GGG-3′ C) 5′-TAC AAG ATA CAG CGG-3′ D) 5′-TAC AAA ATA CAC CGG-3′ E) 5′-TAC AAA ATA CAG AGG-3′ Answer: D Section: 12.2 Skill: Application/Analysis 6) Given the DNA sequence 5′-TAC AAA ATA CAG CGG-3′, which of these sequences represents a frameshift mutation?

A) 5′-TAG AAA ATA CAG CGG-3′ B) 5′-TAC AAA TAC AGC GGG-3′ C) 5′-TAC AAG ATA CAG CGG-3′ D) 5′-TAC AAA ATA CAC CGG-3′ E) 5′-TAC AAA ATA CAG AGG-3′ Answer: B Section: 12.2 Skill: Application/Analysis 7) Mutations can also occur within the regulatory regions of genes. Fragile X is an example of a disorder caused by one such mutation by increasing the length of the long arm of the X chromosome and ultimately decreasing mRNA stability. Which type of regulatory mutation causes Fragile X? A) promoter B) polyadenylation C) splice site D) triplet-repeat expansion E) frameshift Answer: D Section: 12.2 Skill: Application/Analysis 8) What type of mutation is seen here?

Wild type: 5′-TAC AAA ATA CAG CGG-3′ Mutation: 5′-TAC AAG ATA CAG CGG-3′ A) transition

B) transversion C) insertion D) deletion E) nonsense Answer: A Section: 12.2 Skill: Application/Analysis

9) Which type of mutation converts a nucleotide to an alternative structure with the same composition but slightly different placement of hydrogen bonds with a rare, less stable form that causes base-pair mismatch?

A) depurination B) deamination C) tautomeric shift D) transition E) transversion Answer: C Section: 12.3 Skill: Knowledge/Comprehension 10) The compound 5-bromodeoxyuridine (BrdU) is a derivative of uracil, and if BrdU becomes incorporated during DNA replication, it pairs with adenine. This compound is best classified as which type of mutagen?

A) base analog B) deaminating agent C) oxidative agent D) alkylating agent E) intercalating agent Answer: A Section: 12.4 Skill: Application/Analysis 11) You would like to induce a transversion mutation into a sequence of DNA. Which type of chemical mutagen would give you the best chance of inducing the correct mutation without causing transition mutations as well?

A) base analog B) deaminating agent C) oxidative agent D) alkylating agent E) intercalating agent Answer: C Section: 12.4 Skill: Application/Analysis 12) An intercalating agent such as aflatoxin is more likely to cause which type of mutation?

A) transition B) transversion C) frameshift D) depurination E) deamination Answer: C Section: 12.4 Skill: Application/Analysis

13) You have conducted an Ames test on a given compound. Which of the following would be classified as a positive result on the Ames test?

A) His- strain grows on an his- plate.

B) His- strain grows on an his+ plate.

C) His+ strain grows on an his- plate.

D) His+ strain grows on an his+ plate.

E) His+ strain grows on either an his- or an his+ plate.

Answer: A Section: 12.4 Skill: Application/Analysis 14) A strain of E. coli is unable to use the UV repair pathway because of an apparent absence of DNA helicase activity. Which UV repair gene is likely mutated in this strain?

A) uvr-A B) uvr-B C) uvr-C D) uvr-D E) pol I Answer: D Section: 12.5 Skill: Application/Analysis 15) What phenotype would you expect to see in a strain of E. coli with a mutation in the phr gene?

A) decrease in UV-induced mutations B) increase in UV-induced mutations C) decrease in reversion of base-pair substitution mutations D) increase in the methylation of nucleotide bases E) decrease in the methylation of nucleotide bases Answer: B Section: 12.5 Skill: Knowledge/Comprehension 16) Which pathway is affected by an inherited mutation in the ATM gene?

A) nucleotide excision repair pathway B) photoreactivation C) p53 repair pathway D) reactivation repair pathway E) UV repair pathway Answer: C Section: 12.5 Skill: Application/Analysis

17) Which of these statements best describes gene expression in a damaged cell?

A) In a damaged cell, p53 is low and BAX transcription is inactive, so Bcl-2 represses apoptosis.

B) In a damaged cell, p53 is high and BAX transcription is active, so Bcl-2 is repressed and apoptosis is induced.

C) In a damaged cell, p53 is low and BAX transcription is active, so Bcl-2 represses apoptosis.

D) In a damaged cell, p53 is low and BAX transcription is active, so Bcl-2 is repressed and apoptosis is induced.

E) In a damaged cell, p53 is high and BAX transcription is inactive, so Bcl-2 is activated and apoptosis is induced.

Answer: B Section: 12.5 Skill: Knowledge/Comprehension 18) Homologous recombination is initiated by A) Rad51 generating double-stranded DNA breaks.

B) RecA, RecB, and RecC generating single-stranded DNA breaks.

C) Spo11 generating double-stranded DNA breaks.

D) p53 generating single-stranded DNA breaks.

E) ATM signaling double-stranded DNA breaks.

Answer: C Section: 12.7 Skill: Knowledge/Comprehension 19) If one Holliday junction region is resolved by an NS cut and the other by an EW cut, the resulting chromosomes A) are nonrecombinant and carry the combinations A1 and B2 or A2 and B1.

B) are nonrecombinant and carry the combinations A1 and B1 or A2 and B2.

C) are A1 and A2 or B1 and B2.

D) are recombinant and carry the combinations A1 and B2 or A2 and B1.

E) are recombinant and carry the combinations A1 and B1 or A2 and B2.

Answer: D Section: 12.7 Skill: Application/Analysis 20) Gene conversion can be observed in fungi by identifying aberrant ratios in the ascus. Which of the following represent possible aberrant ratios caused when there is no mismatch repair of base-pair mistakes in the heteroduplex region?

A) 4:4 B) 3:1:1:3 C) 5:3 D) 6:2 E) 3:3:2 Answer: B Section: 12.8 Skill: Knowledge/Comprehension

12.2 Short-Answer Questions 1) If you have screened 100,000 pollen grains and identified 5 mutants, what is the mutation frequency?

Answer: 5 × 10-5 Section: 12.1 Skill: Application/Analysis 2) You use several model organisms in your lab's research. Which of your model organisms would you expect to exhibit a higher mutation frequency: E. coli, D. melanogaster, or M. musculus?

Answer: M. musculus Section: 12.1 Skill: Application/Analysis 3) What term describes a gene or region of a genome where mutations occur much more often than average?

Answer: hotspots of mutation Section: 12.1 Skill: Knowledge/Comprehension 4) Localized mutations that occur at a specific or identifiable location on a gene are better known as what?

Answer: point mutations Section: 12.2 Skill: Knowledge/Comprehension 5) What are the two basic types of base-pair substitutions?

Answer: transition and transversion Section: 12.2 Skill: Knowledge/Comprehension 6) What process, which is responsible for many trinucleotide repeat disorders, alters the number of DNA repeats?

Answer: strand slippage Section: 12.3 Skill: Knowledge/Comprehension 7) What two types of spontaneous damage to individual nucleotides are associated with subsequent mutation?

Answer: depurination and deamination Section: 12.3 Skill: Application/Analysis 8) What type of mutation commonly affects CpG islands, altering promoter regions of mammalian genes?

Answer: deamination Section: 12.4 Skill: Knowledge/Comprehension

9) Which type of mutagen has a structure similar to one of the DNA nucleotides and therefore can work its way into DNA, where it pairs with a nucleotide?

Answer: base analog Section: 12.4 Skill: Knowledge/Comprehension 10) What type of mutagen would you use if you wanted to study the effect of pyrimidine dimers in causing cancer?

Answer: UV radiation Section: 12.4 Skill: Application/Analysis 11) Which process is carried out by specialized bypass DNA that can replicate across the gaps induced during replication by UV damage?

Answer: translesion DNA synthesis Section: 12.4 Skill: Knowledge/Comprehension 12) You have developed a new drug and have been asked to test the drug's mutagenicity before putting it on the market. Which test could you use to determine your compound's mutagenic potential?

Answer: Ames test Section: 12.4 Skill: Knowledge/Comprehension 13) Which mechanism can be used to repair UV damage in Drosophila but not in humans?

Answer: photoreactive repair Section: 12.5 Skill: Application/Analysis 14) DNA glycosylase inhibitors are used to study which DNA repair mechanism?

Answer: nucleotide base excision repair Section: 12.5 Skill: Knowledge/Comprehension 15) Which enzyme is the product of the phr gene in E. coli?

Answer: photolyase Section: 12.5 Skill: Application/Analysis 16) The alkylating agent nitrosoguanidine adds methyl groups to nucleotide bases and is repaired by which enzyme that removes the added methyl group and restores nucleotides to their normal form?

Answer: O6-methylguanine methyltransferase Section: 12.5 Skill: Knowledge/Comprehension 17) What is the human cancer syndrome associated with inherited p53 mutations?

Answer: Li-Fraumeni syndrome Section: 12.5 Skill: Knowledge/Comprehension

18) Which repair mechanism is an error-prone system that uses bypass polymerases to replicate short segments of daughter strands whose normal replication is blocked by lesions in template strand DNA? Answer: translesion DNA synthesis Section: 12.6 Skill: Knowledge/Comprehension 19) Which type of double-stranded break repair would you expect to be used by a cell in the G1 phase of the cell cycle?

Answer: nonhomologous end joining (NHEJ) Section: 12.6 Skill: Application/Analysis 20) Which type of double-stranded break repair would you expect to be used by a cell in the G2 phase of the cell cycle?

Answer: synthesis-dependent strand annealing (SDSA) Section: 12.6 Skill: Application/Analysis 12.3 Fill-in-the-Blank Questions 1) A second-site mutation that compensates for the mutation in one gene by mutating a second gene and restoring the wild-type phenotype is also known as a ________ mutation.

Answer: suppressor Section: 12.2 Skill: Knowledge/Comprehension 2) ________ are structures that have the same composition and general arrangement but a slight difference in bonding and placement of a hydrogen.

Answer: Tautomers Section: 12.3 Skill: Knowledge/Comprehension 3) Agents generating mutation-inducing DNA damage are collectively referred to as ________.

Answer: mutagens.

Section: 12.4 Skill: Knowledge/Comprehension 4) The Ames test is designed to identify the rate of ________ that restore the ability of bacteria to synthesize their own histidine, thus eliminating the need for histidine supplementation of the growth medium.

Answer: reversion mutations Section: 12.4 Skill: Knowledge/Comprehension 5) DNA damage signaling processes are essential for regulating the ________ transition within the cell cycle.

Answer: G1-S Section: 12.5

Skill: Knowledge/Comprehension 12.4 Essay Questions 1) You are interested in studying the function of a gene, so you decide to create point mutations to help you identify important residues in your protein.

Which of the adenines in the following sequence could you mutate to create a silent transversion mutation?

5′-ATG GGA TGG CAC CTG-3′ Answer: A to G or C to T Section: 12.2 Skill: Synthesis/Evaluation 2) Which of the adenines in the following sequence could you mutate to create a missense transition mutation?

5′-ATG GGA TGG CAC CTG-3′ Answer: A to T or A to C Section: 12.2 Skill: Synthesis/Evaluation 3) Which bases would you mutate to create a nonsense transition mutation?

5′-ATG GGA TGG CAC CTG-3′ Answer: G to A Section: 12.2 Skill: Synthesis/Evaluation 4) Explain what is meant by deamination. How does this mutation affect a cytosine (C) versus a methylcytosine (5meC)? What effect does deamination have on CpG islands, and how does that affect the mutation frequencies in these regions?

Answer: Deamination is the loss of an amino (NH2) group from a nucleotide base. Each of the DNA nucleotide bases contains an amino group.

Deamination of cytosine is frequently a source of mutation. When C is deaminated, the amino group is usually replaced by an oxygen atom, forming the nucleotide base uracil. DNA mismatch repair mechanisms recognize uracil as an RNA nucleotide base and remove it from DNA. The excised uracil is replaced by cytosine, so the wild-type sequence is restored.

When deamination occurs on a cytosine that has been methylated, it creates thymine and generates a base-pair mismatch between the newly formed thymine on one strand and the previously complementary guanine on the other strand. If mismatch repair enzymes correct the mismatch before the next DNA replication cycle, two outcomes are possible: Either (1) the repair will restore the wild-type G-C base pair, or (2) the repair will generate an A-T base-pair transition substitution. If mismatch repair does not occur before replication, the guanine-containing strand will be used to produce a daughter duplex with the wild-type (G-C) sequence, while the thymine-containing strand will be used to synthesize a daughter duplex containing a G-C to A-T base-pair substitution.

CpG dinucleotides are cytosines that are side by side with guanines in a DNA strand and are joined by a phosphodiester bond. Cytosines in CpG dinucleotides are frequent targets for methylation in mammalian promoters where methylation helps regulate transcription. Experimental evidence shows that CpG

dinucleotides are hotspots of mutation as a result of deamination of methylated cytosine and the production of thymine. This situation could ultimately result in altered transcriptional activity in genes by mutations in the promoters.

Section: 12.3 Skill: Synthesis/Evaluation 5) You accidentally exposed your cell culture dish to radiation, but you are not sure of the precise wavelength of the exposure. How can you use the induced DNA damage to tell whether the radiation was UV or higher energy? What is the difference between the DNA damage caused by UV radiation compared to higher-energy radiation? Which type of damage is the most likely to be the most lethal to your cells?

Answer: Any radiation energy in wavelengths shorter than 380 nm is mutagenic. Ultraviolet radiation is a commonly encountered mutagen that creates pyrimidine dimers in DNA, and these can lead to basepair substitution mutations. If the dimer cannot be removed, it may also lead to replication stalling at the site of the lesion. Replication blockage by pyrimidine dimers may induce reinitiation of DNA synthesis at an adjacent RNA primer site. This reinitiation of replication potentially leaves gaps spanning dozens to hundreds of nucleotides in newly synthesized DNA strands, but the gaps are subsequently filled by translesion DNA synthesis, which is carried out by specialized bypass DNA polymerases that can replicate across the gaps but lack proofreading capabilities.

Higher-energy radiation can induce DNA strand breaks. DNA can be damaged in multiple ways, the most serious being the induction of DNA single-strand or double-strand breaks. These breaks potentially block DNA replication and thus pose a significant threat to the integrity and survival of affected cells.

DNA damage of this type is dealt with by specialized strand-break repair mechanisms. Double-stranded DNA breaks are often the most difficult to repair and thus are the most lethal to cells.

Section: 12.4 Skill: Synthesis/Evaluation 6) A bacterial culture is exposed to UV radiation and is preparing to repair the DNA damage. Describe two mechanisms of repair used by E. coli to repair UV-induced DNA damage. Would you expect the bacteria to be able to repair more or less of the damage if they were to be placed in a dark environment after UV damage?

Answer: Two mechanisms of DNA repair that bacteria can use are photoreactive repair or UV repair.

UV-induced pyrimidine dimers can be directly repaired by photoreactive repair, a DNA repair mechanism found in bacteria, single-celled eukaryotes, plants, and some animals (e.g., Drosophila), but not in humans. Photoreactivation uses photolyase to break the bonds formed during pyrimidine dimerization.

UV-induced damage to DNA can also be fixed by UV repair, which involves excision of a short region of the strand containing the photoproduct, followed by synthesis of new DNA to replace the excised region. In E. coli, the UVR-BC complex cleaves a pair of phosphodiester bonds on the strand containing the photoproduct, releasing a 12-bp fragment of DNA surrounding the mutation. UVR-D, a helicase, binds along with DNA pol I, allowing the polymerase to use the complementary strand to synthesize a replacement for the missing segment. DNA ligase then binds to the region to seal the sugar-phosphate backbone.

Photoreactive repair requires light to power the repair mechanism, so placing the bacteria in dark after the damage has occurred would mean fewer repairs would take place. Also, there would be a decrease in bacteria viability in the dark as compared to the light.

Section: 12.4 Skill: Synthesis/Evaluation

7) Double-strand breaks cause genomic instability and are thus potentially lethal to cells. They also greatly increase the risk of cancer and the chance of chromosome structural mutations. Compare and contrast the two mechanisms of double-strand break repair. Why would one mechanism be used over another?

Answer: Two mechanisms have evolved to carry out double-strand break repair.

Nonhomologous end joining (NHEJ) is an error-prone repair process that repairs double-strand breaks occurring before DNA replication. If a double-stranded break damages a eukaryotic chromosome during G1 of the cell cycle, replication of the damaged chromosome is blocked. Completion of NHEJ produces an intact DNA duplex and allows replication across the repaired region in the upcoming replication cycle, but the repair is often imperfect because resection removes nucleotides that cannot be replaced.

While NHEJ is error prone, it prevents more extensive loss from degradation of unprotected ends.

Mutations can be generated, however, when nucleotides are lost by transcribed genes.

Synthesis-dependent strand annealing (SDSA) is an error-free process that repairs double-strand breaks occurring after the completion of DNA replication. When a double-stranded break (DSB) affects one sister chromatid; the other chromatid is intact. The strand invasion process displaces one strand of the duplex and creates a displacement (D) loop. DNA replication within the D loop synthesizes new DNA strands from intact template strands, and the sister chromatids are reformed by dissociation and annealing of the nascent strand to the other side of the break. By accomplishing the removal of DNA in the immediate vicinity of a double-stranded break and the replacement of the excised DNA with a duplex identical to that in the sister chromatid, SDSA carries out error-free repair of double-stranded breaks.

To summarize, nonhomologous end joining is an error-prone system for the repair of double-strand DNA breaks that occur before S phase. Synthesis-dependent strand annealing takes place after completion of the S phase and uses the identical sister chromatid as a source of template strand sequence to synthesize a replacement duplex in an error-free manner.

Section: 12.6 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 13 Chromosome Aberrations and Transposition 13.1 Multiple-Choice Questions 1) What type of aneuploidy is responsible for Turner syndrome in humans?

A) trisomy 13 B) trisomy 18 C) trisomy 21 D) monosomy XO E) monosomy YO Answer: D Section: 13.1 Skill: Knowledge/Comprehension 2) When nondisjunction occurs early in embryogenesis rather than gametogenesis, what would you expect in the resulting karyotype?

A) monosomy

B) trisomy C) mosaicism D) uniparental disomy E) random X-inactivation Answer: C Section: 13.1 Skill: Knowledge/Comprehension 3) Prader-Willi and Angelman syndromes are caused by which type of chromosomal mutations, both in connection with chromosome 15?

A) monosomy B) trisomy C) mosaicism D) uniparental disomy E) random X-inactivation Answer: D Section: 13.1 Skill: Knowledge/Comprehension 4) In an allopolyploid organism, what is true regarding the fertility of interspecies hybrids?

A) Interspecies hybrids are fertile due to nonhomology of chromosomes.

B) Chromosome doubling and nondisjunction in gametocytes can lead to homologous chromosome pairing, disjunction, and fertile hybrids.

C) Interspecies hybrids will be fertile so long as there is an even number of chromosomes in the offspring.

D) Mitotic nondisjunction results in haploid cells.

E) Meiotic nondisjunction produces three haploid gametes and one diploid gamete.

Answer: B Section: 13.2 Skill: Application/Analysis 5) If a plant is an autotetraploid with the genotype A1A1A2A2, what is the frequency of offspring with the genotype A1A1A1A1 following self-fertilization of the plant?

A) 1/2 B) 1/4 C) 9/16 D) 8/36 E) 1/36 Answer: E Section: 13.2 Skill: Application/Analysis 6) Which type of chromosome breakage results in very small deletions that are too small to identify on a karyotype?

A) acentric deletions B) microdeletions C) terminal deletions D) partial deletions E) interstitial deletions

Answer: B Section: 13.3 Skill: Knowledge/Comprehension 7) Which type of chromosome deletions is caused by two concurrent chromosome breaks (rather than a single break)?

A) acentric deletions B) microdeletions C) terminal deletions D) partial deletions E) interstitial deletions Answer: E Section: 13.3 Skill: Knowledge/Comprehension 8) Heterozygous carriers of chromosome inversions or translocations A) will be viable and completely fertile because no genes were deleted.

B) may or may not exhibit phenotypic abnormalities.

C) may be infertile due to complications during meiosis.

D) Both A and B are correct.

E) Both B and C are correct.

Answer: E Section: 13.4 Skill: Knowledge/Comprehension

9) A region of a chromosome on spanning the centromere is broken and reattached in the reverse direction. This is an example of which type of chromosomal defect?

A) paracentric inversion B) pericentric inversion C) pericentric translocation D) dicentric inversion E) dicentric translocation Answer: B Section: 13.4 Skill: Application/Analysis 10) A chromosome has broken, and a piece of one chromosome is translocated to a nonhomologous chromosome. This is an example of what type of chromosomal alteration?

A) paracentric inversion B) dicentric bridge C) unbalanced translocation D) Robertsonian translocation E) inversion loop Answer: C Section: 13.4 Skill: Application/Analysis 11) Crossover suppression implies that A) there are no crossovers between homologous chromosomes.

B) recombinant chromosomes are absent in the progeny.

C) small inversions have a high frequency of crossover.

D) nonreciprocal translocations prevent equal exchange of chromosomes, resulting in loss of chromosome fragments.

E) Both A and C are correct.

Answer: B Section: 13.4 Skill: Application/Analysis 12) A chromosome contains the following gene order: ABCD•EFGH Which of the following rearrangements represents a pericentric inversion?

A) A B C • D E F G H B) A F G H • B C D E C) A C B D • E F G H D) A F D E • C B G H E) A B C D • H G F E Answer: D Section: 13.4 Skill: Application/Analysis

13) A chromosome contains the following gene order: ABCD•EFGH Which of the following rearrangements represents a paracentric inversion?

A) A B C • D E F G H B) A F G H • B C D E C) A C B D • E F G H D) A F D E • C B G H E) A B C D • H G F E Answer: C Section: 13.4 Skill: Application/Analysis 14) If an organism is an inversion heterozygote, A) homologous chromosomes align during meiosis through formation of an inversion loop at synapsis.

B) homologous chromosomes align during meiosis through formation of a Holliday junction at synapsis.

C) nonhomologous chromosomes pair during meiosis due to the translocation of genes between two different chromosomes.

D) it will experience normal chromosome synapsis during meiosis.

E) it will be unable to complete replication due to DNA damage, and the cell cycle will be arrested.

Answer: A Section: 13.4 Skill: Application/Analysis 15) Humans with the condition of familial Down syndrome exhibit which type of chromosomal aberration?

A) chromosome 21 duplication B) chromosome 21 inversion C) chromosome 14/21 Robertsonian translocation D) chromosome 21 deletion E) trisomy of chromosome 21 Answer: C Section: 13.4 Skill: Knowledge/Comprehension 16) An organism with a genome size of n = 23 experienced a Robertsonian translocation. How many chromosomes would you expect to see in the karyotype of a somatic cell in an affected individual?

A) 21 B) 25 C) 42 D) 44 E) 46 Answer: D Section: 13.4 Skill: Knowledge/Comprehension

17) In balanced translocation heterozygotes, which pattern of chromosome segregation produces viable gamets and progeny?

A) adjacent-1 segregation B) adjacent-2 segregation C) alternate segregation D) Adjacent-1 and alternate segregation are both viable.

E) Adjacent-1, adjacent-2, and alternate segregation are all viable.

Answer: C Section: 13.4 Skill: Knowledge/Comprehension 18) In bacteria, what is contained within the IS elements?

A) transposase gene B) short tandem repeats C) inverted repeats D) Both A and B are correct.

E) Both A and C are correct.

Answer: E Section: 13.6 Skill: Knowledge/Comprehension 19) What sequences or structures within the DNA can be used to identify sites of transposition?

A) target site duplication and direct repeats B) target site duplication and short tandem repeats C) inverted loops D) target site inversion and direct repeats E) chromosome translocation and insertion of transposase gene Answer: A Section: 13.6 Skill: Knowledge/Comprehension 20) AluI is an example of which type of transposable element commonly found in the genome of humans?

A) LINE elements B) SINE elements C) P elements D) Copia elements E) Ty elements Answer: B Section: 13.7 Skill: Knowledge/Comprehension

13.2 Short-Answer Questions 1) During gametogenesis, what percentage of gametes would be aneuploid if the nondisjunction event occurs during meiosis I? Answer: 100% (4/4) Section: 13.1 Skill: Application/Analysis 2) During gametogenesis, what percentage of gametes would be trisomic if the nondisjunction event occurs during meiosis II? Answer: 25% (1/4) Section: 13.1 Skill: Application/Analysis 3) What is the name of the chromosomal region, located on region 21q22.2 that was identified by studying individuals with partial trisomy of chromosome 21?

Answer: DSCR (Down syndrome critical region) Section: 13.1 Skill: Knowledge/Comprehension 4) In Drosophila and Lepidoptera, sex-chromosome mosaicism produces what type of sexually ambiguous phenotype?

Answer: gynandromorphy Section: 13.1 Skill: Knowledge/Comprehension 5) Of the three mechanisms leading to autopolyploidy, which is the most rare event?

Answer: multiple fertilizations Section: 13.2 Skill: Knowledge/Comprehension 6) "Seedless" fruits and vegetables have how many chromosomes and what type of chromosomal distribution?

Answer: odd-numbered allopolyploidy Section: 13.2 Skill: Application/Analysis 7) How many possible genotypes will there be following self-fertilization of a plant with the B1B1B2B2B3B3 genotype?

Answer: 15 Section: 13.2 Skill: Application/Analysis 8) Cri-du-chat syndrome is a human disorder caused by which type of chromosomal defect?

Answer: terminal deletion Section: 13.3 Skill: Knowledge/Comprehension

9) A partial chromosome deletion or duplication can be observed at prophase I homologous chromosome synapsis during meiosis by looking for what structure?

Answer: unpaired loop Section: 13.3 Skill: Knowledge/Comprehension 10) Pseudodominance is used to map genes by using what method?

Answer: deletion mapping Section: 13.3 Skill: Knowledge/Comprehension 11) What structure, seen during synapsis, is indicative of a chromosome inversion?

Answer: inversion loop Section: 13.4 Skill: Knowledge/Comprehension 12) Gametes produced by inversion heterozygotes can produce no viable recombinant chromosomes due to what phenomenon?

Answer: crossover suppression Section: 13.4 Skill: Knowledge/Comprehension 13) What effect does a Robertsonian translocation have on chromosome number?

Answer: reduction (2n - 2) Section: 13.4 Skill: Knowledge/Comprehension 14) What are the two common patterns of chromosome segregation seen in the tetravalent structures found in translocation heterozygotes?

Answer: alternate and adjacent-1 Section: 13.4 Skill: Knowledge/Comprehension 15) Barbara McClintock identified corn mutants with the unstable mutant phenotype with what pattern of color?

Answer: colorless with purple spots Section: 13.5 Skill: Knowledge/Comprehension 16) In McClintock's maize studies, which genetic element is found at the site of chromosome breakage?

Answer: dissociation (Ds) element Section: 13.5 Skill: Knowledge/Comprehension 17) Which enzyme(s) is/are responsible for excising and copying transposable genetic elements from chromosomes and inserting them into new locations?

Answer: transposase Section: 13.5 Skill: Knowledge/Comprehension

18) Of the various genetic elements described by McClintock, which one is classified as an autonomous transposable element?

Answer: Ac Section: 13.5 Skill: Application/Analysis 19) P elements cause which phenomenon responsible for sterility in Drosophila?

Answer: hybrid dysgenesis Section: 13.7 Skill: Knowledge/Comprehension 20) What two features are consistently seen within retrotransposons?

Answer: reverse transcriptase gene and long terminal repeats (LTRs) Section: 13.7 Skill: Knowledge/Comprehension 13.3 Fill-in-the-Blank Questions 1) The phenomenon of ________ in allopolyploids consists of more rapid growth, increased fruit and flower production, and improved disease resistance.

Answer: hybrid vigor Section: 13.2 Skill: Knowledge/Comprehension 2) ________ is rare and occurs most commonly when repetitive regions of homologous chromosomes misalign, resulting in partial deletions and partial duplications.

Answer: Unequal crossover Section: 13.3 Skill: Knowledge/Comprehension 3) ________ is a genetic phenomenon that occurs when a normally recessive allele is "unmasked" and expressed in the phenotype because the dominant allele on the homologous chromosome has been deleted.

Answer: Pseudodominance Section: 13.4 Skill: Knowledge/Comprehension 4) Chromosomal translocation involves chromosome breakage and reattachment of the broken segment to a ________.

Answer: nonhomologous chromosome Section: 13.4 Skill: Knowledge/Comprehension 5) McClintock's transposable genetic element hypothesis was that the unstable mutant phenotype resulted when a ________ created a mutation by its insertion into the ________ allele and led to reversion when the expression of ________ led to its removal.

Answer: transposable element (Ds); C; Ac Section: 13.5

Skill: Knowledge/Comprehension 13.4 Essay Questions 1) Describe two mechanisms for causing uniparental disomy, and explain how this chromosomal defect is involved in Prader-Willi syndrome and Angelman syndrome.

Answer: Uniparental disomy occurs when both copies of a homologous chromosome pair originate from a single parent. Uniparental disomy has two mechanisms of origin. First, the rare mechanism would involve nondisjunction of the same chromosome in both the sperm and egg. Second, the more common mechanism involves nondisjunction in one parent that creates an aneuploid gamete contributing two copies of chromosome 15. The other gamete is normal and contributes a single copy of chromosome 15. Gamete union results in trisomy 15, but through the process of trisomy rescue, one of the extra copies of chromosome 15 is randomly ejected in one of the first mitotic divisions following fertilization. Zygotes with this result have a normal result in a zygote that retains two copies of chromosome 15 from the same parent (i.e., uniparental disomy).

Both Angelman syndrome and Prader-Willi syndrome are usually the result of a partial deletion of the 15q11.12 portion of chromosome 15. Uniparental disomy produces 10-20% of cases of AS and PWS. When uniparental disomy causes AS, both copies of chromosome 15 are from the father, and there is no maternal copy of chromosome 15. In uniparental disomy cases of PWS, both copies of chromosome 15 are from the mother, and there is no copy of the paternal chromosome 15.

Section: 13.1 Skill: Synthesis/Evaluation 2) How is aneuploidy different from polyploidy? What are the mechanisms by which aneuploidy and polyploidy are caused, and what are the consequences of these chromosomal alterations in terms of survival and fertility of the offspring?

Answer: Aneuploidy occurs when one or more chromosomes are lost or gained relative to the normal euploid number. Chromosome nondisjunction is a principal cause of aneuploidy. Nondisjunction can occur in somatic cells or in germ-line cells with similar results. If a single pair of homologous chromosomes fails to properly disjoin in a somatic cell during mitotic cell division, one of the resulting daughter cells carries an extra chromosome (2n + 1), and the other is missing a chromosome (2n - 1).

Mitotic cells of animals that contain the wrong number of chromosomes may suffer reduced viability, although cells with abnormal numbers of chromosomes are common in cancer where other alterations of chromosome number play a central role in cancer cell survival and proliferation.

Polyploidy is the presence of three or more sets of chromosomes in the nucleus of an organism.

Polyploidy in nature can result either from the duplication of euploid chromosome sets from a single species or from the combining of chromosome sets from different species.

Polyploids whose karyotype is comprised of chromosomes derived from a single species are designated as autopolyploids (auto = "self"), and polyploids with chromosome sets from two or more species are called allopolyploids (allo = "different"). Polyploidy is common in plant species but is extremely rare in animals; it is caused by multiple fertilization, mitotic nondisjunction, or meiotic nondisjunction.

Section: 13.2 Skill: Synthesis/Evaluation

3) Compare and contrast the two types of transposable elements found in bacteria. Provide examples of how transposable elements can be both harmful and beneficial to bacteria.

Answer: Bacterial genomes contain two categories of transposable elements: insertion sequence (IS) elements and transposons.

IS elements are simple transposable elements that contain only the sequences and genes necessary for autonomous transposition. Bacterial IS elements encode transposase and are flanked by inverted repeat sequences unique to each insertion sequence.

Transposons, on the other hand, can carry multiple genes and confer new traits on the bacteria that contain them.

The process of transposition is a mutational event that can inactivate or alter normal gene expression.

Both types of transposable element can produce mutations if they are inserted into a wild-type gene.

These mutations are often harmful if they disrupt the function of a normal gene. Conservative and simple bacterial transposons may be beneficial to the cell in that they often carry the gene for transposase and genes for antibiotic or heavy metal resistance.

Section: 13.6 Skill: Synthesis/Evaluation

P elements in Drosophila contribute to the process of hybrid dysgenesis, which occurs when M-cytotype females are crossed with P-cytotype males.

4) What does the model of hybrid dysgenesis predict for the F1 and F2 generations when a P-cytotype male is crossed to an M-cytotype female?

Answer: F1 will be wild type, but sterile.

Section: 13.7 Skill: Synthesis/Evaluation 5) What does the model of hybrid dysgenesis predict for the F1 and F2 generations when an M-cytotype male is crossed to a P-cytotype female?

Answer: F1 and F2 will be wild type and have normal fertility.

Section: 13.7 Skill: Synthesis/Evaluation 6) Provide a description of the cellular mechanisms by which one reciprocal cross leads to hybrid dysgenesis while the other does not.

Answer: The M cytotype is normal, while the P cytotype has multiple P elements and a repressor protein within the cytoplasm to block transposition. When the male has the P cytotype, he has multiple transposable elements but the ovum provides the vast majority of cytoplasm to the offspring relative to the sperm. Thus, there is no transposition repressor protein present in the M-cytotype female, and the P elements create widespread mutation by unrepressed insertional inactivation of genes. When the female is the P cytotype, the embryo will have multiple P elements but will also have large amounts of the transposition repressor protein within the cytoplasm. This represses the widespread insertional inactivation events that lead to hybrid dysgenesis in the reciprocal cross, so the offspring are healthy and fertile.

Section: 13.7 Skill: Synthesis/Evaluation

7) How are retroviruses and retrotransposons similar? Describe the genes encodes by the DNA of the retrovirus and the two DNA elements consistently found in retrotransposons.

Answer: Retroviruses are viruses that infect eukaryotic cells and have a genome consisting of singlestranded RNA. On infection, the RNA is transcribed into double-stranded DNA by reverse transcriptase, and the DNA is then integrated into a chromosome of the infected plant or animal cell. Integrated retroviral DNA encodes three genes, called gag, env, and pol. Gag and env encode proteins that form the retroviral particle, while the pol gene encodes the enzyme reverse transcriptase.

Retrotransposons are transposable elements and are related to retroviruses. They also have the pol gene, and some contain gag, but none have env. Thus retrotransposons are able to synthesize double-stranded DNA from single-stranded RNA and can transpose the DNA throughout the genome, but they do not possess the ability to produce protein particles.

Two constant features of retrotransposons are that they all have a pol gene encoding reverse transcriptase and retrotransposons are flanked by long terminal repeats (LTRs).

Section: 13.7 Skill: Synthesis/Evaluation 8) If you wish to target a retrovirus in a cell but do not want to disrupt retrotransposons,which gene(s) would you target?

Answer: I would target env or gag, since they encode the protein particles that form the retroviral particle and are not required for retrotransposition.

Section: 13.7 Skill: Synthesis/Evaluation 9) How might L1 retrotransposons be useful as a potential gene therapy agent? What are the problems with using this strategy for gene therapy?

Answer: Because the retrotransposons can be inserted into the genome, you could theoretically add a missing gene to the retrotransposon sequence and allow the cell to insert the gene into a chromosome, hopefully restoring functional gene expression to the cell. You have little control over where the gene will be inserted, so it may disrupt the function of another gene in the process. Moreover, just because it is inserted into the genome is no guarantee it will be transcribed and translated properly, because essential gene control elements such as promoters and enhancers will not be in the same location relative to the gene.

Section: 13.7 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 14 Regulation of Gene Expression in Bacteria and Bacteriophage 14.1 Multiple-Choice Questions 1) In negative control, what molecule would you expect to find bound to the operator if there is no transcription?

A) activator B) repressor C) inducer D) corepressor E) RNA polymerase

Answer: B Section: 14.1 Skill: Application/Analysis 2) You want to design a repressor protein mutant. Which protein domain is the best target for preventing binding of the corepressor?

A) DNA-binding domain B) allosteric domain C) promoter domain D) helix-turn-helix domain E) activator binding site Answer: B Section: 14.1 Skill: Application/Analysis 3) The presence of which combination of molecules would lead to active transcription of an operon?

A) activator + corepressor B) activator + repressor C) inducer + corepressor D) repressor + corepressor E) activator + inducer Answer: E Section: 14.1 Skill: Application/Analysis 4) In the lac operon, what acts as the inducer?

A) glucose B) permease C) allolactose D) β-galactosidase E) transacetylase Answer: C Section: 14.2 Skill: Knowledge/Comprehension

5) The enzyme β-galactosidase catalyzes what reaction?

A) allolactose → glucose + lactose B) lactose → glucose + fructose C) lactose → galactose + glucose D) glucose → galactose + lactose E) galactose → glucose + lactose Answer: C Section: 14.2 Skill: Knowledge/Comprehension 6) A bacterium is unable to transport lactose into the cell to be broken down. Which gene is likely mutated in this bacterium?

A) lacZ B) lacY C) lacI D) lacP E) lacO Answer: B Section: 14.2 Skill: Knowledge/Comprehension 7) Which region of the lac operon would you target if you want to disrupt the -10 and -35 consensus sequences?

A) lacZ B) lacY C) lacI D) lacP E) lacO Answer: D Section: 14.2 Skill: Application/Analysis 8) In the presence of glucose, where is the lac repressor bound?

A) lacZ B) lacO C) lacP D) lacI E) The lac repressor is not bound to the operon.

Answer: B Section: 14.2 Skill: Application/Analysis

9) Which structural gene of the lac operon can be mutated without affecting the cell's ability to break down lactose?

A) lacZ B) lacY C) lacA D) lacP E) lacO Answer: C Section: 14.2 Skill: Application/Analysis 10) Which of the following are constitutive mutants of the lac operon?

A) lacZ and lacP B) lacZ and lacI C) lacO and lacI D) lacO and lacZ E) lacI and lacY Answer: C Section: 14.3 Skill: Application/Analysis 11) Which mutants are noninducible?

A) lacIS B) lacIC) lacI+ D) lacOC E) lacO+ Answer: A Section: 14.3 Skill: Application/Analysis 12) Which of these haploid strains produce β-galactosidase constitutively but do not produce permease?

A) I- P+ O+ Z+ Y+ B) I+ P+ O+ Z- YC) I- P+ O+ Z-Y+ D) I+ P+ O- Z+ Y+ E) I- P+ O+ Z+ YAnswer: E Section: 14.3 Skill: Application/Analysis

13) Which of these haploid strains produce permease but do not produce β-galactosidase?

A) I- P+ O+ Z+ Y+ B) I+ P+ O+ Z-Y+ C) I- P+ O+ Z-YD) I+ P+ O- Z+ Y+ E) I- P+ O+ Z+ YAnswer: B Section: 14.3 Skill: Application/Analysis 14) During the attenuation of the trp operon, which stem loop leads to polycistronic mRNA synthesis during tryptophan starvation?

A) 1-3 (antitermination) stem loop B) 3-4 (termination) stem loop C) 1-2 (pause) stem loop D) 2-3 (antitermination) stem loop E) 2-4 (termination) stem loop Answer: D Section: 14.4 Skill: Knowledge/Comprehension 15) In the absence of tryptophan, A) the inactive repressor cannot bind trpO, so operon gene transcription occurs.

B) the active repressor binds trpP, so operon gene transcription is repressed.

C) the inducer cannot bind trpO, so operon gene transcription occurs.

D) the active repressor cannot bind trpO, so operon gene transcription is attenuated.

E) the repressor binds the corepressor, and operon gene transcription occurs.

Answer: A Section: 14.4 Skill: Application/Analysis 16) Which sigma factor, encoded by the rpoH gene, is active at high temperatures?

A) σ70 B) σ32 C) σ45 D) σ37 E) σ50 Answer: B Section: 14.5 Skill: Knowledge/Comprehension

17) Which enzyme is responsible for integration of a temperate phage into a host genome?

A) β-galactosidase B) transposase C) integrase D) transacetylase E) permease Answer: C Section: 14.5 Skill: Knowledge/Comprehension 18) Which region of the lambda phage genome enables the linear chromosome to circularize when it enters a host cell?

A) cohesive (cos) site B) IS10 C) early operators D) integrase (int) gene E) excisionase (xis) gene Answer: A Section: 14.6 Skill: Knowledge/Comprehension 19) If you wanted to prevent a cell from entering the lysogenic pathway, which gene controlled by PRM would you target?

A) cro B) Q C) cI D) cII E) O Answer: C Section: 14.6 Skill: Application/Analysis 20) In λ phage, which protein, the product of the cI gene, blocks the transcription required to initiate the lytic cycle?

A) cro B) repressor C) enhancer D) integrase E) operator Answer: B Section: 14.6 Skill: Application/Analysis

14.2 Short-Answer Questions 1) Regulation of transcription of bacterial genes takes place at which two levels?

Answer: initiation and amount Section: 14.1 Skill: Knowledge/Comprehension 2) What are the two active sites or "domains" on a repressor protein?

Answer: DNA-binding and allosteric Section: 14.1 Skill: Knowledge/Comprehension 3) If a corepressor is inhibited, what effect would you expect to see in the operon?

Answer: Transcription will occur.

Section: 14.1 Skill: Application/Analysis 4) Which proteins facilitate RNA polymerase binding at promoters?

Answer: activator proteins Section: 14.1 Skill: Knowledge/Comprehension 5) Catabolite repression refers to the repression of the lac operon in the presence of which catabolite?

Answer: glucose Section: 14.2 Skill: Knowledge/Comprehension 6) Binding of which complex increases the ability of RNA polymerase to transcribe the lac operon?

Answer: CAP-cAMP Section: 14.2 Skill: Knowledge/Comprehension 7) Which enzyme is the product of the lacZ gene?

Answer: β-galactosidase Section: 14.3 Skill: Knowledge/Comprehension 8) Which part of the lac operon is cis-dominant?

Answer: lacO Section: 14.3 Skill: Knowledge/Comprehension 9) Which part of the lac operon produces a regulatory protein that is trans-acting?

Answer: lacI Section: 14.3 Skill: Knowledge/Comprehension

10) In the absence of both glucose and lactose, the presence of which molecule is capable of inducing basal transcription?

Answer: cAMP Section: 14.3 Skill: Application/Analysis 11) Which protein in the arabinose operon is responsible for both positive and negative regulation of transcription?

Answer: araC Section: 14.3 Skill: Knowledge/Comprehension 12) In the trp operon, where is the attenuator region located?

Answer: trpL Section: 14.4 Skill: Knowledge/Comprehension 13) Which molecule functions as the corepressor in the trp operon?

Answer: tryptophan Section: 14.4 Skill: Knowledge/Comprehension 14) What are the three possible alternative stem loops that can form in mRNA? Answer: pause, antitermination, and termination Section: 14.4 Skill: Knowledge/Comprehension 15) Mutations of trpL decrease the efficiency of transcriptional regulation by disrupting the formation of which stem-loop structure?

Answer: 3-4 (termination) loop Section: 14.4 Skill: Knowledge/Comprehension 16) Bacteria grown at 45°C initiate expression of which two types of proteins?

Answer: heat shock and chaperone Section: 14.5 Skill: Knowledge/Comprehension 17) Antisense control of translation involves the regulation of which enzyme produced by bacterial insertion sequence IS10?

Answer: transposase Section: 14.5 Skill: Knowledge/Comprehension 18) The genetic switch controlling whether a bacterium enters the lytic or lysogenic cycle relies on the binding of which two proteins?

Answer: cro and λ repressor Section: 14.6 Skill: Knowledge/Comprehension

19) What is the process by which a bacterium switches from a lysogenic to lytic cycle?

Answer: induction Section: 14.6 Skill: Knowledge/Comprehension 20) What protein cleaves the λ repressor monomers to inactivate the repressor protein, and is activated

by DNA-damaging agents?

Answer: RecA Section: 14.6 Skill: Knowledge/Comprehension 14.3 Fill-in-the-Blank Questions 1) Most of the regulation of gene expression in bacteria occurs at the ________ level.

Answer: transcriptional Section: 14.1 Skill: Knowledge/Comprehension 2) Changing conformation at the active site as a result of binding a substance at a different site is known as ________.

Answer: allostery Section: 14.1 Skill: Knowledge/Comprehension 3) Expression of a(n) ________ alters gene transcription in E. coli by activating transcription of specialized heat stress response genes.

Answer: alternative sigma factor Section: 14.5 Skill: Knowledge/Comprehension 4) Bacterial translation is inhibited by ________ RNA, which is complementary to a portion of a specific mRNA. Answer: antisense Section: 14.5 Skill: Knowledge/Comprehension 5) In λ phage, entry into the ________ cycle requires transcription of late genes.

Answer: lytic Section: 14.6 Skill: Knowledge/Comprehension

14.4 Essay Questions 1) DNA-binding proteins exert control by binding DNA at specific sequences. Describe how DNA and protein interact at the molecular level. Are proteins able to identify specific DNA sequences, or do they bind to all regions with equal frequency? What motifs are commonly found in DNA-binding proteins?

Answer: DNA-protein interactions occur when the amino acid side chains of the proteins interact with the specific nucleotide bases and the sugar-phosphate backbone of DNA. The proteins make their contact with specific base pairs located in the major groove and the minor groove of the DNA helix using a unique pattern of hydrogen, nitrogen, and oxygen atoms that characterize each base pair. To achieve protein-DNA specificity in these interactions, the protein must simultaneously contact multiple nucleotides.

Protein secondary structures are a common motif in the structures of DNA-binding regulatory proteins, particularly α helices.

Bacterial regulatory DNA sequences frequently contain inverted repeats or direct repeats, and proteins often interact with one of the inverted repeat segments. By far, the most common structural motif in these proteins in bacteria is the helix-turn-helix (HTH) motif. In the HTH motif, two α-helical regions in each of two polypeptides in a homodimer interact with inverted repeat regulatory sequences in DNA. In each of the polypeptides, one α-helical region forms the recognition helix. This is connected to a short string of amino acids forming the "turn," which is connected to a second α-helical region known as the stabilizing helix that functions to hold the two polypeptides together.

Section: 14.1 Skill: Synthesis/Evaluation

2) What is the purpose of the lac operon, and under what circumstances is it activated? Describe the structure of the lac operon, including the repressor protein, structural genes, and regulatory regions of the operon. What is the role of catabolite repression in the lac operon?

Answer: The lac operon is responsible for the production of three polypeptides that permit E. coli to use the sugar lactose as a carbon source for growth and metabolic energy.

Glucose is the preferred energy source of E. coli, but lactose can serve as an alternative carbon source for bacteria. The lac operon acts as a genetic switch to begin breaking down lactose, but only if (1) lactose is present in the cell; and (2) all glucose has been depleted.

The lac operon consists of a multipart regulatory region and a structural gene region containing three genes. The regulatory region contains three protein-binding regulatory sequences: the promoter that binds RNA polymerase, the operator (lacO) that binds the lac repressor protein, and the CAP-cAMP region.

The three structural genes of the lac operon are lacZ, a gene encoding the enzyme β-galactosidase; lacY, which encodes the enzyme permease; and lacA, which encodes transacetylase. The β-galactosidase enzyme cleaves the β-galactoside linkage of lactose to yield glucose and galactose. The enzyme also converts lactose into allolactose, which acts as the inducer for the operon. The permease enzyme product of lacY gene controls entry of lactose into the cell. Transacetylase, the product of lacA, is not essential for lactose utilization, although in bacteria it protects against potentially damaging by-products of lactose metabolism. The regulatory gene, lacI, that produces the lac repressor has its own promoter, which is not regulated and drives constitutive transcription. The lac operon promoter, lacP, contains the consensus sequence sites that are critical for RNA polymerase binding. LacO, which binds the repressor protein produced by lacI, overlaps lacP near the start of transcription.

Section: 14.2 Skill: Synthesis/Evaluation

3) What is the role of cis-acting and trans-acting factors in transcription? Explain how these factors play a role in the lac operon, and discuss how complementation can be used to identify where the mutations have occurred (structural genes versus regulatory regions). Which type of mutations can be complemented by a second gene with normal function? Are IS and OC mutants cis or trans-acting?

Answer: If a factor is trans-acting, genes that are not physically adjacent can be regulated together and are mediated by genes that encode regulatory proteins. The lac repressor is an example of a trans-acting element because mutations can be complemented by a second gene with normal function.

If a factor is cis-acting, the DNA sequence must be adjacent to the genes it regulates. The lac operator is an example of a cis-acting element because mutations cannot be complemented by a second normal element.

Trans-acting factors (and complementation) indicate mutations in structural genes, whereas cis-acting factors indicate regulatory sequences have been mutated.

IS is a trans-acting mutation because the mutation disrupts ligand binding of the repressor to the operator on both copies of the DNA. OC is a cis-acting mutation because only one molecule of DNA is unable to be repressed due to the mutation. The operon is constitutively active on the molecule of DNA that contains the OC allele, but the other copy of the gene is intact and functions properly when the repressor is present.

Section: 14.3 Skill: Synthesis/Evaluation

4) How does the arabinose (ara) operon accomplish both positive and negative transcriptional regulation? Describe the unique DNA structure that forms during one of these mechanisms, and include a diagram to illustrate the structure of the operon in both the presence and absence of arabinose.

Answer:

The arabinose (ara) operon system is unique because a single regulatory protein, araC, carries out both positive and negative transcriptional regulation. In the absence of arabinose, a DNA loop forms and araC will bind to operator sequences araI, araO1, and araO2. A homodimer of araC protein binds at araO1, and monomers of araC bind to araI and araO2. The araC proteins bound at araI and araO2 then link to one another, inducing the formation of a DNA loop. The DNA loop prevents RNA polymerase from binding to Para, thus blocking transcription of the ara operon.

When arabinose is absent from the growth medium, araC acts as a negative control protein to suppress gene transcription. The combination of the transcription-suppressing DNA loop and araC repression means that in the absence of arabinose, the ara operon is turned off by negative transcriptional control of araC protein.

The same protein, araC, also exerts positive transcriptional control on ara gene transcription. In the presence of arabinose and the absence of glucose from the growth medium, araC operates in conjunction with CAP-cAMP to transcribe araB, araA, and araD in the ara operon.

When arabinose is present, it binds to araC proteins at all three operator sites. This breaks the connection between araC proteins at araI and araO2, and it destroys the DNA loop that suppresses ara operon transcription. After the DNA loop is broken, a second araC protein-arabinose complex binds at araI, producing araC-arabinose dimers at araI and araO1 that flank the CAP binding site. As with transcriptional activation of the lac operon, CAP must complex with cAMP and bind at the CAP binding site to activate ara operon gene transcription. The formation of the CAP-cAMP complex occurs when glucose is absent. The large complex comprised of dimers of araC-arabinose must bind at araI and

araO1, and the CAP-cAMP complex must bind at the CAP binding site, for RNA polymerase to bind efficiently and initiate transcription at Para. With these complexes in place, the ara operon is transcribed, and the B, A, and D proteins necessary to metabolize arabinose are produced.

Section: 14.3 Skill: Synthesis/Evaluation 5) Describe the difference between inducible and repressible operons. How does attenuation differ from inducible or repressible operons?

Which of the two inducible operons you've studied would be active if a cell has the following food sources? a. only lactose b. only glucose c. only arabinose d. lactose and glucose e. glucose and arabinose f. lactose and arabinose Answer: Operons like lac and ara that are involved in catabolism of alternative energy sources are typically inducible, since they are called upon only when glucose is depleted and the alternative sugar is available. In contrast, repressible operons are involved in anabolic pathways (pathways that synthesize compounds needed by the cell) and can be regulated by negative feedback mechanisms that operate through activity of the end product of the pathway to block operon gene transcription. Some repressible operons can also fine-tune the transcription to help achieve the steady-state amount of a given compound. This lets the cell make more product when needed, but stop when the cell has reached its necessary amount. So, whereas inducible and repressible are simple on-off systems, attenuation allows a wider range of transcriptional activation levels. a. only lactose = lac operon b. only glucose = none c. only arabinose = ara operon d. lactose and glucose = none e. glucose and arabinose = none f. lactose and arabinose = lac and ara operons would likely both be active until the cell has enough carbon sources to meet its needs.

Section: 14.4 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 15 Regulation of Gene Expression in Eukaryotes 15.1 Multiple-Choice Questions 1) Regulatory regions of a eukaryotic gene all contain which of the following sequences, which act only on one copy of the chromosome?

A) zinc fingers B) cis-acting regulatory sequences C) homeodomains D) trans-acting regulatory sequences E) leucine zippers

Answer: B Section: 15.1 Skill: Knowledge/Comprehension 2) You have identified a mutation in a gene which also seems to decrease transcription of another gene 2000 bp away from the mutation site. What regulatory sequence, which may be found within another gene, has likely been mutated in this instance?

A) core promoter B) proximal elements C) enhancer sequence D) homeodomain motif E) upstream activator sequence Answer: C Section: 15.1 Skill: Knowledge/Comprehension 3) In yeast, if you want to prevent the Gal4 regulatory protein from binding near each of the GAL genes, which sequence element would you target?

A) core promoter B) proximal elements C) enhancer sequence D) homeodomain motif E) upstream activator sequence Answer: E Section: 15.1 Skill: Knowledge/Comprehension

4) In the presence of galactose, you unexpectedly find that transcription is still blocked. Assuming Gal80 is present and functional, which other protein may be mutated to prevent transcriptional activation?

A) Gal2 B) Gal3 C) Gal4 D) Mig1 E) Gal10 Answer: B Section: 15.1 Skill: Knowledge/Comprehension 5) Galactose is absent, but you find active transcription of the GAL genes. Which protein is likely mutated or absent to allow for constitutive transcription of the GAL genes?

A) Gal2 B) Mig1 C) Gal4 D) Gal10 E) Gal80 Answer: E Section: 15.1 Skill: Knowledge/Comprehension 6) Which protein binds to the silencer sequence and promotes transcriptional silencing in the presence of glucose?

A) Gal2 B) Mig1 C) Gal4 D) Gal10 E) Gal80 Answer: B Section: 15.1 Skill: Knowledge/Comprehension 7) If a mouse inherits a deletion in the SHH enhancer, what effect would you expect to see?

A) enhanced limb development B) abnormal limb development C) position effect variegation D) increased β-globin production E) abnormal β-globin production (thalassemia) Answer: B Section: 15.1 Skill: Knowledge/Comprehension

8) For the following gene, MspI and HpaII restriction sites and their positions are indicated here. Recall that HpaII is methylation sensitive, while MspI is not.

If this gene is being actively transcribed, how many fragments would you expect from this restriction digest?

A) 1 B) 3 C) 4 D) 5 E) 6 Answer: E Section: 15.2 Skill: Application/Analysis 9) For the following gene, MspI and HpaII restriction sites and their positions are indicated here. Recall that HpaII is methylation sensitive, while MspI is not.

If this gene is methylated, which of the following combinations of DNA fragments would you expect from this restriction digest?

A) 5 bands = 100 bp, 200 bp, 1000 bp, 1500 bp, and 1800 bp.

B) 4 bands = 100 bp, 200 bp, 800 bp, and 900 bp.

C) 3 bands = 200 bp, 500 bp, and 1300 bp.

D) 1 band = 2000 bp.

E) No bands = DNA will be completely digested.

Answer: B Section: 15.2 Skill: Application/Analysis 10) Which sequences would likely have both a nucleosome-depleted region (NDR) and a poly A/T tract to attract transcriptional activators to a transcription start sequence?

A) chromatin modifiers B) enhancers C) open promoters D) closed promoters E) insulators Answer: C Section: 15.2 Skill: Application/Analysis

11) Molecular biologists can determine experimentally whether a region of DNA contains closed chromatin or open chromatin by assessing the sensitivity of the region to A) nucleosomes.

B) histone deacetylase.

C) RNA polymerase II. D) methyltransferase.

E) DNAse. Answer: E Section: 15.2 Skill: Knowledge/Comprehension 12) Chromatin remodeling involves both sliding and relocating of the nucleosomes. Which eukaryotic chromatin remodeling complex is likely involved?

A) ISWI complex B) SWR1 complex C) SWI/SNF complex D) SHH complex E) Mig1 complex Answer: C Section: 15.2 Skill: Knowledge/Comprehension 13) If you want to affect chromatin packaging, which amino acid could you mutate to affect both histone acetylation and methylation patterns?

A) arginine B) methionine C) histidine D) lysine E) asparagine Answer: D Section: 15.2 Skill: Application/Analysis 14) A region of chromatin has recently become DNAse I hypersensitive. Which enzyme has been activated to cause this change in chromatin structure?

A) histone methyltransferase B) histone acetylase C) histone deacetylase D) phosphatase E) kinase Answer: B Section: 15.2 Skill: Application/Analysis

15) Prader-Willi syndrome is a genetic disorder involving a partial deletion of chromosome 15q on the paternal chromosome. When both copies of a gene (or chromosome) are functional but only one is expressed, this is an example of A) position effect variegation.

B) X inactivation.

C) genomic imprinting.

D) histone acetylation.

E) chromatin modifications.

Answer: C Section: 15.2 Skill: Knowledge/Comprehension 16) When CpG islands are unmethylated, A) chromatin in the promoter region is open, allowing access by transcription factors and RNA polymerase.

B) chromatin in the promoter region is closed, preventing transcription factors and RNA polymerase from binding.

C) chromatin in the enhancer region is closed, so they are unable to bind regulatory proteins to initiate transcription.

D) DNAse hypersensitivity in that region of the chromosome is lost.

E) genes downstream of the CpG islands cannot be expressed, because the promoter region is blocked by histones.

Answer: A Section: 15.2 Skill: Application/Analysis 17) The imprinting control region (ICR) involved in genomic imprinting on chromosome 15 is what type of sequence?

A) enhancer B) insulator C) silencer D) promoter E) intronic Answer: B Section: 15.2 Skill: Knowledge/Comprehension 18) During RNAi, what do miRNAs target for destruction?

A) heterochromatic regions of DNA B) CpG islands C) mRNAs D) histones E) ribosomes Answer: C Section: 15.3 Skill: Synthesis/Evaluation

19) Many types of cancer are known to overexpress the receptor protein tyrosine kinase. Which molecular technique can be used to reduce expression of an oncogene in vitro?

A) PCR B) DNAse sensitivity assay C) RNAi D) Southern blotting E) western blotting Answer: C Section: 15.3 Skill: Knowledge/Comprehension 20) Inherited mutations in the Xist promoter lead to defects in which epigenetic process?

A) RNAi B) X inactivation C) genomic imprinting D) position effect variegation E) methylation of CpG islands Answer: B Section: 15.3 Skill: Knowledge/Comprehension 15.2 Short-Answer Questions 1) Which molecules bind regulatory sequences of DNA to encourage positive regulation of transcription?

Answer: activator proteins Section: 15.1 Skill: Application/Analysis 2) Which protein complex directs DNA bending into loops that contact RNA polymerase and transcription factors bound at the core promoter or with protein complexes bound to proximal promoter elements?

Answer: enhanceosome Section: 15.1 Skill: Application/Analysis 3) What are three proteins you would expect to find bound to the core promoter region in eukaryotes?

Answer: TBP, GTFs, and Pol II Section: 15.1 Skill: Knowledge/Comprehension 4) In gal+ yeast, what is the product of the galactose utilization pathway?

Answer: glucose-1-phosphate Section: 15.1 Skill: Knowledge/Comprehension

5) In the GAL gene system, which protein acts as an activator protein through its transcription-initiating effect?

Answer: Gal4 Section: 15.1 Skill: Knowledge/Comprehension 6) In the GAL gene system, which protein binds to the activation domain of the activator protein, ultimately blocking transcription in the absence of galactose?

Answer: Gal80 Section: 15.1 Skill: Knowledge/Comprehension 7) Gal4 is mutated such that it still binds to the DNA but cannot interact with Gal80. What effect would you expect to see in the absence of galactose?

Answer: transcriptional activation Section: 15.1 Skill: Application/Analysis 8) Gal80 is mutated such that it cannot interact with Gal3. What effect would you expect to see in the presence of galactose?

Answer: Transcription is blocked.

Section: 15.1 Skill: Knowledge/Comprehension 9) Eukaryotes use these highly specialized enhancer elements, which regulate the transcription of multiple genes packaged in complexes of closely related genes (e.g., β-globin).

Answer: locus control regions (LCRs) Section: 15.1 Skill: Knowledge/Comprehension 10) What are the protein-binding sequences that direct enhancers to interact with the intended promoter and that block communication between enhancers and other promoter?

Answer: insulator sequences Section: 15.1 Skill: Knowledge/Comprehension 11) For the following gene, which mutant likely has lost its enhancer?

Mutant # Deletion Region % Transcription WT None 100% 1 1-200 150% 2 250-400 100% 3 500-800 50% 4 950-1100 0% Answer: mutant 3 (region 500-800) Section: 15.1 Skill: Application/Analysis

12) For the following gene, which type of regulatory sequence has likely been deleted in mutant 1?

Mutant # Deletion Region % Transcription WT None 100% 1 1-200 150% 2 250-400 100% 3 500-800 50% 4 950-1100 0% Answer: silencer Section: 15.1 Skill: Application/Analysis 13) For the following gene, you notice the following results.

Mutant # Deletion Region % Transcription (lungs) % Transcription (kidneys) WT None 100% 100% 1 1-200 150% 150% 2 250-400 100% 100% 3 500-800 50% 50% 4 950-1100 0% 100% What type of sequence has been mutated in mutant 4?

Answer: tissue-specific (lung) promoter Section: 15.1 Skill: Application/Analysis 14) What are the two mechanisms by which chromatin remodelers can move nucleosomes?

Answer: nucleosome sliding and repositioning Section: 15.2 Skill: Knowledge/Comprehension 15) In yeast, which gene would you expect to be activated during phosphate starvation?

Answer: PHO5 Section: 15.2 Skill: Knowledge/Comprehension 16) In vertebrates and plants, what types of modifications to the N-terminal tails of histones lead to heterochromatin formation?

Answer: deacetylation and methylation Section: 15.2 Skill: Knowledge/Comprehension 17) Which of the four nucleotides are typically methylated in "islands," resulting in closed chromatin structure?

Answer: cytosines Section: 15.2 Skill: Knowledge/Comprehension 18) What effect does methylation of CpG islands have on human promoters?

Answer: reduced transcription Section: 15.3

Skill: Knowledge/Comprehension 19) Which protein complex binds dsRNA fragments to generate ssRNAs for RNAi? Answer: RNA-induced silencing complex (RISC) Section: 15.3 Skill: Knowledge/Comprehension 20) Which enzyme cuts dsRNA into 21-25 bp fragments during RNAi? Answer: Dicer Section: 15.3 Skill: Knowledge/Comprehension 15.3 Fill-in-the-Blank Questions 1) Gal4 binding leads to the formation of a multiprotein complex known as ________, which is an enhanceosome that forms after the activator protein binds.

Answer: mediator Section: 15.1 Skill: Knowledge/Comprehension 2) In the GAL gene system, ________ are cis-acting regulatory elements, and ________ protein is a trans-acting regulatory protein.

Answer: UASG; Gal4 Section: 15.1 Skill: Knowledge/Comprehension 3) Demethylation and acetylation lead to open chromatin structure and are associated with ________ regions of genomes.

Answer: euchromatic Section: 15.2 Skill: Knowledge/Comprehension 4) ________ in Drosophila results from the movement of the transcriptionally active w+ allele into the centromeric region of the fruit-fly X chromosome.

Answer: Position effect variegation (PEV) Section: 15.2 Skill: Knowledge/Comprehension 5) In animals, the ________ enzyme cuts pri-mRNA to produce mi-RNA. Answer: Drosha Section: 15.3 Skill: Knowledge/Comprehension

15.4 Essay Questions 1) How does transcription inhibition differ between prokaryotes and eukaryotes? Describe the two mechanisms used by eukaryotes to direct the enhancers toward certain promoters and away from others.

Answer: Prokaryotes typically use a repression mechanism in which repressor proteins inhibit transcription by binding to operator sequences that overlap promoters. This blocks the binding of RNA polymerase, thus inhibiting the gene or operon.

In eukaryotes, this mechanism of transcription inhibition is not seen. Instead, eukaryotic repressors inhibit transcription through other mechanisms.

1. Eukaryotic repressors can bind to silencer sequences, cis-acting regulatory sequences that block transcription by directly preventing enhancer-mediated transcription. For example, yeast use the proteins Mig1 and Tup1 to bind to the silencer site when glucose is present, thus inhibiting transcription of the galactose utilization pathway.

2. Eukaryotic insulator sequences can block the action of the enhancer. Insulators are cis-acting sequences located between enhancers and promoters of genes that are to be insulated from the effects of the enhancer. Insulators direct enhancers to interact with the intended promoter and block communication between enhancers and other promoters. This mechanism likely involves allowing the formation of DNA loops containing enhancers and their intended promoter targets while preventing the formation of DNA loops containing an enhancer and a promoter that is not its intended target.

Section: 15.1 Skill: Synthesis/Evaluation

2) Both cis-acting sequences and trans-acting proteins affect gene expression in eukaryotic chromosomes. What are the three basic mechanisms by which trans-acting proteins access specific regulatory DNA sequences in eukaryotic chromosomes? Describe how chromatin is altered in each of these processes.

Answer: 1. Some regulatory sequences are not tightly bound by histones, which allow open access to the regulatory DNA. These sequences include the ―linker‖ sequences between nucleosomes and sequences with specific characteristics that prevent histones from binding efficiently.

2. Chromatin remodelers can enzymatically change the distribution or composition of nucleosomes.

These remodeler enzymes can act by reorganizing nucleosomes to repress transcription, sliding nucleosomes along the chromosome or removing them to activate transcription, or changing histone composition to activate genes.

3. Chromatin modifiers can enzymatically modify histones by adding or removing methyl or acetyl groups at specific amino acid residues. Adding acetyl groups is associated with gene activation (euchromatin), while removing acetyl groups and/or adding methyl groups are associated with gene repression (heterochromatin). These chromatin-modifying enzymes are recruited to specific sites in chromatin by trans-acting factors that bind to specific DNA sequences.

This combination of activities determines the relative access of trans-acting transcription factors to cisacting DNA sequences in particular cells, at different times of organismal development, and under certain physiological conditions. Thus chromatin remodelers and chromatin modifiers mediate the reversible transition from inactive heterochromatic DNA to active euchromatic DNA. Section: 15.1 Skill: Synthesis/Evaluation 3) Explain what effect histone deacetylase inhibitors would have on gene expression and why they may be an effective anticancer therapeutic because of their impact on tumor suppressor genes.

Answer: Histone acetylation interferes with the interaction between DNA and the nucleosomes, leading to less compact, more transcriptionally active chromatin. Histone deacetylases remove acetyl groups, causing chromatin to become condensed and transcriptionally silenced. Histone deacetylase inhibitors (HDI) cause hyperacetylation of histones, which increases gene expression.

Many cancer cells show increased growth because they have decreased or lost expression of tumor suppressor genes. HDIs can inhibit the proliferation of tumor cells by reactivating transcription of cell cycle control genes and inducing cell cycle arrest, differentiation, and apoptosis of cancer cells. Since HDIs cause hyperacetylated histones to accumulate, this leads to transcriptional activation of some genes and repression of other genes.

Section: 15.2 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 16 Forward Genetics and Recombinant DNA Technology 16.1 Multiple-Choice Questions 1) Before Muller's discovery that radiation induces mutation, scientists had to work on mutations that were found solely by phenotype differences in natural populations. Which of the features of Drosophila made it a fortuitous choice for Morgan and his colleagues?

A) having a long life cycle

B) especially high rate of mutation C) large number of easily visible phenotypes D) well-known biochemical pathways E) the number of genes with only two alleles Answer: C Section: 16.1 Skill: Synthesis/Evaluation 2) Genes for regulatory elements, such as creb, were found to be important in animal learning. Which of the following would increase the probability that such an element is studied throughout the animal kingdom?

A) its unique sequence in Drosophila B) the finding that creb is highly conserved C) the number of repeats of the creb gene D) the interaction of creb with cAMP E) the finding that creb activates odor perception Answer: B Section: 16.1 Skill: Application/Analysis 3) Among the attributes of model organisms, which of the following characteristics of C. elegans makes it useful for mutagenesis?

A) hermaphrodite reproduction B) adults with ~1000 cells C) self-fertilization of eggs with stored sperm D) lack of a Y chromosome to determine gender E) adults that mature in about 2 days Answer: E Section: 16.1 Skill: Application/Analysis

4) A researcher wants to mutagenize an organism. He is not in need of a large number of mutants but is more concerned with being able to find and then to amplify the mutated sequence. Which of the following would therefore be more useful?

A) ionizing radiation B) UV radiation C) transposon insertion D) a chemical mutagen E) search for spontaneous mutant Answer: C Section: 16.1 Skill: Application/Analysis 5) If mutagenizing a number of Drosophila results in several observable phenotypes but also in several lethal mutations, does this mean that recessive lethal mutations cannot be studied further?

A) No; the lethal allele can be recovered from heterozygous siblings.

B) No; the lethal alleles can be sequenced from the other embryos.

C) Yes; lethal alleles can be studied only if they are recessive.

D) Yes; lethal alleles can be studied only if they are conditional.

E) Yes; unless the lethal allele can be supplemented with a missing gene product.

Answer: A Section: 16.1 Skill: Application/Analysis 6) If a mutagenesis screen provides some number (e.g., 24) of alleles shown by complementation analysis to be in the same gene, which of the following is true?

A) These mutants can be used to identify the role of the gene in metabolism.

B) The gene being studied must be present in more than one copy per haploid genome.

C) The gene, in whole or in part, must be involved in transformation.

D) The gene must be highly conserved in evolution.

E) The gene must be involved in regulating a signal pathway.

Answer: A Section: 16.1 Skill: Application/Analysis 7) In 1978, Arber, Smith, and Nathans won the Nobel Prize in Physiology and Medicine for work with restriction enzymes. Which of the following can still be performed without using these enzymes?

A) PCR of whole genomes B) genetic mapping C) sequence cloning D) vector packaging E) making recombinant DNA Answer: B Section: 16.2 Skill: Application/Analysis

8) Some restriction enzymes cut at specific and unambiguous sequences (e.g., NotI at GCGGCCGC). As such, the distance between two such sites in a large random DNA sequence is 48. However, the sequence recognized by XhoII is said to be ambiguous (Pu-GATC-Py, where Pu = any purine and Py any pyrimidine). What, therefore, is the expected distance between two XhoII sites in a large random sequence?

A) 4 × 4 × 4 × 4 B) 1 × 4 × 4 × 4 × 4 × 1 C) 2 × 4 × 4 × 4 × 4 × 2 D) 46 E) 42 × 22 Answer: C Section: 16.2 Skill: Application/Analysis 9) You have a DNA sequence of 1600 bp. To characterize it, you decide to make a restriction map.

Cutting with the first enzyme, you show by electrophoresis that the fragments produced are 400 and 1200 bp respectively. After using the second enzyme, you recover 700-bp and 900-bp fragments. After treatment with both enzymes, fragments were 300, 400, and 900 bp. Which is the correct map?

A) -900 ------/400--/3--B) -400/--900-----/300 --C) -300-/--400--/--900---D) -400-/300-/---900----E) -400-/-100/--300--/900 Answer: D Section: 16.2 Skill: Application/Analysis 10) In producing a recombinant plasmid to be used to clone a given donor insert, it is possible to cut both the donor and plasmid with the same restriction enzyme, resulting in complementary sticky ends.

Assuming plenty of plasmid DNA is available, why is further selection necessary before the introduction of the plasmid into a cellular system?

A) Contamination will have introduced other donor inserts.

B) Some donor inserts will be single stranded and deteriorate.

C) Some donor inserts will be sensitive to particular antibiotics.

D) Some donor pieces will remain uninserted.

E) Some donor strands will be inserted with an incorrect orientation.

Answer: E Section: 16.2 Skill: Application/Analysis

11) In forensics, the laws in the United States and elsewhere allow the establishment of databases of PCR amplification of short tandem repeats (STRs) of the Y chromosome. Which of these statements would provide an appropriate basis for making such databases divided by ethnicity?

A) Different heritage groups have different levels of such polymorphisms.

B) It would be easier to evaluate suspects of different ethnicities.

C) It would show detectives which ethnic category they should be looking for.

D) Most crimes of certain types are found among certain groups.

E) Y polymorphisms increase with the probability of criminal behavior.

Answer: A Section: 16.2 Skill: Synthesis/Evaluation 12) In selecting recombinant bacteria, cells are chosen that are resistant to a specific antibiotic. How are the bacteria made resistant?

A) They are pre-selected for the experiment on this basis.

B) The antibiotic resistance is encoded in the vector.

C) The antibiotic resistance gene is encoded on the donor insert.

D) Resistance is activated by the recombination event.

E) Resistance is activated when the cells are provided with the antibiotic Answer: B Section: 16.2 Skill: Application/Analysis 13) Which of the following vectors would be most efficient to use to amplify fragments ~300 kb in length?

A) BACs B) plasmids C) YACs D) lambda phage E) cosmids Answer: C Section: 16.2 Skill: Application/Analysis 14) One use of PCR has been to monitor the trafficking of endangered organisms. Suppose that one suspect was accused of illegally importing a specific animal, but the suspect asserts that the young animal had been bred in this country from two resident zoo animals. How would you best proceed?

A) Use blood cells to obtain DNA from the animals and look for sequences found only in illegally traded ones.

B) Use DNA obtained from biopsy of the suspect organism and use a probe for infectious agents found more frequently in its native country.

C) Use DNA obtained from the possibly imported animal and treat with restriction enzyme; use a probe for disease genes common in its native country.

D) Use DNA from hair follicles of supposed zoo parents and of the supposed offspring to compare sequences amplified with primers for a highly polymorphic region E) Amplify highly polymorphic regions from zoo and native populations to compare with the suspicious one.

Answer: D Section: 16.2

Skill: Application/Analysis 15) Current sequencing technologies have major advantages over the dideoxy method because they produce more sequence in less time and at less cost. Which of the following would provide further advantage?

A) being able to produce finer analysis of smaller sequences B) the ability to produce primers for increasingly long sequences C) the ability to sequence noncoding regions D) sequencing very rare DNAs from degraded samples E) sequencing without the need to amplify the DNA Answer: E Section: 16.3 Skill: Synthesis/Evaluation 16) In the construction of a YAC library, each chromosome has the advantage of being able to hold very long DNA sequences. However, besides selectable marker(s) the YAC vector must minimally include which of the following?

A) a bacterial origin (ori) B) cos site(s) and origin C) centromere, telomeres, and yeast origin D) centromere and origin E) centromere and telomeres Answer: C Section: 16.4 Skill: Application/Analysis 17) Whether libraries are cDNA or genomic, whether amplification is based on PCR or cloning, and whether screening is done with northern or Southern blotting or probes, all are based on which of the following?

A) complementarity B) complementation C) reverse transcription D) reverse translation E) gene expression Answer: A Section: 16.4 Skill: Application/Analysis

18) Because of redundancy in the genetic code, some amino acids can have more than one possible codon. Use the table above and select which of the answer choices is a possibility for the following amino acid sequence: valine, glycine, glycine, cysteine, tyrosine A) GUU GGC GGA UGA UAG B) GUA GGG GGG UGU UAC C) GUG GGG GGA UAC UAC D) GUA GGG GGC UGU UAC E) GAU GGC GGC UGU UAU Answer: D Section: 16.5 Skill: Application/Analysis 19) Which of the following would make a human disorder-causing allele or gene easiest to locate?

A) a genomic library for the chromosome on which the gene is located B) a cDNA library for the chromosome on which the gene is located C) a group of 10 known candidate genes D) a transposon known to segregate with the gene E) a single patient with a chromosomal translocation associated with the gene Answer: E Section: 16.5

Skill: Synthesis/Evaluation 20) In a disease associated with a nucleotide repeat, anticipation is often noticed in subsequent generations, usually associated with a more severe phenotype. Which of the following is a more likely cause of anticipation?

A) There is more frequent unequal crossing over.

B) Subtle phenotypic nuances are more often sought by a physician.

C) Family members are exposed to the same mutagens.

D) Highly repeated sequences are expressed more often.

E) The sequences have greater stability in later generations.

Answer: A Section: 16.5 Skill: Synthesis/Evaluation 16.2 Short-Answer Questions 1) Identifying an autosomal recessive mutation in a mutagenic screen in Drosophila requires identification of a mutant in the F3 generation. If testing for a sex-linked recessive lethal mutation (e.g., cn l + using a balancer chromosome such as cnCyO), in which generation can lines with mutations be identified?

Answer: F2 Section: 16.1 Skill: Application/Analysis 2) Conditional alleles result in phenotypic differences in permissive as opposed to restrictive conditions.

What do these conditions have in common?

Answer: environmental differences or environmental variations Section: 16.1 Skill: Knowledge/Comprehension 3) What kind of analysis is needed to determine whether two mutations are in the same gene or in different genes?

Answer: complementation Section: 16.1 Skill: Knowledge/Comprehension 4) In a given situation, reduced or missing function in gene A results in a viable but noticeable phenotype. The same is true for gene B. However, when both gene A and gene B products are reduced or missing in the same organism, the result is lethality. What is this called?

Answer: synthetic lethality or synthetic enhancement Section: 16.1 Skill: Knowledge/Comprehension 5) Each restriction enzyme comes from a particular bacterial strain. Why does a restriction enzyme not destroy the bacterial strain's own DNA? Answer: Its own DNA has been modified or the bases have been modified.

Section: 16.2 Skill: Knowledge/Comprehension

6) What genetically determines the difference in frequency of restriction fragments from two different enzymes used on the same genome?

Answer: length (sizes) of recognition sequence or number of bases in recognition sequence Section: 16.2 Skill: Knowledge/Comprehension 7) In a simple restriction mapping study, a sequence was cut by BamHI into one large and one small fragment. Cutting the same sequence with XhoI also results in two fragments. What can be done next to begin to map these sites?

Answer: double digest or use both enzymes together Section: 16.2 Skill: Application/Analysis 8) To produce a molecular clone requires amplification of a sequence via PCR or what other process?

Answer: using a vector (to amplify this sequence) in an organism Section: 16.2 Skill: Knowledge/Comprehension 9) How can blunt-ended DNA fragments be altered to allow them to be incorporated into a plasmid with sticky ends?

Answer: addition of oligonucleotides or linkers Section: 16.2 Skill: Knowledge/Comprehension 10) Which approach for sequencing large DNA molecules uses fragmentation that produces random but overlapping pieces?

Answer: shotgun (sequencing) Section: 16.2 Skill: Knowledge/Comprehension 11) Which type of DNA library (genomic or cDNA) would include introns and promoters?

Answer: genomic Section: 16.4 Skill: Knowledge/Comprehension 12) What technique of genomic library construction allows the genome to be cut by a given restriction enzyme at some but not all recognition sequences?

Answer: partial digestion or using less of the enzyme Section: 16.4 Skill: Knowledge/Comprehension 13) In a cDNA library, not all expressed genes will be represented. On what does this depend?

Answer: on the tissue the mRNA comes from or on the source of mRNA Section: 16.4 Skill: Knowledge/Comprehension

14) How can you use the known human cDNA gene sequence for cytC that you already have in an experiment(s) to identify and compare the sequences from other vertebrates?

Answer: Use human cDNA as a probe.

Section: 16.5 Skill: Application/Analysis 15) What method can you use to identify a gene when you know the amino acid sequence of a homologous protein?

Answer: reverse translation Section: 16.5 Skill: Knowledge/Comprehension 16) Suppose a scientist wants to identify a gene in a particular plant species by transposon tagging, but he is initially stumped because he finds no active transposons. He hypothesizes, however, that he can introduce a transposon from another plant species. Why might this be possible?

Answer: transgenes (are active in most species/organisms) Section: 16.5 Skill: Application/Analysis 17) In positional cloning, the researcher begins with a phenotype. To move toward identification of its DNA sequence, she must begin with what step?

Answer: genetic mapping Section: 16.5 Skill: Knowledge/Comprehension 18) DNA markers that are found to segregate with a mutation of interest can be used to determine the mutation's distance from the markers by looking at what?

Answer: recombination frequency Section: 16.5 Skill: Knowledge/Comprehension 19) For a contiguous sequence to be most useful in molecular mapping of a particular mutation, what should the contiguous sequence include?

Answer: two markers on either side of the mutation Section: 16.5 Skill: Knowledge/Comprehension 20) In mapping the HD gene associated with Huntington disease, eventually researchers narrowed down their hunt to four candidate genes within the mapped region. How did they identify one of these as the definitive HD gene?

Answer: polymorphisms (in HD patients) Section: 16.5 Skill: Knowledge/Comprehension

16.3 Fill-in-the-Blank Questions 1) A genetic strategy that starts with knowing the DNA sequence and works toward identifying an associated phenotype is known as ________.

Answer: reverse genetics Section: 16.1 Skill: Knowledge/Comprehension 2) In the process of identifying mutations in animals, dominant mutations can be found after one generation of mating between the mutagenized male organism and wild-type female. To identify the non-sex-linked (autosomal) recessive mutants requires ________ generations.

Answer: three Section: 16.1 Skill: Knowledge/Comprehension 3) A balancer chromosome includes a recessive lethal allele, a marker dominant allele, and ________.

Answer: inversion(s) Section: 16.1 Skill: Knowledge/Comprehension 4) A hypothetical restriction sequence is 5′-TCTAGA-3′. The palindromic sequence is 3′-________-5′.

Answer: AGATCT (must be in this orientation) Section: 16.2 Skill: Application/Analysis 5) Specific sequences in the lambda genome that interact with coat proteins to package genomes into phage particles are known as ________ sites.

Answer: cos (cohesive end sequences) Section: 16.2 Skill: Knowledge/Comprehension 16.4 Essay Questions 1) Why do we so frequently use mutant alleles to determine structure and/or function of the wild type?

Answer: Mutant alleles create alternate versions of an existing protein that can provide important information about the normal function of the protein. When the wild type function is lost, you can determine the phenotype of the organism and gain insights into the role of the protein. Similarly, if two different mutants phenocopy one another, you can identify redundant genes or members of the same gene pathway. You may find that mutations in one region of the protein may show no phenotypic effect, while other regions are highly important to normal protein function. Thus, you can begin to map the important conserved functional domains within your protein of interest if you have mutant alleles with different mutations in the same proteins.

Section: 16.1 Skill: Synthesis/Evaluation

2) Describe two or more ethical issues arising from sequencing the genomes of individuals.

Answer: Most of the concerns that have been voiced about sequencing the genomes of individuals involve issues of privacy and confidentiality. Knowing your genetic makeup can have implications for your insurance companies, as well as your employers, if sequencing identifies a genetic predisposition to a serious disease. Similarly, you may find identification of a particular gene in an individual may have unintentional implications for family members who also have a high risk of having the same genetic mutations. Often, the genetic mutations represent a predisposition to a particular disease and are not necessarily 100% penetrant. Thus, simply having the genetic marker does not mean that you will always develop the disease or disorder since many genes are multi-factorial and often influenced by environmental factors as well. Knowing you have a predisposition to a disease may help you make healthier choices and avoid other known risk factors, but they may also create additional stress on the individual and a fatalistic attitude about developing the disease. So, the decision to sequence one's genome is an important decision and can have very different ramifications, depending on the individual and what they do with this knowledge. Genetic counselors can help the individual address these issues and make an educated decision about whether sequencing is the appropriate decision in their specific instance.

Section: 16.3 Skill: Synthesis/Evaluation 3) Why is only a fraction of an organism's genes represented in any cDNA library?

Answer: cDNA libraries are comprised of complementary DNAs that are reverse transcribed from existing mRNA transcripts. Thus, only genes that are being actively transcribed in a given organism will be represented in the cDNA library, providing an accurate snapshot into the genes being expressed in this particular organism or tissue.

Section: 16.3 Skill: Synthesis/Evaluation 4) Should we allow (or urge) testing of minor age children or teens for adult-onset disorders for which there is no treatment or cure?

Answer: Answers may vary, depending on the situation. If knowing that the child will be affected in the future can help the parents prepare the child physically, psychologically, and financially, it may be beneficial for long-term planning for the family. However, genetic counseling is encouraged in these situations to provide the support the family needs to take the information and know how to proceed in the best interest of the child.

Section: 16.5 Skill: Synthesis/Evaluation 5) Would it be possible and/or appropriate to design a genetic screen to look for a gene(s) involved in homosexuality?

Answer: Answers may vary, but may include similar issues about privacy and confidentiality that may be an issue for any genetic disorder. Patients may be interested to identify a genetic component to better understand their psychology and physiology, but it is important to remember that homosexuality is likely to be a multifactorial predisposition controlled by a combination of many different genes as well as environmental factors. Thus, the chances of identifying a single gene responsible for homosexuality are slim and this complicates any attempt to design a wide-scale screening effort.

Section: 16.5 Skill: Synthesis/Evaluation

Genetics: An Integrated Approach (Sanders) Chapter 17 Applications of Recombinant DNA Technology and Reverse Genetics 17.1 Multiple-Choice Questions 1) In nature, what is the order of events in processing the product of the gene for insulin?

A) cleavage of 35 amino acids from the middle of the polypeptide to make two chains that are then joined by disulfide bonds B) cleavage of 24 amino acids from the N terminus to form proinsulin that is subsequently cleaved to make A and B chains C) removal of 5′ leader sequence from the mRNA, and splicing out of a large central intron to produce the mRNA that codes for the final form of polypeptide D) removal of large mRNA sequences by RNA editing, followed by splicing of the products of the two insulin genes E) transcription of the two insulin genes A and B, translation, and posttranslational joining of the two polypeptides Answer: B Section: 17.1 Skill: Knowledge/Comprehension 2) Which of the following is a possible advantage for the environment of having farmers use plants that genetically produce Bt toxin?

A) Plants with genes for the toxin are eaten less often by insects, and therefore less insecticide is needed.

B) Plants that produce Bt toxin must be grown from new seeds each season, and therefore fewer of them are planted.

C) Bt toxin is advantageous to many species of invertebrates that grow in the same fields.

D) Harvesting plants with these genes is less disturbing to the soil layers where they are planted.

E) Plants with these genes must be sprayed with the organism, but need much less than other plants.

Answer: A Section: 17.1 Skill: Application/Analysis 3) In using Agrobacterium tumefaciens to transfer genes into plants, what is transferred from bacterium to plant?

A) Ti plasmid B) T-DNA C) opines D) only the recombining sequence E) the entire bacterium Answer: B Section: 17.1 Skill: Knowledge/Comprehension

4) If a gene in Drosophila is expressed only when an imaginal disc (A) begins to differentiate, it is referred to as a homeotic gene. Recessive mutations in such genes often result in loss of function so that the structure does not develop. If the gene is expressed, but in a different imaginal disc (B), what is the result?

A) loss of function of disc B's primary gene B) gain of function of several inappropriate genes in disc A C) gain of function in disc B at the "wrong" location D) deregulation of gene expression throughout disc A E) deregulation of gene expression throughout disc B Answer: C Section: 17.2 Skill: Application/Analysis 5) In reverse genetics, what is the correct order in which the experimenter proceeds?

A) silencing the genes in question using RNAi, followed by generating gain-of-function alleles B) screening genes via PCR, followed by deletion of the flanking sequences C) selection of a phenotypic alternative for the gene in question and sequencing its DNA D) induction of and selection for a mutant allele with a known sequence, followed by screening for mutations of interest E) random bombardment of the DNA with a known mutagen, followed by observation of offspring for newly acquired traits Answer: D Section: 17.3 Skill: Application/Analysis 6) You have evidence that a gene, VTE4, of known sequence and well-understood transcriptional regulation, is available from another lab. The product of this gene enables seeds to produce a significantly increased amount of vitamin E from a substrate already produced in Brassica napus, the plant that gives us canola oil. To produce canola oil with increased vitamin E, how would you proceed?

A) Select Brassica plants with the highest titer of the vitamin, isolate the genes responsible, and cause these genes to duplicate.

B) Use a yeast system to engineer yeasts that produce the vitamin, and get manufacturers to add this to the canola oil during bottling.

C) Introduce the gene and necessary cis elements into Brassica, along with a reporter gene, then measure the vitamin E produced.

D) Clone VTE4 in A. tumefaciens and use these bacteria to produce transgenic Brassica.

E) Introduce VTE4 into tomato plants or another food that humans eat more than Brassica.

Answer: C Section: 17.5 Skill: Application/Analysis

7) In Drosophila, mutation screening usually involves use of a balancer chromosome that includes three elements: a set of overlapping inversions, an easily recognized dominant mutation, and a recessive lethal mutation that prevents balancer homozygotes from surviving. Which one, or combination, of these elements is necessary and sufficient to suppress crossovers?

A) inversions B) dominants C) lethal recessive D) inversions plus dominant E) lethal plus inversions Answer: A Section: 17.1 Skill: Application/Analysis 8) Suppose you have just discovered a Drosophila gene that has a sequence similar to a gene of interest in humans. You need to have mutants of the Drosophila gene because A) you need to see whether fly mutants have a defect in the same system.

B) you need to see whether the mutant produces the same phenotype.

C) you need to see whether the Drosophila mutant disrupts the same metabolic pathway.

D) you want to compare the mutant Drosophila sequence to each mutant human sequence.

E) you need to have as many alleles in Drosophila as you have found in humans.

Answer: C Section: 17.1 Skill: Synthesis/Evaluation 9) EMS (ethylmethane sulfonate) is mutagenic because A) it breaks phosphodiester binds.

B) it adds alkyl groups to bases.

C) it forms cross-links between DNA strands.

D) it makes DNA more susceptible to radiation.

E) it causes chromosomal deletions.

Answer: B Section: 17.1 Skill: Knowledge/Comprehension 10) Besides the obvious fact that chemical mutagens are dangerous to handle, another major disadvantage to their use is that A) most of these mutations are immediately lethal.

B) most chemically induced changes are detectable only by sequencing.

C) chemical mutagens activate transposons.

D) two or more generations of further matings are required to isolate them.

E) they alter all A-T and C-G base pairing.

Answer: B Section: 17.1 Skill: Application/Analysis

11) The reasons for using transposon-based mutagenesis include all of the following except A) disrupting the coding sequence.

B) disrupting a regulatory sequence.

C) creating improper splice sequences.

D) addition of alkyl groups to bases.

E) blocking of enhancer-promoter interactions.

Answer: D Section: 17.1 Skill: Synthesis/Evaluation 12) Enhancer trapping might occur if you create a library of insertions bearing enhancer-less GAL4 genes. Their expression will be controlled by A) also inserting a viral enhancer.

B) also inserting a GAL4 reporter.

C) including a bacterial origin sequence.

D) enhancers mapping near the insertion site.

E) including an enhancer activator.

Answer: D Section: 17.3 Skill: Application/Analysis 13) P element transposition requires the enzyme transposase, which acts at the ends of the P element. To find transposons in a germ line, which of these is best?

A) use of a "crippled" P element that cannot transpose itself but can produce the enzyme B) use of a "crippled" P element that can be transposed if there is another source of transposase C) use of a P element that does not share the genome of the germ line D) use of restriction sites that flank the inserted P element E) use of DNA ligase to incorporate the P element Answer: A Section: 17.3 Skill: Application/Analysis 14) If you wish to rid an experimental organism of the function of a gene, you can use RNAi. What would you inject into the organism?

A) co-suppressor with double-stranded (ds) DNA B) ssRNA complementary to cDNA C) ssRNA complementary to mRNA D) dsRNA homologous to the gene E) dsRNA including transposase Answer: D Section: 17.3 Skill: Application/Analysis

15) Suppose you have engineered a plasmid vector in order to produce a functional animal protein in E. coli. But the product you collect, while it has the correct amino acid sequence, is in fact nonfunctional.

What might you try that would have a greater chance of success?

A) a eukaryotic expression vector that has sequences for regulating transcription B) using a host other than E. coli that does not have the same codon bias C) a eukaryotic expression vector that includes sequences for controlling posttranslational effects D) treatment of the posttranslation proteins to add signal sequences E) using yeast artificial chromosomes that can hold longer sequences Answer: D Section: 17.1 Skill: Application/Analysis 16) If genes are introduced into a yeast plasmid to serve as a shuttle vector, the plasmid has the ability to A) move genes from human to yeast.

B) replicate in yeast or in bacteria.

C) degrade selected yeast genes.

D) recombine with yeast nuclear genes.

E) degrade other yeast plasmids.

Answer: B Section: 17.1 Skill: Knowledge/Comprehension 17) When a transgene is used in mice to knock out a specific function, two problems that might occur are the unintended introduction of multiple copies, or abnormal gene expression. The second of these problems is related to A) position effect.

B) homologous recombination.

C) vector insufficiency.

D) mitotic recombination.

E) unequal crossing over.

Answer: A Section: 17.1 Skill: Application/Analysis 18) To introduce dsRNA into C. elegans, it is enough just to dip the worms into media that includes E. coli containing the dsRNA because A) E. coli infect C. elegans and then the RNA is released.

B) E. coli break up into pieces due to the presence of the worms.

C) C. elegans feed on the transgenic bacteria.

D) C. elegans emit toxins that paralyze the bacteria.

E) E. coli emit chemicals that cause pores to open in the nematode body wall.

Answer: C Section: 17.2 Skill: Knowledge/Comprehension

19) Hedgehog (Hh), and many other signaling pathways, were originally discovered in Drosophila.

Hedgehog was so named because of bristles all over the early embryo, resembling a hedgehog. It works, with other genes, to develop the anterior-posterior axis of the embryo.

The mutant develops bristles all over the embryo instead of just in a stripe toward the anterior. Thus Hh is a segment polarity gene. The hedgehog gene family in vertebrates is involved in development of central and peripheral nervous system, skeleton limbs, lungs, skin and germ cells. Shh (Sonic hedgehog) is responsible for major development of brain, spinal cord, and skeleton and is received at the membrane via the patched receptor (PTCH). In the absence of Shh, PTCH inhibits the activity of the pathway, while binding of Shh allows signal transduction to go forward.

In humans, holoprosencephaly is a developmental disorder involving incomplete cleavage of the forebrain, leading to the abnormalities of brain and facial structures. From the above, which abnormalities of genes must be involved?

A) some other member of the Hh gene family B) Shh only C) PTCH only D) Shh in combination with PTCH E) either Shh or PTCH Answer: E Section: 17.5 Skill: Application/Analysis 20) Positional cloning identified the human homologue of the Drosophila patched (PTCH) gene as a candidate gene for the human disorder NBCCS or Gorlin syndrome. Screening of the patched coding region revealed a wide spectrum of mutations in NBCCS patients. Most of these mutations were predicted to result in premature protein truncation. What phenomenon would the researchers look for in the patients?

A) haploinsufficiency for the protein B) recessive familial pattern of inheritance C) predisposition to developmental abnormalities of the brain D) incomplete formation of the forebrain E) Shh mutations Answer: A Section: 17.5 Skill: Application/Analysis

17.2 Short-Answer Questions 1) In producing an E. coli plasmid to be an expression vector, an experimenter includes genes for selectable markers, a promoter for the expression of the introduced genes. She uses a cDNA sequence to ensure the insert contains no introns. The gene inserted is transcribed, but not translated. What sequence did she omit from the plasmid?

Answer: Shine-Dalgarno (sequence) Section: 17.1 Skill: Application/Analysis 2) One way to genetically engineer S. cerevisiae is by homologous recombination. If a given sequence is thus introduced and is integrated via two regions of homology, is it more probably introduced as a linear or a circular entity?

Answer: linear Section: 17.1 Skill: Application/Analysis 3) If a gene is replaced with a heterologous sequence such that the gene is then nonfunctional, what is the result usually called?

Answer: a knockout Section: 17.1 Skill: Knowledge/Comprehension 4) Agrobacterium tumefaciens is often used in the production of transgenic plants because of its Ti plasmid. This plasmid is usually "disarmed" by deleting its tumor genes along with the genes for producing what products that would have been advantageous to the bacterium?

Answer: opines Section: 17.1 Skill: Knowledge/Comprehension 5) If a plant species cannot readily have a new gene integrated into it via a Ti plasmid, it must then be subjected to different methods. These often include removing the cell wall and inserting DNA, after treating the membrane with heat or electricity to do what to the membrane?

Answer: depolarize Section: 17.1 Skill: Application/Analysis 6) In the formation of transgenic animals, DNA is usually inserted into eggs or embryos. If some cells of the embryo do not receive the introduced DNA, the result is an organism with cells of different genotypes; the organism is known as what?

Answer: a chimera Section: 17.1 Skill: Knowledge/Comprehension 7) For a transgenic animal line to be produced from a chimera, the transgene must be carried by what?

Answer: germ line or germ cells Section: 17.1 Skill: Knowledge/Comprehension

8) By what method is T-DNA integrated into the plant genome?

Answer: nonhomologous recombination Section: 17.1 Skill: Knowledge/Comprehension 9) In the pUC plasmids, a polylinker region that includes multiple restriction sites is embedded in the lacZ component. If a sequence of interest is inserted into this region, what is the effect on lacZ? Answer: inactivation Section: 17.1 Skill: Application/Analysis 10) If a researcher is cloning a sequence by using pUC and uses the plasmid to transform bacteria, which are then plated on X-gal medium, would the recombinant plasmids be found in blue or white colonies or both?

Answer: white Section: 17.1 Skill: Application/Analysis 11) The GFP gene is useful as a reporter gene because the cells and tissues with it emit bioluminescence when treated with what, that functions as a substrate?

Answer: UV light Section: 17.2 Skill: Knowledge/Comprehension 12) If an experimenter wants to use the GFP method but needs to detect the presence or absence of several proteins at the same time, he can take advantage of mutational variants of GFP that emit what?

Answer: other colors Section: 17.2 Skill: Application/Analysis 13) What is the function of imaginal disc cells during the larval stage of Drosophila development?

Answer: none Section: 17.2 Skill: Knowledge/Comprehension 14) What kind of gene silencing is induced by short double-stranded RNAs? Answer: RNA interference/RNAi Section: 17.3 Skill: Knowledge/Comprehension 15) In an organism, X, it is found to be impossible to generate loss-of-function mutants for most genes.

There is an alternative method of introducing random mutations that can later be screened for the gene(s) in question. This can be accomplished by using insertions due to what structures?

Answer: transposons or T-DNA Section: 17.3 Skill: Application/Analysis

16) Knockout libraries are available already prepared for some organisms such as Drosophila. These libraries are originally made by what method?

Answer: reverse genetics Section: 17.3 Skill: Knowledge/Comprehension 17) In nature, RNAi protects cells against infection by what kind of pathogen?

Answer: dsRNA viruses or double-stranded RNA viruses Section: 17.3 Skill: Knowledge/Comprehension 18) If a mutagen such as EMS is used to produce numerous mutagenized cell lines for an organism, PCR can be used to select for a particular gene. However, what feature of this selected sequence is used to find the ones with mutations in that gene?

Answer: heteroduplexes or heteroduplex DNA Section: 17.3 Skill: Application/Analysis 19) In the case of Jesse Gelsinger, a young man who died during gene therapy, the therapy involved delivery via a viral vector. Was this germ-line or somatic therapy?

Answer: somatic Section: 17.4 Skill: Application/Analysis 20) In grape cultivation, growers often find that heterozygous forms are more desirable. To propagate the heterozygotes without having to interbreed in each generation, what method can they use?

Answer: cloning Section: 17.5 Skill: Application/Analysis 17.3 Fill-in-the-Blank Questions 1) If recombination results in random integration of a DNA sequence, so that it produces a nonhomologous region, the recombination is ________.

Answer: illegitimate Section: 17.1 Skill: Knowledge/Comprehension 2) If the process used to investigate a genetic condition begins with knowing a DNA sequence and aims at finding the phenotype it causes, this process is known as ________.

Answer: reverse genetics Section: 17.5 Skill: Knowledge/Comprehension 3) Suppose a gene from a cow is introduced into a frog. This results in a ________ transgene.

Answer: heterologous Section: 17.1 Skill: Knowledge/Comprehension

4) Suppose a mutagen is used on an inbred population of mice to establish cell lines, each including a different mutant allele, which can then be tested for function. This technique is known as ________.

Answer: TILLING Section: 17.3 Skill: Knowledge/Comprehension 5) In species X, the UUU codon for phenylalanine is used significantly more frequently than other phenylalanine codons, and knowing this allows the scientist to manipulate translation efficiency. This unequal frequency is known as ________.

Answer: codon bias Section: 17.1 Skill: Knowledge/Comprehension 17.4 Essay Questions 1) Design an experiment that uses enhancer trapping to drive the expression of a gene of interest.

Answer: Enhancer trap constructs include a reporter gene that is fused to a weak promoter that, by itself, is insufficient to drive expression of the reporter gene. The construct also requires a mobile element, typically a P element, which allows this basal transcription unit to be inserted into the genome.

When the reporter gene is integrated near an enhancer region, the enhancer is "trapped" and the reporter gene is expressed, which allows you to identify spatial regulation by enhancers. In E. coli, a common enhancer trap is P[lacZ], while yeast often uses a P[GAL4] enhancer trap. Enhancer trapping also uses a visible marker to help with easy identification of new insertions. For example, the white gene, which is responsible for wild-type red eye color in Drosophila, is disrupted by an insertion event, creating a white-eyed phenotype. Drosophila also has the benefit of having a large number of readily available fly stocks containing GAL4 insertions, which act as the "drivers." Using one of these stocks, plus a second fly stock containing a UAS DNA sequence followed by your gene of interest, you can simply cross these two lines of flies. This eliminates the need to directly clone transgenic flies with the enhancer linked to your gene and saves a lot of time and resources, making enhancer trapping a relatively easy process.

Section: 17.4 Skill: Application/Analysis 2) Discuss the ethical and societal reasons for and against germ-line gene therapy.

Answer: Answers may vary, but it is important to distinguish between somatic and germline gene therapy. Somatic gene therapy will treat the disorder in the individual, but they still have the potential to pass the same mutation to their progeny. Germline therapy corrects the mutation in all cells, essentially changing not only the individual but all subsequent generations. There may be differing opinions about the ethics of the two types of therapy, since somatic gene therapy can be performed on individuals old enough to consent to treatment, whereas germline gene therapy can affect all offspring before fertilization has even occurred.

Section: 17.5 Skill: Synthesis/Evaluation

3) In the 1990s, gene therapy was seen as a probable cure for many genetic diseases. List three or four reasons that it has so far been of limited success.

Answer: Gene therapy is technically challenging, particularly in the mechanisms of gene delivery and activation in the correct target tissue. The gene must be targeted to the correct tissue to ensure proper wild-type expression. If the gene is targeted to the wrong tissue, it would be inefficient, but could also result in severe health problems by introducing a gene into a tissue which normally does not express that gene product. Similarly, if a gene is intended to be targeted to one particular tissue and is, instead, introduced into the germline, it may be passed on to offspring with unintended consequences. Instead of expressing the gene only in the targeted tissue, it would now affect all cells in the patient's offspring, which may prove to be fatal or have severe side effects.

Within the individual, gene therapy may inadvertently result in the activation or deactivation of other genes based on where the DNA incorporates into the genome. The viral vector itself can also elicit a severe immune reaction in some individuals, which was the case in early gene therapy trials. In addition, if a construct has been incorporated, it is difficult to control levels of gene expression and the amount of expression may be higher or lower than anticipated.

Section: 17.5 Skill: Synthesis/Evaluation 4) What characteristics of plants allow them to be cloned relatively easily, whereas cloning in animals is rare and difficult to achieve?

Answer: There are several reasons why cloning is easier in plants than animals. First, plants can be easily propagated through vegetative reproduction or apomixis. In addition, plant cells are often totipotent, which makes the cloning process much easier because there is no need to "reprogram" the nucleus. Animal cloning can be accomplished through parthenogenesis in rare instances, but is primarily accomplished through reproductive cloning and somatic cell nuclear transfer. This process requires the transfer of a nucleus from a donor cell to a recipient egg, and is often complicated by the incomplete reprogramming of the genome because donor cells are no longer totipotent as in plant cells. Cell differentiation in animals therefore limits the success rate of somatic cell nuclear transfer and the formation of viable organisms from such techniques.

Section: 17.5 Skill: Synthesis/Evaluation 5) Transgenic fusions between a gene and a reporter are used to determine whether a gene is expressed.

Which of these components allows the researcher to work with living specimens, and why?

Answer: The reporter genes are extremely helpful for tracking gene expression, but it is important that the reporter product does not harm the organism. For instance, lacZ is often used to screen cells in culture through blue-white screening, but requires the substrate X-Gal to produce the blue product.

Injecting X-gal into a living organism may have toxic effects and is, therefore, not commonly used when researchers want to examine living organisms. Alternatively, reporter genes such as green fluorescent protein (GFP) are not toxic and require no substrate. They have been used as markers in the cloning of transgenic pigs and had no toxic effects. The other concern is the effect of the reporter gene fusion on wild-type function of the protein. Because protein conformation and proper folding of the protein is so important to the function of the protein, fusing another gene product to the protein can have unintended consequences for protein function. It may be necessary to test N-terminal versus C-terminal fusions, as well as using protein reporters that are different sizes. Additionally, some organisms and microenvironments may not be conducive to certain reporter genes. Fluorescent protein reporters such as GFP are not.

Section: 17.2 Skill: Application/Analysis

Genetics: An Integrated Approach (Sanders) Chapter 18 Genomics: Genetics from a Whole-Genome Perspective 18.1 Multiple-Choice Questions 1) In the 2009 article "Genomic Footprints of a Cryptic Plastic Endosymbiosis in Diatoms" (Science 324: 1724-26), Moustafa et al. described their search for possible gene transfer from organelles to the nucleus in diatom genomes. They assumed that diatomic plastids came from red algae. However, they found that most were from green algae. Which is most likely?

A) Diatoms once had green algal endosymbionts.

B) Red algae had green algal endosymbionts.

C) Red plastids were replaced by green plastids.

D) Cyanobacteria imported green plastid genes.

E) All red algal genes mutated to green algal genes.

Answer: A Section: 18.2 Skill: Synthesis/Evaluation 2) Celiac disease (CD) is a common, usually inherited human disorder of intestinal inflammation that is triggered by eating gluten, a major protein in wheat and other cereal grains. It is the most frequent food intolerance. One of the three genes involved is for SH2B3, a protein involved in innate, nonspecific immune response to bacterial pathogens. Celiac sufferers frequently have a particular allele of this gene.

It is so common that which of the following might be hypothesized?

A) The gene must be common to all primate genomes.

B) The gene must be common in most mammalian genomes.

C) The allele must have been positively selected for during a period of especially frequent bacterial infections.

D) The wild-type allele must have been selected against during periods of frequent bacterial infections.

E) The gene must code for a T-cell receptor for a common bacterial component.

Answer: C Section: 18.2 Skill: Synthesis/Evaluation 3) Although whole-genome duplications that result in polyploidy are fairly common in plant species, which of the following are common in other eukaryotes?

A) duplication of whole chromosome arms B) segmental duplications of small gene regions C) deletion of an entire haploid set of chromosomes D) whole chromosome loss E) miscellaneous aneuploidies in living members of a species Answer: B Section: 18.2 Skill: Knowledge/Comprehension

4) To study the maximum amount of genome diversity in the human species, which of the following provides the most information?

A) karyotyping B) SNP variation C) DNA fingerprinting D) sequencing of one or more specific genes E) ortholog comparisons Answer: B Section: 18.2 Skill: Knowledge/Comprehension 5) When authors discuss whole genome-wide examination of genes, to what do they refer?

A) protein expression patterns B) patterns of protein immunoprecipitation C) RNA sampling techniques D) comparison of conserved sequences E) use of DNA microarrays Answer: E Section: 18.3 Skill: Knowledge/Comprehension 6) Which of the following describes genome tiling arrays?

A) use of any DNA microchips B) use of any RNA microchips C) use of several microchips that together contain all sequences of a genome D) use of an entire series of SNP comparisons, including at least one per chromosome E) use of high-throughput technology Answer: C Section: 18.3 Skill: Knowledge/Comprehension 7) What is the purpose of the two-hybrid system?

A) to study the nature of two hybrid organisms at a time B) to study the interaction between an enhancer and a promoter C) to explore a cDNA library to find interacting DNA sequences D) to discover how and whether two proteins interact E) to locate the chromosomal position of hybridized genes Answer: D Section: 18.3 Skill: Knowledge/Comprehension 8) Which of the following defines the essential gene set of an organism such as a species of yeast?

A) genes that, when mutated, reduce the growth of a heterozygous organism B) genes whose deletion alleles are not found among haploid offspring C) the fewest genes that, when mutated, result in slow growth in optimal environments D) genes whose mutations result in haploinsufficiency E) genes that, when mutated, are either lethal or conditionally lethal in special environments Answer: B Section: 18.3

Skill: Application/Analysis 9) When Warnecke and colleagues were looking for genes that are involved in the digestion of lignocellulose by termites, they needed genes A) from the sequences of the termite genome.

B) of the primary strain of bacteria that inhabit the termites' bodies.

C) for the bacterial enzymes found in the lignocellulose-digesting part of the gut.

D) for the long-term maintenance of lignin-digesting microbial organisms.

E) that can be expressed in a lignocellulose-rich environment.

Answer: C Section: 18.3 Skill: Application/Analysis 10) The splicing regulatory protein (SR) identifies exons to be included in mRNA. How might SR proteins regulate alternative splicing?

A) There are alternative binding regions for these proteins within each exon.

B) There are alternative SR proteins in different tissues at different stages of development.

C) The SR proteins lose the ability to bind to exon splice suppressors.

D) The SR proteins recruit the splicing small RNAs (U1, U2,…) or unusual ones.

E) The spliceosome complex can bind only to regions with bound SR proteins.

Answer: B Section: 18.3 Skill: Application/Analysis 11) Transcomplementation is a name sometimes given for a test to see whether a gene from one organism can "rescue" a loss-of-function mutant of a different organism with a homologous gene. To which of the following processes might this term apply?

A) replacement of a nonfunctional globin gene with a globin pseudogene B) substitution of a drought-resistant gene from potato plants with a drought-resistant gene from a cactus species C) introduction of a human WEE1 gene into a yeast culture with wee1- mutants D) replacement of a cdc25- with a cdc13+ gene to rescue a cell cycle dysfunction E) use of Drosophila heat shock gene promoters to regulate heat shock genes in fugu Answer: C Section: 18.3 Skill: Application/Analysis 12) Subtelomeres are blocks of homologous DNA sequences close to the ends of chromosomes. In a worldwide population of humans, when sequences of chromosomes 4q and 10q subtelomeres were analyzed only 17 4q and 8 10q haplotypes were found among African, European, and Asian groups.

Which of the following might be among the results or conclusions of this analysis?

A) finding that few or no interchromosomal exchanges can occur among these chromosomes B) a conclusion that all haplotypes were present before the human population migrated from Africa C) reporting that haplotype polymorphism shows no response to natural selection D) finding greater than 80% variation between homologous ends of chromosomes in humans E) a conclusion that assigning chromosomal origin of repeat arrays by using specific restriction enzymes is possible only in a population with the same ethnic origin Answer: B Section: 18.2

Skill: Application/Analysis 13) Scientists studying the causes of a type of progressive hearing loss known as otosclerosis want to identify the genetic contribution to the disorder. By which of the following are they most likely to find a causative gene or genes?

A) a genome-wide association study with a large number of SNPs, using pooled DNA samples B) genome-wide linkage analysis studies of families whose members have otosclerosis C) a candidate gene case-control association study with genes known to be involved in ear development D) comparison of gene expression among otosclerosis patients E) allelic heterogeneity studies for a gene expected to be involved Answer: A Section: 18.3 Skill: Application/Analysis 14) It has been predicted that genes may be involved in smoking initiation and/or current smoking behavior. Previously, large twin and family studies have provided evidence of this and an estimate of heritability (genetic contribution) of 37-55% for initiation and 46-59% for current smoking, depending on gender. A large group of researchers from several institutions composed a pathway-or-systems-based series of studies including SNP analysis, cellular location and protein interaction, replication with other samples, statistical analysis, and bioinformatics. What is a likely result?

A) The study is unlikely to produce meaningful results because it is too diffuse in its methods.

B) The samples used also correlate with behaviors like depression, often associated with smoking, and they may be finding depression-associated genes.

C) Two specific genes, one for a transporter protein and one for an opioid receptor, are clearly shown to be causative in all members of each sample.

D) A group of genes that tend to code for proteins with similar biological function have been identified to contribute significantly.

E) Genome-wide association studies are of no practical use in studying human behaviors.

Answer: D Section: 18.1 Skill: Synthesis/Evaluation 15) Which of the following would likely exclude a sequence from being considered as an active gene?

A) a promoter with signals identical to the promoter of another gene B) exons in different reading frames from one another C) an exon followed by 3′ processing signals D) only one exon with an in-frame stop codon E) introns missing the GT-AG splice junctions appropriate for this gene Answer: E Section: 18.1 Skill: Application/Analysis

16) In an experiment designed to analyze the mRNAs present in a cell at a given time in the life cycle, it was found that about 50% of the mRNAs are of one type, less than 20% represent 8 other mRNAs, and the rest consist of several thousand different mRNAs. What does this information indicate?

A) Only the mRNA expressed at the highest level is of any importance to this cell type.

B) The pattern of expression of the low-level mRNAs is what makes the cell type unique.

C) The low-level expressed genes are from organelles.

D) Those genes expressed at low level are likely to be "housekeeping" genes found in many cell types.

E) The genes most specific to this cell type and life stage are the 8 found at an intermediate level.

Answer: D Section: 18.3 Skill: Synthesis/Evaluation 17) Recombination between two Alu elements results in deletions. In humans these have often been associated with specific disorders. In a recent study, researchers compared Alu recombination-mediated deletions of humans and chimpanzees. They identified 492 human-specific deletions of this type. Why is this finding important?

A) It shows that humans are not as genetically close to chimpanzees as researchers previously thought.

B) It shows how much the chimpanzee genome has changed since the divergence of these two lineages.

C) It shows that the time of divergence of the two lineages is much older than was previously thought.

D) It shows that more deletions have occurred in humans than in chimps.

E) It demonstrates the importance of Alu recombination in shaping genome divergence.

Answer: E Section: 18.2 Skill: Application/Analysis 18) A report of a newly identified syndrome in families of the Old Order Amish showed that the lethal condition that resulted from numerous, multisystem abnormalities is inherited as an autosomal recessive.

SNP-based mapping showed that a portion of a region on chromosome 6p, namely the ICK gene (intestinal cell kinase), was homozygous in all autopsied patients but heterozygous in normal members of the affected families. The mutation seems so far to be specific only to the Amish. Which of the following would be a reasonable next step in studying this mutation?

A) sequencing to identify the abnormality in the ICK gene B) use of a microarray to see whether the gene is expressed in heterozygotes C) kinase activity assays in adult affected members of the families D) development of a gene-based test for assessing carrier risk E) development of a gene-based test for prenatal diagnosis Answer: A Section: 18.1 Skill: Application/Analysis

19) A significant polymorphic inversion on chromosome 17, the MAPT inversion, is identified by marker SNPs that can now be used to screen large numbers in a population. The inverted region is large and extends for approximately 900 kb. The two haplotypes (one inverted and one not) are each associated with increased risk of a number of neurological conditions. How can the inverted region be used to find the most recent common ancestor of the inversion?

A) by typing SNPs from humans and from other primates B) by assessing SNPs from a large number of human populations C) by looking for heterozygotes for the two kinds of chromosome 17 D) by fertility studies on populations with each of the haplotypes E) by using a BAC clone library to identify the ends of the inversion Answer: B Section: 18.2 Skill: Synthesis/Evaluation 20) In looking for a gene responsible for albinism in humans, a group of researchers decided that at least one form of this condition was due to a lack of the enzyme tyrosinase. With this decision as a starting point, which of the following series of steps was most feasible?

A) Sequence the amino acids of the protein, use rabbits to make an antibody to the protein, purify and fluorescently label the protein, and hybridize the antibody to a cDNA library.

B) Sequence the protein, use the code table to figure out the best RNA sequence for it, synthesize the RNA, and use the RNA as a probe in cells from albino individuals.

C) Radioactively tag the enzyme and use it as a probe to identify the biochemical reaction in which it functions, work out the metabolic pathway involved, and find cells in albinos that cannot carry out these reactions.

D) Sequence the protein, work out the cDNA that codes for it, label the cDNA code, and use a FISH study to localize the gene on a karyotype.

E) Sequence the appropriate region from albino and non-albino cells and compare the sequences or do subtractive hybridization to see where there is a deletion.

Answer: A Section: 18.3 Skill: Application/Analysis 18.2 Short-Answer Questions 1) Suppose that you have been able to partially sequence a significant number of cDNA clones from an as yet unsequenced insect genome. These clones are useful as what sequences that can help annotation?

Answer: ESTs Section: 18.1 Skill: Application/Analysis 2) Often, in the absence of experimental data, computerized algorithms are used to predict gene structures from large sequences. What is this approach known as?

Answer: bioinformatics Section: 18.1 Skill: Knowledge/Comprehension

3) Suppose that you use the sequence of the lacI gene, whose function you know is to encode a transcription factor, to find similar sequences in another bacterial strain. Are you more likely to find similar operons or similar transcription factors?

Answer: transcription factors Section: 18.1 Skill: Application/Analysis 4) In humans, a large number of genes associated with immune function share sequence similarity and are evolutionarily related. These would be known as what kind of group?

Answer: a gene family Section: 18.1 Skill: Application/Analysis 5) A researcher has produced groups of contigs that are linked via paired end sequences, although these include some sequence gaps. What are these groups called?

Answer: scaffolds Section: 18.1 Skill: Knowledge/Comprehension 6) Genes can be assigned, relatively speaking, to the categories "essential" and "nonessential." In the identification of essential genes, a transposon can be inserted into a gene to inactivate it. The gene would then be identified as essential if the inactivation results in what?

Answer: lethality or death Section: 18.1 Skill: Application/Analysis 7) In a simple bioinformatics exercise, students are asked to locate the beginning of an ORF among the six possible reading frames for a hypothetical DNA molecule. The first step in their process should involve looking for what sequence?

Answer: ATG Section: 18.1 Skill: Application/Analysis 8) The BLAST program from NIH enables a researcher to look for homologous genes. It does so in general by searching for identity in what property?

Answer: nucleotide sequence Section: 18.1 Skill: Knowledge/Comprehension 9) In 2000, when then President Clinton along with Francis Collins of the Human Genome Project and Craig Venter of Celera announced the completion of a "draft" of the human genome, the event did not, in fact, represent true completion because most of what type of sequences were not included?

Answer: repetitive Section: 18.1 Skill: Knowledge/Comprehension

10) If a gene product is not a protein but a noncoding RNA, the sequence might be initially hypothesized from comparative genomics but must be confirmed how?

Answer: by experimenting or experimentally Section: 18.1 Skill: Knowledge/Comprehension 11) Very many related genes can be recognized by conserved protein domains. To what gene structures do these domains correlate?

Answer: exons Section: 18.1 Skill: Knowledge/Comprehension 12) Species whose ancestors might once have been free living may now be present as parasites. Would their genomes now tend to be larger, smaller, or approximately the same in size and complexity?

Answer: smaller Section: 18.1 Skill: Knowledge/Comprehension 13) Interspecific genome studies show us conserved sequences, but intraspecific comparisons identify what?

Answer: (sequence) differences or polymorphisms Section: 18.2 Skill: Knowledge/Comprehension 14) Suppose a geneticist sequences a specific gene region in two species, A and B, previously known to be closely related. He specifically wants to see which of the two species is closer to an ancestral species.

The sequence comparison showed that this region in species B has an extra intron, unlike other near relatives, but that otherwise A and B are virtually the same for this region. Based on this data alone, he could reason that which species was more likely ancestral?

Answer: A Section: 18.2 Skill: Application/Analysis 15) An Alu sequence is a kind of retrotransposon, and approximately 1.4 million of them are found in the human genome. As they move about in the genome, they generate mutation by what mechanism?

Answer: insertion (into intron/exon) Section: 18.2 Skill: Application/Analysis 16) An inherited human disorder, familial dysautonomia, results from a nucleotide mutation in the gene IKAP that is expressed in the nervous system. The decreased IKAP protein leads to abnormal development, and the resulting disease is usually fatal by age 30. The nucleotide change alters splicing.

If this change affects only the nervous system and not the immune system, in which the gene is also expressed, what feature must be found in this gene?

Answer: alternative splicing Section: 18.2 Skill: Application/Analysis

17) A gene in a sheep-like species has been found to differ from that in related species by the inclusion of a variable number of tandem repeats of a short sequence. These repeats or duplications arise from what process?

Answer: unequal crossing over Section: 18.2 Skill: Application/Analysis 18) Mammalian genomes all have a number of genes for globin polypeptides. Several of these genes are functional, but several are never expressed. What are the nonfunctional ones known as?

Answer: pseudogenes Section: 18.2 Skill: Knowledge/Comprehension 19) Conserved noncoding sequences (CNGs) are usually sequences involved in what kinds of functions?

Answer: regulatory or regulation Section: 18.2 Skill: Knowledge/Comprehension 20) The karyotypes of several primate species, such as chimpanzee and gorilla, show two nearly telocentric chromosomes as opposed to the single large metacentric chromosome 2 in humans. Does this demonstrate chromosome breakage or fusion at the time of lineage divergence?

Answer: fusion Section: 18.2 Skill: Application/Analysis 18.3 Fill-in-the-Blank Questions 1) If an entire genome is fragmented and then a large number of pieces are sequenced, the approach is known as the ________ approach.

Answer: shotgun Section: 18.1 Skill: Knowledge/Comprehension 2) The whole-genome shotgun approach (WGS) was first used on which eukaryotic model organism?

Answer: Drosophila Section: 18.1 Skill: Knowledge/Comprehension 3) A genome sequence is considered to be complete when all its ________ sequences are included, as well as reasonable amounts of repetitive sequences.

Answer: euchromatic Section: 18.1 Skill: Knowledge/Comprehension 4) Sequencing of DNA from a community of organisms rather than from a single individual is known as ________.

Answer: metagenomics Section: 18.1 Skill: Knowledge/Comprehension

5) The process of finding the location of genes and functional (e.g., regulatory) sequences and their functions within a genome sequence is called ________.

Answer: annotation Section: 18.1 Skill: Knowledge/Comprehension 18.4 Essay Questions 1) Now that small molecular-level deletions and subtle chromosomal rearrangements can be discovered using advanced cytogenetic techniques and microarrays, disorders whose causes had not previously been known can have their basis identified. However, since most of these are not widely publicized, being individually rare, it has been difficult to identify patients with shared chromosomal abnormalities who share phenotypic features. A new database, known as DECIPHER, is now available to search for patients with like phenotypes and/or like chromosomal abnormalities. If you are the clinical geneticist who has a patient whose data is entered into this database, suggest three or more ways in which this could be useful either to you or to the patient.

Answer: Many patients suffering from rare disorders harbor submicroscopic deletions or duplications that affect the copy number of dosage-sensitive genes or disrupt normal gene expression. However, because these aberrations are extremely rare, the web-based database DECIPHER can be used to help identify submicroscopic chromosomal imbalance, inversions, and translocations. DECIPHER also maps common SNPs found in normal populations, which is extremely useful for narrowing down the list of possible aberrations responsible for a given syndrome. DECIPHER also aids genetic counseling by retrieving relevant information from a variety of bioinformatics resources and can help link phenotypes to a particular chromosome rearrangement. By compiling information worldwide, rare cases that share phenotype and structural rearrangements can be linked together and cohorts can be used to aid with gene mapping. This will ultimately help with identification of new syndromes and furthering understanding of gene function by describing previously unrecognized syndromes.

Section: 18.1 Skill: Synthesis/Evaluation 2) You are interested in sequencing the gene for Tangier disease (OMIM 205404). The disorder is rare, so it is feasible to obtain only very few samples. The approximate chromosomal location of the gene is known. How would you set about deciding on a model organism to use to study this disease, and what would you do once you had decided? (An outline of your procedure after the organism is chosen will be sufficient).

Answer: If the chromosome location of the gene is known, you use homologous genes in a model organism to help you identify specific sequence information. For instance, if the gene you are interested in studying has a known homolog in mouse and Drosophila, you could choose one or both organisms and identify a conserved region of the gene. Using that sequence, you can design primers to help you sequence the human version of the gene. After identifying the human sequence, you can also use model organisms to help you identify residues necessary for proper protein function. Using site-directed mutagenesis, you can generate mutant versions of the protein and look for changes in its function and phenotypes that resemble the human disorder in your model organism. This will ultimately give you an animal model for Tangier disease. Sometimes the mutation does not cause the same disorder or symptoms seen in humans, depending on your model organism, so it may also be necessary to screen your organism to determine if it is the best choice of animal to study the disease.

Section: 18.1 Skill: Synthesis/Evaluation

3) How and why is the two-hybrid system used to find where two proteins interact? Use a real example.

Answer: Two-hybrid screening is used to detect protein-protein or protein-DNA interactions by testing for physical binding. You will look for the activation of a reporter gene when a transcription factor binds to an upstream activating sequence. In a two-hybrid screen, the transcription factor is broken into two fragments: the binding domain (BD), which binds to the UAS, and the activation domain (AD), which activates transcription of the reporter gene. For instance, the yeast two-hybrid screen uses a protein that is translationally fused to a GalL4 BD and another protein to a Gal4 AD. If the proteins interact, the two Gal4 elements are brought together and transcription occurs.

Section: 18.3 Skill: Synthesis/Evaluation 4) Under what circumstances is a metagenomic analysis preferable to analysis of a gene or genes from a single species?

Answer: Metagenomic analyses are preferred when you are interested in studying the diversity in a given environment, and they enable researchers to look for sequences found in organisms that make up a small part of a biotic community. Analysis of a gene from a single species is helpful when you wish to study the role of that specific gene, but does not provide insight into genetic diversity or homologous genes in other species.

Section: 18.3 Skill: Synthesis/Evaluation 5) Being able to do a BLAST search, among other things, assumes that you and every other researcher has access to known gene sequences. Some would prefer, however, that such sequence information should be "owned" by their discoverers. What are the pros and cons of such a position?

Answer: The "pros" of sequence ownership include, but are not limited to, the ability to use the sequence to finance further research into treatments that specifically target that region of DNA (the development of "designer drugs"). It also could help fund diagnostic tools for rapid identification of novel mutations or mapping of common SNPs in a given population. "Cons" would include the fact that charging people to use your sequence information may severely limit studies on your gene and may delay development of treatments or research on this particular gene or mutation on the basis of what is profitable. This would particularly affect research on rare mutations, which may not be as profitable in terms of developing new treatments when research funding is limited and labs cannot afford to pay.

Section: 18.1 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 19 Cytoplasmic Inheritance and the Evolution of Organelle Genomes 19.1 Multiple-Choice Questions 1) Random segregation of organelles during cell division is likely to result in an increased rate of which of the following?

A) incomplete penetrance B) dominant loss-of-function mutations C) dominant gain-of-function mutations D) chromosomal duplication mutations E) chromosomal deletion mutations Answer: A

Section: 19.1 Skill: Synthesis/Evaluation 2) The human egg has about 2000 mtDNA, but somatic cells have a range of hundreds to thousands.

Which of the following is most likely to account for the difference?

A) Heteroplastic cells can give rise to homoplastic cells.

B) Large mitochondria in oocytes can divide into many smaller mitochondria, and smaller mitochondria can fuse.

C) Specific tissue types replicate organelle DNA far more frequently than they do nuclear DNA. D) Mutant mitochondria replicate at different frequencies than do wild type.

E) Haploid organelle genomes can replicate to form diploid or polyploidy genomes.

Answer: B Section: 19.1 Skill: Application/Analysis 3) The direction of shell coiling in snails such as Limnea peregra is determined by the egg's genotype, which affects orientation of the spindle in early cleavage divisions. How could the egg genotype cause the directional orientation of a spindle in the embryo?

A) by products of mitochondrial genes that pull the centrioles toward the mitochondrial position B) by attachment of spindle fibers to mitochondria at the edges of the cell C) by maternally directed placement of mRNAs and proteins at appropriate positions in the egg cytoplasm D) by the release of maternal mitochondrial genomes into the egg cytoplasm E) by zygotic gene activation due to maternal mitochondrial genes Answer: C Section: 19.3 Skill: Application/Analysis

4) Extrachromosomal inheritance includes organelle inheritance, infectious inheritance, and maternal effects that is, anything transmitted through the cytosol. In infectious heredity, the infecting pathogen is cytoplasmic. What other non-Mendelian characteristic(s) must also occur?

A) infection of specific organelles B) transfer of the pathogen's genes to the host nucleus C) transfer of the pathogen's genes to the organelles D) infection passing from maternal cytoplasm to egg cytoplasm E) infection passing from maternal nucleus to egg nucleus Answer: D Section: 19.1 Skill: Application/Analysis 5) Sigma virus infection in Drosophila melanogaster causes flies to become paralyzed and then die in high concentrations of CO2. The virus is passed on extrachromosomally. Two isogenic strains of Drosophila cultures are maintained: R (resistant to CO2 or wild type) and S (susceptible to CO2, infected with sigma). After crosses between male R and female S flies, subsets of offspring of each gender are tested for CO2 sensitivity. Which of the following results do you expect?

A) All the offspring are sensitive.

B) All the offspring are female.

C) All the offspring are male.

D) All the offspring are sensitive, but only males pass it on.

E) Only male offspring backcrossed to the female parent are sensitive.

Answer: A Section: 19.1 Skill: Application/Analysis 6) In certain Latin American countries where, during periods of political volatility, children were removed from their families and the parents murdered (then referred to as the "disappeared"), the best possibility of restoring the children to members of their larger family have used mtDNA matching. Why and how?

A) Remains of the children's mothers can be used to match their mtDNA with that of their children.

B) Children's mitochondria will have genomes that are 1/4 identical to each of their grandparents.

C) Children's mitochondria will have genomes that are 1/2 identical to their maternal grandparents.

D) Children's mitochondria will have genomes that are identical to 1/2 of those of their maternal grandmothers.

E) Children's mtDNA will be identical to those of their first cousins.

Answer: D Section: 19.2 Skill: Application/Analysis 7) The noncoding regions of mammalian mitochondrial genomes have few to no repair mechanisms, but no real selective pressure. Which of the following are consequences of this condition?

A) a rate of mutation that is much higher than that of nuclear sequences B) a high rate of loss of mitochondrial genomes relative to nuclear sequences C) a high rate of natural selection for genomes identical to those of the mother D) a lower rate of mutations that remain in maternal lineages over evolutionary time E) a rate of mutation that can increase with successive generations in a maternal line Answer: A Section: 19.2

Skill: Application/Analysis 8) In Leber's disease (LHON), degeneration of the optic nerve results in blindness due to a defect in NADH dehydrogenase affecting electron transport. If this is due to a mutation in a mitochondrial gene, which combination of effects would you expect to see when doing a pedigree analysis of a family with LHON? i. variable expressivity ii. incomplete penetrance iii. progressive blindness over time iv. maternal inheritance v. diminished muscle mass A) i and ii only B) ii and iii only C) iii and iv only D) i, ii, iii, and v only E) i, ii, iii, and iv only Answer: E Section: 19.2 Skill: Application/Analysis 9) Myoclonal epilepsy and ragged red fiber disease (MERFF) is a human condition named for the ragged red fibers of its skeletal muscle cells and myoclonic epilepsy. Individuals with this disorder have a mutation in a mitochondrial gene for a tRNA, specifically that for lysine. Affected individuals are heteroplastic. Why? A) Interference with translation slows protein formation and therefore maintenance of mutation-bearing organelles.

B) Having homoplastic organelles would no doubt be lethal due to curtailed translation.

C) Lysine is required in translation of every one of the mitochondrial genes.

D) Mutations of this gene affect individuals before birth.

E) Mutations of this gene are inherited as dominant characteristics.

Answer: B Section: 19.3 Skill: Application/Analysis 10) Some members of the species D. bifasciata produce mostly female offspring if reared at or below 21°C. The causative agent of this condition (called sex ratio) is a protozoan infection that is lethal to most male larvae. How can you demonstrate that it is both inherited and extrachromosomal?

A) Inject ooplasm of affected females into normal females and then mate these with normal males.

B) Inject cytoplasm from somatic cells of an affected female into normal females and then mate them with normal males.

C) Mate affected females with normal males and assess daughter somatic cells for presence of the protozoan.

D) Perform reciprocal crosses of affected and normal individuals and examine the offspring for infection.

E) Perform reciprocal crosses of affected and normal individuals and look for affected male offspring.

Answer: A Section: 19.3 Skill: Application/Analysis

11) In 1918, the tsar and tsarina of Russia and their five children were presumably assassinated and buried in an unmarked grave. Later, several women came forward and claimed that they were in fact Anastasia, one of the daughters whose bones had not been found. More recently, forensic scientists have extracted DNA, including mtDNA, from the skeletal remains. Which of the following samples of mtDNA had to be analyzed to discern which, if any, of the claimants was Anastasia?

A) the tsar, tsarina, and the other assassinated children B) each of the claimants or their living children C) the tsarina and the claimants D) the tsar and tsarina only E) the tsar and tsarina and living relatives of each of them Answer: C Section: 19.2 Skill: Application/Analysis 12) A consequence of organelle heredity is that sequence changes in mtDNA vary more than nuclear DNA in response to reduction in population size. Which of the following does this imply about human evolutionary history?

A) Current human populations must have a much higher mutation rate than ancient populations.

B) Mitochondrial variants in people descending from originally small populations show enough diversity to be used to construct an evolutionary tree.

C) The original human population must have grown to be quite large before its members began migrating to other lands and other continents.

D) Nuclear gene sequencing must be used to resolve the question of human ancestral origins.

E) A single human population, living about 200,000 years ago, could not have given rise to all current human populations.

Answer: B Section: 19.5 Skill: Synthesis/Evaluation 13) Two databases, OMIM and Mitomap, are especially useful in exploring human mitochondrial mutations. If you are interested in learning whether aminoglycoside-induced deafness is inherited maternally, and what gene(s) is involved, which of the following would give you the most information?

A) the Mitomap position map with gene names B) the sequence with specifically altered amino acid coding C) the OMIM description of symptoms and diagnosis D) the OMIM description of the TRMU gene at 22q13 E) the Mitomap loci of genes associated with deafness Answer: E Section: 19.1 Skill: Application/Analysis

14) In a cross between a petite yeast mutant and a wild-type strain, results showed some petite and some wild-type outcomes. Which of these statements must be true of this petite mutation?

A) It is suppressive.

B) It is a nuclear or segregational mutation.

C) It is a neutral mitochondrial mutation.

D) It replicates faster than wild type.

E) It has small deletions of mitochondrial DNA. Answer: B Section: 19.2 Skill: Application/Analysis 15) Chlamydomonas, although a green alga, is sensitive to an antibiotic, erythromycin. If mt- matingtype alga that is sensitive is mated with an mt+ cell that is resistant, and if the mt+ strain donates chloroplasts, what results do you expect and why?

A) All progeny are sensitive because this is the dominant trait.

B) Approximately half the progeny are sensitive because this is the recessive trait.

C) All the progeny are sensitive because this is the recessive trait.

D) All progeny are resistant because cpDNA is responsible.

E) Approximately half the progeny are sensitive because cpDNA is responsible.

Answer: D Section: 19.2 Skill: Application/Analysis 16) Suppose there is (hypothetical) evidence that a gene in mice produces a substance that induces twitchiness in hind leg muscles. A female mouse of a true-breeding twitchy strain is mated with a male of a true-breeding non-twitchy strain. All progeny are twitchy. Which of these procedures would you use to provide evidence for your hypothesis that twitchiness is due to mtDNA? A) sequence the mtDNA of the twitchy mother B) breed twitchy male progeny with normal females to show no paternal inheritance C) breed twitchy female with twitchy male F1 mice to verify there is maternal inheritance D) backcross twitchy female offspring with normal fathers E) sequence the mtDNA of the twitchy male offspring Answer: B Section: 19.3 Skill: Synthesis/Evaluation

17) The following is a hypothetical pedigree for a family, some of whose members have symptoms of Leigh syndrome, caused by an mtDNA mutation that reduces ATPase activity.

Which of the following in the youngest generation might transmit the condition in future?

A) any of the women B) any of those affected C) any of the members of this generation D) only those with an affected mother E) any of the affected women Answer: E Section: 19.3 Skill: Application/Analysis 18) The 16-kb mitochondrial genome encodes 13 proteins. How can there be more than 13 mitochondrially inherited conditions?

A) Some conditions are due to nuclear genes.

B) Many of the conditions are due to translation errors.

C) There are different kinds and numbers of mutations per gene.

D) Each parent may contribute different mutations.

E) Exon shuffling allows different polypeptides to form.

Answer: C Section: 19.3 Skill: Synthesis/Evaluation 19) Organisms have been created by nuclear transfer cloning, in which a diploid nucleus from a somatic cell of an adult animal is injected into an enucleated egg cell to create an embryo. Among the many species thus cloned, there has been a prevalence of mitochondrial defects. Which of the following is a most likely explanation?

A) The transferred nucleus may destroy the egg mitochondria.

B) The egg cell may be heteroplastic before injection.

C) The somatic cell's mitochondria may have been transferred as well.

D) The transferred nucleus may have mutations that alter mitochondrial function.

E) The egg nucleus may have had one or more mutations that alter mitochondrial function.

Answer: D Section: 19.3 Skill: Synthesis/Evaluation

20) You are a genetic counselor working with a phenotypically normal young couple who have been referred because the wife has a family history of a mitochondrial disorder that results in considerable dysfunction. What is the most important consideration you can suggest?

A) to refrain from having children that are hers biologically B) that she be tested to find out what percentage of her mitochondria bear the mutation C) that both husband and wife have mtDNA testing for the sequence in question D) that her affected mother's DNA be tested E) that they consider having only sons Answer: B Section: 19.3 Skill: Synthesis/Evaluation 19.2 Short-Answer Questions 1) In any mammal, the phenotype for a mitochondrial trait is based on what maternal characteristic?

Answer: genotype Section: 19.2 Skill: Knowledge/Concepts 2) Is the rate of mutation in mitochondrial DNA of mammals greater or less than that of the species' nuclear DNA? Answer: greater than Section: 19.2 Skill: Knowledge/Concepts 3) Allan Wilson showed that the mtDNA of Africans is more diverse than that of other human populations. Does this finding provide evidence for the multiregional (MRE) model or the recent African origin (RAO) model?

Answer: RAO Section: 19.2 Skill: Knowledge/Concepts 4) Mitochondrial genome transcription, translation, and replication occur where?

Answer: (in the mitochondrial) matrix Section: 19.3 Skill: Knowledge/Concepts 5) In one or two words, give two characteristics of mitochondrial genomes.

Answer: Any two of the following: circular no histones/chromatin anchored (to the mitochondrial membrane) variable in size Section: 19.3 Skill: Knowledge/Concepts

6) What name is given to the alteration of an RNA sequence after transcription?

Answer: RNA editing Section: 19.4 Skill: Knowledge/Concepts 7) Do the protein-coding parts of organelle genes more closely resemble those of Bacteria (Eubacteria) or Archaea?

Answer: Bacteria Section: 19.5 Skill: Knowledge/Concepts 8) What are the closest "relatives" of chloroplasts still found as free-living organisms today?

Answer: cyanobacteria Section: 19.5 Skill: Knowledge/Concepts 9) Sequencing of genomes of eukaryotes, both mitochondrial and nuclear, have revealed evidence of sequence transfer from one to the other. Are such transfers ancient, recent, or both?

Answer: both Section: 19.5 Skill: Knowledge/Concepts 10) If a chloroplast coding sequence has been transferred to the nuclear genome, but the protein product is needed in the chloroplast, what must be added to the original polypeptide sequence?

Answer: signal sequence Section: 19.5 Skill: Knowledge/Concepts 11) Among the discoveries that Correns and Bauer (1908) made in plants was that, sometimes, reciprocal crosses resulted in what phenotype only?

Answer: maternal Section: 19.1 Skill: Knowledge/Concepts 12) When a cell divides, chloroplasts are randomly distributed. What is this called?

Answer: replicative segregation Section: 19.1 Skill: Knowledge/Concepts 13) Three of the four modes of inheritance of organelle genes are • maternal inheritance as in mammals • paternal inheritance as in gymnosperms • selective degradation of one source during mating.

What is the fourth?

Answer: biparental inheritance Section: 19.2 Skill: Knowledge/Concepts

14) What number of membranes is typical of both mitochondria and chloroplasts, and which of them anchors the organelle's genome?

Answer: 2; inner Section: 19.3 Skill: Knowledge/Concepts 15) The protein-coding genes of chloroplasts are involved in what part of photosynthesis?

Answer: oxidative phosphorylation Section: 19.4 Skill: Knowledge/Concepts 16) DNA transfer has occurred many times between organelle and nuclear genomes. Two pairs of sequences are analyzed. Pair A has more similarities than pair B. Which of these is the most ancient transfer?

Answer: Pair B Section: 19.5 Skill: Application/Analysis 17) Variation in numbers of mutated chloroplast genomes can give rise to green, white, or variegated leaves. What color leaves do you expect in offspring of pollen from a white-leafed plant and ovules from a green-leafed plant?

Answer: green Section: 19.1 Skill: Application/Analysis 18) In mammals, the number of mitochondria per cell is highly variable. Which of the following would have the largest number: skin cells, red blood cells, or skeletal muscle cells?

Answer: skeletal muscle (cells) Section: 19.1 Skill: Application/Analysis 19) Mating in Chlamydomonas occurs between haploid mt+ and mt- individuals. The mt- chloroplast genome is selectively degraded 95% of the time. In one such mating, the mt+ genome is strR and mt- is strS. Will offspring live or die if they are exposed to streptomycin?

Answer: live Section: 19.2 Skill: Application/Analysis 20) In plants, cytoplasmic male sterility (CMS) is maternally inherited due to mutations in mitochondria.

However, these can revert to normal due to a nuclear gene for restoration of fertility (RF). The recessive rf/rf condition does not restore fertility. If the female parent of a male plant is CMS rf/rf and the male parent is normal (N) Rf/Rf, will male progeny be sterile or fertile?

Answer: fertile Section: 19.3 Skill: Application/Analysis

19.3 Fill-in-the-Blank Questions 1) Any form of inheritance that does not follow Mendelian patterns and that involves most of the cytoplasm being contributed to the embryo by one of the parents is called ________.

Answer: extrachromosomal (inheritance) Section: 19.1 Skill: Knowledge/Concepts 2) The organelles involved in organelle heredity are ________ and ________.

Answer: chloroplast, mitochondria (either order) Section: 19.1 Skill: Knowledge/Concepts 3) If a cell includes a mixture of variable numbers of wild-type and mutant organelles, this condition is known as ________.

Answer: heteroplasmy Section: 19.1 Skill: Knowledge/Concepts 4) Mitochondrial mutations in mammals are passed to the offspring by the ________ parent.

Answer: female or maternal Section: 19.2 Skill: Knowledge/Concepts 5) In the 1950s, mitochondrial genomes were discovered by differential staining for ________.

Answer: DNA Section: 19.1 Skill: Knowledge/Concepts 19.4 Essay Questions 1) How can you establish that, in a newly described species of plant, a particular phenotype is due to a chloroplast rather than a nuclear gene?

Answer: Mutations in chloroplast genes violate Mendel's Law of Segregation in that they show uniparental inheritance patterns. In addition, the irregular segregation of chloroplasts during cell division can result in variable segregation patterns, producing a variegated phenotype due to unequal chloroplast distribution in leaf cells.

Section: 19.4 Skill: Application/Analysis 2) What makes it possible for human mitochondria, which have only 22 sequences that code for tRNA, to translate mitochondrial polypeptides that require 32 tRNA? Answer: Thanks to the third-base wobble, many of the tRNAs can recognize codons using the first two bases of the codon. In addition, many nuclear genes can be imported into mitochondria and the exchange of genetic information between the nucleus and mitochondria is required for proper mitochondrial function. Therefore, some of the tRNAs are read using these nuclear-encoded tRNAs if the mitochondria does not transcribe those particular tRNAs. Section: 19.3 Skill: Synthesis/Evaluation

3) If the mother and grandmother of a given family both exhibit a half dozen phenotypic effects of a mitochondrial disorder, how is it possible that the mother's three children each have different phenotypes from one another and different degrees of severity?

Answer: Phenotypes are directly dependent on the number or percentage of mutated mitochondria per cell. Thus, cells that inherit more of the mitochondria with the mutation will show a more severe phenotype than those with enough normal mitochondria to maintain the wild-type function of the gene.

If, during embryonic development, the cells that gave rise to a particular organ or tissue are derived from a cell with more of the mutant mitochondria, that individual will inherit the phenotype associated with that particular tissue defect. Similarly, if the cell that gave rise to a particular tissue or organ had mostly normal mitochondria, that individual may be phenotypically normal with regards to that tissue or organ.

Section: 19.2 Skill: Synthesis/Evaluation 4) Among current bacteria, those with the most in common with mitochondria are the Rickettsia (that cause typhus). What does this mean about Rickettsia's place in the evolution of endosymbiosis?

Answer: Because Rickettsia more closely resembles mitochondria compared to other bacteria, this may indicate that Rickettsia is a less-derived, ancestral bacterial species. It could also indicate that a close relative to Rickettsia gave rise to the symbiotics that ultimately became mitochondria.

Section: 19.5 Skill: Synthesis/Evaluation 5) What is the argument that, although human mitochondria all descend from a single ancestral population, this may not be true for nuclear genomes?

Answer: By comparing the genomes of organelles with proteobacteria, there is a strong degree of conservation between the genes of mitochondria and those of the proteobacteria. Human-like mitochondria can be traced to a time before the evident evolution of humans. Thus, mitochondria from one human to the next have not diverged. Over time, the nuclear genomes of organisms retained their own informational genes, and many endosymbiont operational genes were transferred to the nuclear genome and often replaced their functional equivalents. Unlike the cases of the mitochondria and chloroplasts, where the endosymbionts can be traced to specific lineages of Bacteria, the putative ancestor for the human nuclear genome may have come from several lineages after the endosymbiosis event.

Section: 19.5 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 20 Developmental Genetics 20.1 Multiple-Choice Questions 1) In Drosophila, a developmental mutation called ssa (spineless aristapedia) results in flies with mini leg parts on the antennae. What kind of gene is this mutation?

A) gap gene B) pair-rule gene C) homeotic gene D) segment polarity gene E) maternal effect gene

Answer: C Section: 20.2 Skill: Application/Analysis 2) A pair-rule gene in Drosophila called runt controls some of sex and nervous system formation. In the mouse, runt controls blood cell formation, osteogenesis, and genital structures. In humans, the analog of runt, when mutated, causes children to lack collar bones, and the opening at the top of the skull fails to close. What kind of experiment could tell you whether these genes are conserved in function?

A) sequencing of the runt genes from each of the three species of organisms B) in vivo experiments inserting the copy of the moue gene into a segmented fly embryo C) in vitro experiment that inserts mouse runt gene into cells from an unsegmented Drosophila embryo D) sequencing the proteins encoded for by each of the species' runt genes E) use of immunological methods to identify the target cells of the species' runt transcription factors Answer: C Section: 20.2 Skill: Synthesis/Evaluation 3) An Antennapedia (Antp) mutant in Drosophila has legs in place of antennae. This is caused by A) activation of the Antp gene in the head region instead of in the thorax.

B) the product of an abdominal segment gene being expressed in the head.

C) a chromosomal error that shifts this Antp gene to the anterior end of the complex.

D) the interaction of an inhibitor gene of the bithorax complex with the Antp mutation.

E) loss of function of the Antp gene.

Answer: A Section: 20.2 Skill: Application/Analysis 4) Human Hox genes are in four clusters: HOXA, HOXB, HOXC, and HOXD, for a total of 39 genes.

Besides conserving the clustering of genes, evolution has also conserved A) their chromosomal locations.

B) the order of the genes within them.

C) their expression along the dorsal-ventral axis.

D) the mutations to which they give rise.

E) their bilateral symmetry.

Answer: B Section: 20.2 Skill: Application/Analysis 5) Segmentation genes in Drosophila act in which of the following orders?

A) segmental, helical, spherical B) pair-rule, transdeterminal, gap genes C) gap, pair-rule, segment polarity D) pair-rule, segment polarity, gap genes E) pair-rule, segment polarity, Hox genes Answer: C Section: 20.2 Skill: Application/Analysis 6) As a Drosophila larva molts and becomes a pupa, most of the larval tissues die and adult structures

are formed from cells in clusters known as imaginal discs that have remained unchanged throughout the larval stages. Therefore, the disc cells are A) maternal.

B) zygotic.

C) embryonal.

D) differentiated.

E) transcriptionally activated.

Answer: C Section: 20.2 Skill: Application/Analysis 7) In an organism such as a sea urchin, administration of actinomycin D prevents RNA synthesis.

However, early development up to gastrula formation can take place in the presence of actinomycin D but gastrulation cannot. This implies that A) development beyond gastrulation requires actD. B) actD is part of the normal environment of a sea urchin and serves as a developmental signal molecule.

C) actD receptors begin to form only on cell surfaces of the gastrula.

D) maternal effect genes regulate development until gastrulation.

E) immature sea urchin embryos produce actD to regulate further development.

Answer: D Section: 20.2 Skill: Application/Analysis 8) Suppose that a homeotic gene from Drosophila is introduced into an Arabidopsis embryo with a mutation in one of its homeotic flower genes. What effect would you expect and why?

A) rescue (or negating) of the phenotype of the mutant because the fly Hox gene supplies the missing protein B) correction of the sequence in the mutant by recombination with the Drosophila gene sequence C) expression of the Drosophila homeotic gene correcting the mRNA of the mutant plant D) a mixture of roughly equal numbers of Drosophila and plant gene product, producing an intermediate phenotype E) no effect, because Hox genes are different from MADS box genes Answer: E Section: 20.4 Skill: Application/Analysis

9) In Drosophila, germ cells develop from pole nuclei isolated from the rest of the embryo at the posterior end. How are they "protected" from the effects of posterior concentrations of zygote-produced mRNAs and proteins?

A) Their genes have unresponsive promoters.

B) The cytoplasm at the pole is separated by a compartmentalizing membrane.

C) The pair-rule, gap, and segmentation genes are not expressed in these cells.

D) Homeotic genes in these cells are inhibited by zygotic genes from the mother.

E) Paternal effects operate on the embryo only in the pole cells.

Answer: B Section: 20.2 Skill: Application/Analysis 10) In early vertebrate embryonic development, a number of pathways, such as Wnt, Hedgehog, and Notch, regulate differentiation and organ formation. They operate by A) formation of new transcription factors at each stage.

B) transcription activation due to interaction among transcription factors.

C) cell signaling mediated by receptor binding.

D) establishing anterior-posterior and dorsal-ventral symmetry.

E) stimulating MADS box genes.

Answer: C Section: 20.3 Skill: Application/Analysis 11) There is much controversy in the news both scientific and popular about the use of embryonic stem cells for research. These cells are grown as cell lines in culture, created by taking the inner cell mass from 5-day-old embryos. What is the center of the controversy?

A) the fact that the embryos are mostly those unused by in vitro clinics B) the knowledge that these cell lines have reverted spinal cord injuries in mice C) the unproven usefulness of stem cells to treat human disorders D) the realization that these cells are pluripotent rather than totipotent E) the difference of opinion about whether these embryos are persons Answer: E Section: 20.1 Skill: Synthesis/Evaluation 12) Polycomb group (PcG) mutants cause gain-of-function of Antennapedia (Antp) and other homeotic genes by causing some of these genes to be expressed in tissues where they are usually repressed. PcG has a domain that binds to methylated histone H3. This finding supports which of the following hypotheses?

A) that PcG binds to Antp genes and serves as a transcription activator B) that PcG mutants have phenotypes like those of Antp mutants C) that PcG regulates genes by altering chromatin remodeling D) that PcG initiates repression of Antp genes in tissues E) that PcG is a repressor molecule like those found in bacteria Answer: C Section: 20.2 Skill: Synthesis/Evaluation

13) Mutations of regulatory elements of developmental genes may have either dominant or recessive effects. If a mutation results in a silencer element, that negatively regulates, which do you expect?

A) a recessive effect in which cis elements are up-regulated B) a recessive effect in which cis elements are down-regulated C) a dominant effect in which the silencer down-regulates gene activity D) a dominant effect in which gene activity is up-regulated E) a dominant effect in which the silencer is transactivating Answer: C Section: 20.2 Skill: Synthesis/Evaluation 14) The abdominal segments of Drosophila have no appendages. Loss of function of bithorax genes results in appendages forming in these abdominal segments. Therefore, A) bithorax mutations are necessarily recessive.

B) bithorax genes normally activate appendage-forming genes.

C) ultrabithorax usually acts to oppose bithorax genes.

D) bithorax normally activates a gene that negatively regulates appendages in the abdomen.

E) expression of bithorax wild-type genes suppresses all appendage formation.

Answer: D Section: 20.2 Skill: Synthesis/Evaluation 15) Since mice embryos with loss-of-function Hox mutations result in homeotic alterations, then introduction of such a mutation might cause A) ribs on normally rib-less vertebrae.

B) pointed rather than rounded ears.

C) a change in fur coloration along the dorsal midline.

D) behavioral alterations causing limb shaking.

E) paralysis of hindlimbs but not forelimbs.

Answer: A Section: 20.2 Skill: Application/Analysis 16) Drosophila mutations such as bcd, if homozygous in the mother, produce offspring with lethal defects, no matter what the genotype of the offspring. For some other genes, the mutant phenotype can be "rescued" by introducing a wild-type allele. Why the difference?

A) Non-rescuable mutations are of paternal origin.

B) Rescuable mutations act as dominants.

C) bcd is a maternal effect mutation.

D) Rescue requires recombination between the maternal and introduced alleles.

E) bcd is a Hox gene.

Answer: C Section: 20.2 Skill: Application/Analysis

17) Aniridia is a human condition in which the eye has no iris. The protein encoded by the gene responsible for aniridia is very nearly identical to the protein product of the fly Eyeless protein. What experiment could provide evidence that the two genes are functionally equivalent?

A) Introduce the aniridia mutation into Drosophila embryos to look for iris formation.

B) Sequence the Eyeless and aniridia DNA sequences and regulators.

C) Use the Eyeless mRNA as a probe in other invertebrate and non-mammalian species.

D) Introduce the mouse wild-type sequence into the fly egg to see whether mouse eyes develop.

E) Introduce the mouse aniridia wild-type sequence into the fly to see whether fly eyes develop.

Answer: E Section: 20.2 Skill: Application/Analysis 18) Use of double mutants has been invaluable in studying developmental pathways. In C. elegans, sex is determined by the ratio of sets of autosomes to X chromosomes. A diploid male has one X, and diploid hermaphrodites have two X's. The loss-of-function gene her-1 produces XO worms that are hermaphrodites, but XX worms are unaffected. Mutation of a second gene, tra, causes the XX to develop as male. Given this information, what would be the phenotype of a tra, her-1 double mutant?

A) a male nematode B) a female nematode C) a hermaphrodite D) an intersex variant E) lethality Answer: A Section: 20.3 Skill: Application/Analysis 19) Holoprosencephaly is a highly variable condition in mammals, including humans. Its effects range from an absence of brain areas, mid-face disorders, and cyclopia to slight dental abnormalities. One form, HPE3, is caused by Shh mutations. Familial cases are inherited as dominants. Which of the following do you expect?

A) HPE3 is actually a multigene complex.

B) Each type of malformation is due to a different HPE3 allele.

C) HPE3 expression is modified by genetic background.

D) HPE3 individuals also share a common environmental exposure.

E) All mutations in Shh that result in HPE3 are due to frameshifts.

Answer: C Section: 20.5 Skill: Application/Analysis

20) A homeotic mutant flower has an arrangement (from outside in) of carpels, stamens, stamens, carpels instead of the usual sepal, petals, stamens, carpels. In general, Class A genes specify sepals, Class A and B genes specify petals, Class B and C genes specify stamens, and Class C specifies carpels.

This mutant must therefore have a mutation in A) a Class A gene.

B) a Class B gene.

C) a Class C gene.

D) both A and B. E) both B and C. Answer: A Section: 20.4 Skill: Application/Analysis 20.2 Short-Answer Questions 1) An embryonic stem cell has the potential to differentiate into any cell or tissue type. Is it differentiated, pluripotent, or totipotent?

Answer: totipotent Section: 20.1 Skill: Knowledge/Comprehension 2) Which of the groups of early developmental genes in Drosophila act across segment boundaries to produce parasegments?

Answer: pair rule (genes) Section: 20.2 Skill: Knowledge/Comprehension 3) Cell autonomous genes affect only those cells in which they are expressed. Which group of early developmental genes are not cell autonomous?

Answer: segment polarity (genes) Section: 20.2 Skill: Knowledge/Comprehension 4) Homeotic genes in virtually all animals are said to show colinearity between the spatial arrangement of the tissues they act on as well as what other feature?

Answer: gene location Section: 20.2 Skill: Knowledge/Comprehension 5) The establishment of the axes of the body plan occurs during which stage of Drosophila development?

Answer: embryo or embryonic Section: 20.2 Skill: Knowledge/Comprehension

6) If a cell from the blastoderm of the Drosophila embryo is committed to form part of an abdominal segment of the adult, such as the sixth one, and this cell is transplanted into the thorax-committed region of another embryo, will it develop properties of a thoracic or an abdominal segment?

Answer: abdominal Section: 20.2 Skill: Application/Analysis 7) Certain mutations in non-Hox genes actually produce homeotic mutants due to a failure to modulate chromatin acetylation appropriately. This influence on cellular "memory" is called what?

Answer: epigenetic Section: 20.2 Skill: Knowledge/Comprehension 8) In the development of C. elegans, an embryo of 1090 cells then loses 131 cells. This is an example of what process?

Answer: apoptosis Section: 20.3 Skill: Knowledge/Comprehension 9) Cell signaling during much of C. elegans development occurs along concentration gradients. What kind of signaling does this represent?

Answer: inductive or induction Section: 20.3 Skill: Knowledge/Comprehension 10) During the evolution of development in animals, it is found that many regulatory genes have been co-opted for the development of different structures, such as limbs or digits. This is an example of what current area of research?

Answer: evo-devo Section: 20.4 Skill: Knowledge/Comprehension 11) Which structure in plants acts most like embryonic stem cells in animals?

Answer: meristem Section: 20.5 Skill: Knowledge/Comprehension 12) Certain species of lilies, if eaten by pregnant ewes, can cause the ewes to produce lambs with severe birth defects because the plants' cyclopamine can block embryonic response to Shh. This type of birth defect is a phenocopy of mutations in which genes?

Answer: Shh Section: 20.5 Skill: Application/Analysis 13) A highly conserved protein domain of 60 amino acids is found in a large number of animals. This domain allows the proteins in question to bind to what?

Answer: regulatory DNA (sequences) Section: 20.2 Skill: Knowledge/Comprehension

14) The maternal effect bicoid protein is a transcription factor that binds to the promoter of a zygotic gene to activate its expression. What is this second gene?

Answer: hunchback Section: 20.2 Skill: Knowledge/Comprehension 15) Which group of genes in Drosophila embryos must be mutated if the result is elimination of a significantly sized, contiguous region of segmentation?

Answer: gap genes Section: 20.2 Skill: Application/Analysis 16) Mouse eye genes can be inserted into Drosophila and can regulate the flies' eye formation.

Differences in size and shape of a zebra are controlled by a gene also found in zebrafish. What do these examples illustrate?

Answer: conserved genes or conservation (of genes) Section: 20.5 Skill: Application/Analysis 17) The Antennapedia complex and the bithorax complex of genes all encode transcription factors have a 180-bp DNA-binding domain. What is this domain called?

Answer: homeobox Section: 20.2 Skill: Knowledge/Comprehension 18) In Arabidopsis thaliana, a model organism in plant genetics, there are three classes of genes that control the development of floral structures: A, B, and C. Mutations of these genes cause organs to form in inappropriate places. What category of genes are they?

Answer: homeotic Section: 20.4 Skill: Knowledge/Comprehension 19) In Drosophila and other animals, Hox genes encode transcription factors that control development; but in Arabidopsis, a different family of transcription factors is produced with a different set of conserved amino acids. These are encoded by what group of genes?

Answer: MADS (box genes) Section: 20.4 Skill: Knowledge/Comprehension 20) What characteristic of a cellular blastoderm in Drosophila embryogenesis distinguishes it from the syncytial blastoderm?

Answer: cell membranes or cell compartmentalization Section: 20.2 Skill: Knowledge/Comprehension

20.3 Fill-in-the-Blank Questions 1) A mutation in a fly causes an antenna to develop where a leg should be. This is an example of a ________ mutation.

Answer: homeotic Section: 20.1 Skill: Application/Analysis 2) Molecules whose concentration and position in an embryo determine some aspect of development are called ________.

Answer: morphogens Section: 20.1 Skill: Knowledge/Comprehension 3) The process by which the presence and activity of neighboring cells determine an aspect of development is called ________.

Answer: induction Section: 20.1 Skill: Knowledge/Comprehension 4) Most of the genes controlling development encode either signal molecules or ________.

Answer: transcription factors Section: 20.1 Skill: Knowledge/Comprehension 5) Coordinate genes in an organism such as Drosophila determine the ________ of the early embryo.

Answer: polarity Section: 20.2 Skill: Knowledge/Comprehension

20.4 Essay Questions 1) Use a description of C. elegans development to distinguish clearly between differentiation and determination.

Answer: The developmental potential of a cell describes the range of different cell types it has the potential to become. As development progresses, the potency of individual cells decreases until, ultimately, their genetic fate is determined. Thus, determination implies a stable change at the level of gene expression that means a cell is committed to a certain developmental fate. Differentiation is the process that follows determination in which the cell develops its cell-specific expression profile.

Differentiation results in distinct cell types with clearly identifiable functions, such as a muscle or nerve cell.

The C. elegans example from the text involved vulva formation. The vulva forms during the last larval stage, from six precursor cells called vulval precursor cells (VPCs). Three of these larval cells give rise to structures of the vulva itself, the 1° cell, and two 2° cells of the vulva. The other three differentiate into hypodermis (3° cells). Initially, each of the six VPCs has the potential to differentiate along any of the pathways–1°, 2°, or 3°. If the anchor cell is destroyed, no vulva will form, because all six VPCs differentiate with a 3° fate and become hypodermis. Thus, these six cells are determinate, while the anchor cell must be present to induce VPCs to differentiate with 1° or 2° fates and thus form the vulva.

Because all the VPCs are capable of the same kinds of differentiation, they are said to form an equivalence group.

Section: 20.1 Skill: Synthesis/Evaluation 2) What are Hox and MADS genes, and what do they have in common?

Answer: Hox genes are animal genes sharing a homeobox and sharing a common ancestor that regulate developmental processes. MADS are plant genes whose binding domain is a MADS box. Hox and MADS genes have similar functions in that they both encode transcription factors, but they are not homologous genes. In addition, unlike Hox genes, MADS-box genes are not clustered in the genome.

Section: 20.2, 20.4 Skill: Synthesis/Evaluation 3) How and when does the Antennapedia complex change antennae into legs?

Answer: The Antennapedia complex is normally expressed in parasegments 4 and 5, where it is necessary for thoracic structures that produce legs. If it is expressed inappropriately in head segments (usually in parasegment #3), it promotes development of legs instead of antennae.

Section: 20.2 Skill: Synthesis/Evaluation 4) In some forms of cancer, the expression of certain homeotic genes has been altered. Is this more likely to be a cause or an effect of the cancer? Explain.

Answer: Either answer can be acceptable. If expression of a homeotic gene causes cancer, it could be causing the cell to increase cell division, which may cause the cell to ignore signals to stop growing. If expression is an effect of the cancer, it may be that an inappropriate cancer-promoted transcription factor could cause the expression of a homeotic gene out of its normal sequence. In either instance, the result is that the cell is expressing genes normally seen during fetal development and cells have a higher propensity to become malignant when these genes are reactivated in adult tissues.

Section: 20.4 Skill: Synthesis/Evaluation

5) What genetic evidence can demonstrate that some developmental events are due to chemical signaling?

Answer: In many instances, a developmental event requires that nearby tissue is intact and actively releasing a chemical signal. The inductive signals may be lost when an "anchor cell" is ablated, and further development is inhibited when the chemical signal is lost. In addition, mutations that cause that inductive chemical signal to be disseminated to additional cells can result in multiple structures forming, indicating that it is the presence of the chemical and its activation of gene pathways that encourages developmental events, resulting in the formation of specific cell structures.

Section: 20.3 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 21 Genetic Analysis of Quantitative Traits 21.1 Multiple-Choice Questions 1) Which of the following is most typical of a case of quantitative inheritance?

A) a ratio of 9:3:3:1 B) continuous variation in phenotypes C) discontinuous variation in genotypes D) lack of a clearly dominant phenotype E) a trait with little to no genetic influence Answer: B Section: 21.1 Skill: Knowledge/Concepts 2) If you know the proportion of F2 individuals that are of parental phenotypes, what formula will tell you how many gene pairs are influencing the trait?

A) 1/4n B) N1/4 C) VE + Vg D) H2/h2 E) 2n + 1 Answer: A Section: 21.1 Skill: Knowledge/Concepts 3) As an H2value approaches 0.0, which of the following must be true?

A) Almost all of the phenotype is VG. B) Almost all of the phenotype is VE. C) Identical (MZ) twins are virtually all discordant.

D) Fraternal (DZ) twins are virtually all concordant.

E) h2 = VA/VP. Answer: B Section: 21.3 Skill: Knowledge/Concepts

4) QTL mapping resembles linkage analysis primarily because A) it uses statistics to analyze its data.

B) it uses the location of a known DNA marker.

C) it uses the three-point mapping method.

D) the distances are reported in cM. E) the results are quantifiable.

Answer: B Section: 21.4 Skill: Knowledge/Concepts

5) Which of the following best predicts the response of a trait to artificial selection?

A) VA B) VG C) H2 D) h2 E) standard deviation Answer: D Section: 21.3 Skill: Knowledge/Concepts 6) A recent study investigated the heritability of lean body mass (LBM) that had been previously reported as between 52 and 84% but without known specific genes. The authors used a genome-wide association (GWA) scan to identify two SNPs: rs16892496 and rs7832552 within the thyrotropinreleasing hormone receptor gene that were significantly associated with LBM. Since the gene in question involves muscle metabolism, which of the following would you predict would be associated with LBM in a given population?

A) high LBM associated with higher risk of fracture in the elderly B) high LBM associated with memory loss in the middle aged C) low LBM associated with mobility limitation in the elderly D) low LBM associated with the growth spurt of puberty E) variation of LBM for members of different geographic populations Answer: C Section: 21.4 Skill: Synthesis/Evaluation 7) If lean body mass (LBM) and SNPs in the gene for a G-coupled receptor for a hormone secreted by the hypothalamus are associated, and the binding of this hormone is the first step in the release of thyroxin (necessary for skeletal muscle development), then increases in mutations in the hormone receptor gene would be expected to A) increase LBM. B) increase secretion by the hypothalamus.

C) decrease all thyroxin-mediated function.

D) increase body mass index (BMI).

E) alter the QTLs involved.

Answer: C Section: 21.4 Skill: Application/Analysis

8) Parkinson disease (PD) is a neurodegenerative disease that has been associated with a significant number of different genes, therefore showing it to be genetically heterogenic. One recent study in 278 families found evidence of two such possible loci. Evidence for linkage to chromosome 18q11 was presented by Gao et al. in 2009. Which of the following would be considered reputable evidence for this?

A) lack of quantitative difference for the occurrence of both in individuals B) use of persons from multiple ethnic and geographic backgrounds in the study C) finding the locus to be in Hardy-Weinberg equilibrium in the population D) finding the Lod score for association between the chromosomal location and PD to be 4.1 E) analysis of all other possible SNPs with no such association to chromosome 18 Answer: D Section: 21.2 Skill: Application/Analysis 9) Once a chromosomal position for a putative gene associated with Parkinson disease is verified, which of the following is a next step?

A) developing a FISH screening test to identify pre-Parkinson patients B) determining whether this locus is mutated in all Parkinson patients C) using a genome-wide linkage SNP panel to identify more minor genes D) ascertaining all living and at-risk individuals who are family members of the affected test subjects E) fine mapping of the linkage region to identify particular variants for influencing risk Answer: E Section: 21.3 Skill: Synthesis/Evaluation 10) Alterations in the ratio of CD4:CD8 lymphocytes are associated with risk of HIV infection progressing to AIDS, of autoimmune diseases, and of progression of some cancers. In searching for predictors of these effects in normal individuals, researchers have found two independent QTLs in the region of the genome known as the major histocompatibility complex (MHC), one in a region regulating CD8 and the other in a region regulating CD4. What does this indicate for the general population?

A) Mutation in one or both of these genes causes HIV and AIDS. B) Autoimmune diseases such as type 1 diabetes are caused by one of these genes.

C) Variation in one or the other of these genes may increase the risk of autoimmune diseases.

D) Only one mutation can be the cause of rheumatoid arthritis.

E) The probability of an autoimmune disease demonstrates that additive genes are involved.

Answer: C Section: 21.2 Skill: Application/Analysis 11) It is clearly established that obesity has greatly increased in children over the last several decades in the United States. While several genes have been identified that increase the potential for obesity, environmental influences are also involved. Which of the following is most likely?

A) There is little or no gene-environment interaction.

B) Phenotypic variation is discontinuous.

C) There is a small overlap of phenotypic ranges of different genotypes.

D) There is a wide range of phenotypes with each genotype.

E) There is a demonstrable threshold of environmental effects.

Answer: D Section: 21.1

Skill: Synthesis/Evaluation 12) A hypothetical condition in some domesticated fowl is identified phenotypically by patchy loss of feathers on the animals' backs. Although several factors, genetic and environmental, are found to be involved, only about 3% of any population with any of these factors are found with feather loss. Which of the following is a likely cause?

A) continuous variation B) threshold of liability C) low mutation frequency D) high number of alleles in one of the genes E) high number of environmental effects Answer: B Section: 21.1 Skill: Application/Analysis 13) An animal breeder asks your advice about which of a number of traits could most effectively be selected for in his herd. Which would you most correctly suggest?

A) a trait with high VE B) a high VA relative to VP C) a high VP relative to VG D) a low h2 value E) a low selection coefficient Answer: B Section: 21.3 Skill: Application/Analysis 14) A group that is experimenting with selection in an insect species due to exposure to a newly developed insecticide, cpd AW143, has found that, after 10 generations of exposure, the population is now bimodal in its response to the drug. Close to 30% of the population are completely normal while 70% live only about 1/6 of the normal lifetime. This is an example of A) threshold effect.

B) stabilizing selection.

C) additive genes.

D) directional selection.

E) disruptive selection.

Answer: E Section: 21.3 Skill: Application/Analysis 15) GWA looks at DNA marker distribution relative to that of supposed QTLs. Therefore, which of the following would be true of linkage disequilibrium?

A) There is a higher frequency association between marker and a QTL. B) The GWA score is lower than expected.

C) The Lod score is negative.

D) It is found in a highly inbred strain.

E) It is increased in homozygous individuals.

Answer: A Section: 21.4 Skill: Application/Analysis

16) Mean platelet volume (MPV) in humans increases with heart attacks and strokes and is used to predict recurrence as well as mortality. MPV has high heritability. A recent GWA study found MPV to be associated with three SNPs on chromosomes 12, 3, and 17 respectively. These three QTLs accounted for 4-5% of MPV variance. Which of the following is a reasonable explanation?

A) Other loci may also account for significant variance of MPV. B) These SNPs must be translocated or transposed alleles.

C) Monitoring of these SNPs should identify those at highest risk.

D) The three SNPs, when sequenced, should show high conservation.

E) The three SNPs will be found in genes responsible for stroke.

Answer: A Section: 21.4 Skill: Application/Analysis 17) Until recently, only apolipoprotein E (APOE) polymorphisms have been found to correlate with late-onset (nonfamilial) Alzheimer disease (AD); however, these account for less than 50% of the VG. New GWA studies have found a chromosome 12 risk locus. What method will likely be used by AD researchers going forward?

A) assessment of chromosome 12 for areas of linkage disequilibrium for AD B) association of APOE alleles with chromosome 12 SNPs C) sequencing of candidate genes on chromosome 12 and search for sequence similarity D) genotyping of a large cohort of AD patients and matched controls using a microchip E) analysis of genotypes of a large number of AD patients who have none of the expected APOE alleles Answer: D Section: 21.4 Skill: Application/Analysis 18) Although no specific gene alteration has yet been demonstrated as a cause for schizophrenia (SZ), it is known to have a strong genetic component. A recent study has found a strong association between SZ and a locus at chromosome 10q22-23 in Ashkenazi Jewish and Taiwanese Han populations. Which of the following is probable?

A) Other populations will have the same association with 10q22-23.

B) The neuregulin-3 gene at this locus is mutated in all or most SZ cases.

C) SZ exhibits significant genetic heterogeneity in different populations.

D) The region on 10q must include several genes associated with SZ. E) People with SZ being characterized differently in various medical communities results in this association between the Jewish and Han populations.

Answer: C Section: 21.4 Skill: Synthesis/Evaluation

19) It is estimated that 4.6% of SNPs and about 17% of all genes can be defined as pleiotropic. Which of the following might be a significant lesson for a consumer of genetic information based on this finding?

A) A test identifying a gene for a disease might actually be a test for a different disorder instead.

B) Any variant that we gain information about might in the future be found to associate with a more stigmatizing condition.

C) A physician might be treating the major condition but not notice even more deleterious outcomes.

D) The classification of my genetic disorder might be incorrect.

E) Sequencing of the candidate gene might find the wrong expected functional significance.

Answer: B Section: 21.4 Skill: Synthesis/Evaluation 20) An increased number of patients have recently been diagnosed with celiac disease, which you may know of as gluten sensitivity (OMIM 212750). It is a multifactorial disorder of the small intestine influenced by genes and environment, and it is characterized by malabsorption due to inflammation of the mucosa after eating wheat gluten or related proteins. It is now thought to occur in 1 out of 120 to 300 people. Genes found to be associated with it are in the HLA region of chromosome 6, namely DR3 and DQw2. Which of the following is more probable?

A) It is an autoimmune disorder.

B) It is caused by exposure to gluten in infancy.

C) It has high broad-sense heritability.

D) Its frequency varies greatly by population.

E) It is found primarily in inbred populations.

Answer: A Section: 21.3 Skill: Application/Analysis 21.2 Short-Answer Questions 1) While most additive polygenic traits show continuous distribution, some can be divided into two or more categories based on their contribution to genetic liability. Such traits are said to have what separation between affected and unaffected members of a population?

Answer: threshold Section: 21.1 Skill: Knowledge/Concepts 2) Sometimes the DNA marker allele and a QTL occur more frequently than would otherwise be expected, either because they interact or because of natural selection. What term describes this?

Answer: linkage disequilibrium Section: 21.3 Skill: Knowledge/Concepts 3) QTLs are most often mapped relative to DNA markers by using what scores as measurement?

Answer: Lod (scores) Section: 21.4 Skill: Knowledge/Concepts

4) If we suppose that a specific QTL is genetically linked to a DNA marker at a recombination frequency of 10%, then the parental type combinations in gametes would be expected at what frequency?

Answer: 20% each or 0.2 each or 40% for both types Section: 21.4 Skill: Application/Analysis 5) If the odds ratio for a particular QTL/DNA marker pair (e.g., Q10M10) is 2.2, this indicates that the organism is how many times more likely to get Q10M10 gametes?

Answer: 2.2 Section: 21.4 Skill: Application/Analysis 6) What is the statistically acceptable Lod score that provides evidence of genetic linkage?

Answer: +3 or 3 Section: 21.4 Skill: Knowledge/Concepts 7) After the location of a genome region likely to include a QTL, what must be found in this region?

Answer: candidate gene Section: 21.4 Skill: Knowledge/Concepts 8) What two measures describe the distribution of a trait in a sample?

Answer: variance and standard deviation Section: 21.4 Skill: Knowledge/Concepts 9) Although a polygenic trait is influenced by the effect of several genes, one or more of these may be major genes; but many others with lesser effects are known as what type of gene?

Answer: modifier genes Section: 21.1 Skill: Knowledge/Concepts 10) A homozygous plant has sepals of 30-cm flowers while another homozygote of the same species has 40-cm flowers. The F1 flowers are all 35 cm. In the F2, out of 600 plants the smallest flowers were 30 cm and the largest were 45cm; all the others were of intermediate sizes. If all the genes involved in flower size are additive, how many gene pairs are there?

Answer: 4 Section: 21.1 Skill: Application/Analysis 11) A strain of a cereal grain can be either dark tan, medium-dark tan, medium tan, light tan, or cream colored. When a dark tan and a cream plant are crossed, all F1 are medium tan. The F2 are distributed in a ratio of 1:4:6:4:1 from darkest to lightest. How many genes are involved in this coloration?

Answer: 2 Section: 21.1 Skill: Application/Analysis

12) A strain of a cereal grain can be either dark tan, medium-dark tan, medium tan, light tan, or cream colored. When a dark tan and a cream plant are crossed, all F1 are medium tan. The F2 are distributed in a ratio of 1:4:6:4:1 from darkest to lightest. How many additive alleles are needed to produce the darkest phenotypes?

Answer: 4 Section: 21.1 Skill: Application/Analysis 13) For the following traits, the concordance (occurrence of the same phenotype) of MZ and DZ twins are given as percentages. Which trait shows the most genetic as opposed to environmental influence?

Answer: B Section: 21.1 Skill: Application/Analysis 14) For a specific desirable trait in a breed of sheep, VP is 28.6, VC is 11, and VA is 7.2. What is its broad sense heritability?

Answer: 0.38 Section: 21.3 Skill: Application/Analysis 15) Calculate the narrow sense heritability (h2) for the following data on a trait in turkeys: VA = 11.0 VE = 94.4 VP = 114.2 VD = 13.8 Answer: 0.096 Section: 21.3 Skill: Application/Analysis 16) Which component of variance in a trait such as stamina in horses is the best indicator of whether the trait will respond to selection?

Answer: VP Section: 21.2 Skill: Application/Analysis 17) Of mean, median, and modal, what value is the most common value in a given population?

Answer: modal Section: 21.2 Skill: Knowledge/Concepts

18) Of the variance and the standard deviation for a set of data about a trait, which represents the square of the difference between each of the values and the mean of the sample, divided by the degrees of freedom?

Answer: variance Section: 21.2 Skill: Knowledge/Concepts 19) Which will exhibit the most shared maternal effects (uterine environment, etc.): identical or fraternal twins?

Answer: identical Section: 21.2 Skill: Knowledge/Concepts 20) If H2 for club foot is estimated as 80%, would you expect a small difference such as 30 versus 27 between MZ and DZ twins, or a large difference, such as 30 versus 2?

Answer: large difference or 30 versus 2 Section: 21.2 Skill: Application/Analysis 21.3 Fill-in-the-Blank Questions 1) If a trait results from the action of several genes, it is called a ________ trait.

Answer: polygenic Section: 21.1 Skill: Knowledge/Concepts 2) If a trait results from genetic variation as well as environmental factors, it is known as a ________ trait.

Answer: multifactorial Section: 21.1 Skill: Knowledge/Concepts 3) In some traits, different phenotypes result from genes whose alleles each contribute a specific increment to the whole. The phenotypes then have a ________ phenotypic distribution.

Answer: continuous Section: 21.1 Skill: Knowledge/Concepts 4) What is meant by a QTL? Answer: quantitative trait locus Section: 21.3 Skill: Knowledge/Concepts 5) A GWA study relies on the correlation between a QTL and a ________.

Answer: SNP or DNA marker Section: 21.3 Skill: Knowledge/Concepts

21.4 Essay Questions 1) Recent work in Drosophila demonstrated that the genes known as the chromatin remodeling complex co-localize at loci for hsp genes that respond to heat shock. Outline a procedure you could use to substantiate this finding.

Answer: A genome-wide association study (GWAS) can be used to tie the presence of a sequence variant of a DNA marker to a QTL influencing a specific phenotype. Thus, the relationship between an inherited genetic marker variant and the phenotype is by "association," which means organisms that carry a particular variant are more likely to have a certain phenotype than are organisms that carry a different variant. The assessment of association is quantitative; that is, it expresses the percentage of organisms with a genetic marker that also display a certain phenotype versus the percentage that have the phenotype but not the genetic marker. The advantage of GWAS over other QTL mapping approaches is that GWAS can scan the entire genome for QTLs by statistically testing for marker. Thus, GWAS can be used to correlate known genes or SNPs with hsp loci.

Section: 21.4 Skill: Synthesis/Evaluation 2) Describe how quantitative traits can be explained in Mendelian terms.

Answer: The multiple-gene hypothesis is the proposal that alleles at each of the contributing genes obeyed the principles of segregation and independent assortment and had an additive effect in the production of phenotypic variation. Often, these effects can be attributed to having multiple genes (polygenic inheritance), or some traits can have additive effects such that several mutations can have differing effects than single mutations.

Section: 21.1 Skill: Synthesis/Evaluation 3) Distinguish between broad sense heritability and narrow sense heritability, and explain how either or both of these measures can be used to inform those using artificial selection.

Answer: Narrow sense heritability refers to the proportion of total phenotypic variation that can be attributed to the additive effects of genes, whereas broad sense heritability is the proportion of total phenotypic variation caused by all genetic effects. However, narrow sense heritability is the only component of variation that natural selection can act on and, thus, provides more accurate selection information.

Section: 21.3 Skill: Synthesis/Evaluation 4) Given a list of the percentage of monozygotic and dizygotic twins in concordance for a selection of traits, describe how you would evaluate the relative input of environmental versus genetic factors.

Answer: If the differences in concordance are higher for MZ twins, this would suggest a higher genetic component because they share the genome. If there is no difference in concordance between MZ and DZ twins, this could likewise indicate that environmental factors (either in utero or after birth) may contribute to these particular traits.

Section: 21.2 Skill: Synthesis/Evaluation

5) Many human behavioral traits have both genetic and environmental components. Bipolar disorder has long been known to be familial, and more recently a number of genes and/or SNPs have been found to be involved. What are three or more reasons that this research has been hampered?

Answer: Answers will vary, but several examples are included here. Variations in the diagnosis criteria for bipolar disorder may make differing degrees of the disorder difficult to distinguish. In addition, there may be certain variants, or genes with major versus minor input, so the combination of mutations in an individual may affect the severity of their phenotype. This also may affect the age of onset variability, which can make diagnosis difficult and could be affected by environmental factors such as hormones, stress, etc. There may also be variable response to medications, which could be a function of the particular mutation or the other genes that the individual possesses that may allow them to respond better to a particular treatment.

Section: 21.1 Skill: Synthesis/Evaluation Genetics: An Integrated Approach (Sanders) Chapter 22 Population Genetics and Evolution 22.1 Multiple-Choice Questions 1) You want to calculate the frequency of the allele causing sickle cell disorder in your patients, and you know the genotypes of every individual. Assuming the population is not in Hardy-Weinberg equilibrium, which method can be used to calculate allele frequency?

A) allele-counting method B) genotype proportion method C) square root method D) binomial expansion E) chi-square test Answer: B Section: 22.1 Skill: Application/Analysis 2) In the same population of patients you examined for sickle cell allele frequencies, you now want to test whether the disease symptoms are related to blood type. Since blood type is a codominant trait and the alleles are easily identifiable by the phenotype (blood type) of the individuals, which method is most helpful for identifying blood type allele frequency?

A) allele-counting method B) genotype proportion method C) square root method D) binomial expansion E) chi-square test Answer: A Section: 22.1 Skill: Application/Analysis 3) Phenylketonuria (PKU) is extremely common in Ireland, affecting approximately 1 in 4500 live births. What is the frequency of PKU carriers?

A) 0.022% B) 98.5%

C) 2.9% D) 1.49% E) 2.2% Answer: C Section: 22.1 Skill: Application/Analysis

4) In a group of 500 people, the frequency of genotype NN is 40%. How many individuals would you expect to have the MN genotype?

A) 68 B) 232 C) 300 D) 200 E) 258 Answer: B Section: 22.1 Skill: Knowledge/Comprehension 5) If an X-linked recessive trait has a frequency of 5% in males, what percentage of females are carriers?

A) 9.5% B) 5% C) 95% D) 4.2% E) 10% Answer: A Section: 22.2 Skill: Application/Analysis 6) Given the following genetic profile, how many individuals would remain after natural selection?

Genotype A1A1 A1A2 A2A2 Frequency 0.55 0.20 0.25 Number 2750 1000 1250 Relative fitness (w) 1.0 0.75 0.50 A) 5000 B) 2750 C) 4125 D) 1250 E) 2250 Answer: C Section: 22.3 Skill: Application/Analysis 7) Mexican hairless dogs are heterozygous (Hh), carrying a single copy of the dominant lethal allele. In a given population, the Hh genotypic frequency is 60% and the frequency of each homozygote is 20%.

Assuming w = 1 for the Hh and hh dogs, what is the estimated genotypic frequency of the lethal HH genotype after reproduction?

A) 20% B) 14% C) 10% D) 39% E) 46% Answer: B Section: 22.3 Skill: Application/Analysis

8) Recessive mutations are subject to natural selection only when they are in the homozygous recessive genotype. What does this mean in terms of allele frequency?

A) Recessive mutations persist in a population at a frequency somewhat greater than mutation frequency.

B) Recessive mutations persist in a population at a frequency far less than the mutation frequency.

C) Recessive mutations persist in a population at a frequency equal to the mutation frequency.

D) The frequency of the mutant allele will ultimately approach 0 if there is no heterozygote advantage.

E) The frequency of the mutant allele will be unchanged because recessive mutations are in HardyWeinberg equilibrium.

Answer: A Section: 22.4 Skill: Knowledge/Comprehension 9) Which aspect of population genetics can lead to changes in allele frequencies in a new mixed population following migration?

A) natural selection B) mutation C) gene flow D) founder effect E) bottleneck effect Answer: C Section: 22.5 Skill: Knowledge/Comprehension 10) Which aspect of population genetics refers to chance fluctuations of allele frequencies resulting from sampling error and is more prominent in small, isolated populations?

A) founder effect B) mutation C) gene flow D) gene drift E) bottleneck effect Answer: D Section: 22.5 Skill: Knowledge/Comprehension 11) All Old Order Amish families from Lancaster County with Ellis-van Creveld syndrome can trace their genealogies to Mr. and Mrs. Samuel King, who immigrated to Lancaster County in 1744. This is an example of A) founder effect.

B) mutation.

C) gene flow.

D) Hardy-Weinberg equilibrium.

E) bottleneck effect.

Answer: A Section: 22.2 Skill: Knowledge/Comprehension

12) Both the founder effect and the bottleneck effect are mechanisms that produce large allele frequency sampling errors in a small population due to A) random mutation.

B) gene flow.

C) gene drift.

D) lethal alleles.

E) random mating.

Answer: C Section: 22.6 Skill: Knowledge/Comprehension 13) Cheetahs exhibit significant inbreeding and loss of heterozygosity due to overhunting and loss of habitat experienced at some point in their evolutionary history. Which genetic mechanism occurs when a large population is substantially reduced at random, acting independently of natural selection?

A) founder effect B) mutation C) gene flow D) Hardy-Weinberg equilibrium E) bottleneck effect Answer: E Section: 22.6 Skill: Knowledge/Comprehension 14) The coefficient of inbreeding (F) can be used to estimate the proportion of loci that are A) fixed in a population.

B) homozygous identical by descent.

C) heterozygous by descent.

D) lethal alleles.

E) recessive.

Answer: B Section: 22.7 Skill: Knowledge/Comprehension

15) Given the following pedigree, what is the coefficient of inbreeding for individual V-1?

A) 1/2 B) 1/4 C) 1/8 D) 1/16 E) 1/64 Answer: C Section: 22.7 Skill: Application/Analysis 16) The process by which an original species is transformed into a new species over an extended period of time that spans many generations is known as A) anagenesis.

B) cladogenesis.

C) morphogenesis.

D) neogenesis.

E) founder effect.

Answer: A Section: 22.8 Skill: Knowledge/Comprehension 17) Physical separation of a segment of a large population by a physical barrier that prevents gene flow can lead to A) sympatric speciation.

B) allopatric speciation.

C) hybrid sterility.

D) temporal isolation.

E) gene flow.

Answer: B Section: 22.8 Skill: Knowledge/Comprehension 18) Which of the following is an example of a postzygotic mechanism for reproductive isolation?

A) behavioral isolation B) gametic isolation C) habitat separation D) mechanical isolation E) hybrid sterility Answer: E Section: 22.8

Skill: Knowledge/Comprehension 19) When populations share a single habitat but are isolated by genetic or postzygotic mechanisms that prevent gene flow, what process can cause populations to diverge?

A) random mating B) allopatric speciation C) sympatric speciation D) random mutation E) gene duplication Answer: C Section: 22.8 Skill: Knowledge/Comprehension 20) In CODIS, what type of genetic marker is used to predict population genotype frequencies?

A) dsRNAs B) PCR products C) CpG islands D) STRPs E) FISH chromosome patterns Answer: D Section: 22.8 Skill: Knowledge/Comprehension 22.2 Short-Answer Questions 1) What type of statistical analysis is used to compare observed and expected results in order to evaluate the validity of an estimate based on the Hardy-Weinberg equilibrium?

Answer: chi-square test Section: 22.1 Skill: Knowledge/Comprehension 2) When looking at the MN blood group with three alleles, how many degrees of freedom will be used to compare observed versus expected results?

Answer: 1 Section: 22.1 Skill: Knowledge/Comprehension 3) If the frequency of hemophilia, an X-linked recessive disease, is 10% in males, what is the frequency of females with hemophilia?

Answer: 1% Section: 22.2 Skill: Application/Analysis 4) If the frequency of hemophilia, an X-linked recessive disorder, is 10% in males, what is the frequency of carriers in the population (both males and females)?

Answer: 0% of males and 18% of females Section: 22.2 Skill: Application/Analysis

5) If the frequency of PKU, an autosomal recessive disease, is 1% in males, what is the frequency of females with PKU? Answer: 1% Section: 22.2 Skill: Application/Analysis 6) If the frequency of PKU (an autosomal recessive disease) is 1%, what percentage of the population are completely unaffected (homozygous dominant)?

Answer: 93.7% Section: 22.2 Skill: Application/Analysis 7) A relative fitness value of w = 1 means an organism exhibits what trait?

Answer: highest reproductive success Section: 22.3 Skill: Knowledge/Comprehension 8) If the relative fitness value is w = 0.6, what is the selection coefficient used to identify the differences between the fitnesses of organisms with different traits?

Answer: 0.4 Section: 22.3 Skill: Knowledge/Comprehension 9) What is the ultimate source of genetic variation in populations?

Answer: mutation Section: 22.4 Skill: Knowledge/Comprehension 10) When examining the effects of forward and reverse mutation rates, what equation is used to calculate the change in the frequency of A2, or Δq?

Answer: Δq = µp - vq Section: 22.4 Skill: Knowledge/Comprehension 11) In a theoretical population where f(A1) = 0.95 and f(A2) = 0.05, µ = 1 × 10-6, and v = 1 × 10-7, what is Δq?

Answer: 9.45 × 10-7 Section: 22.4 Skill: Application/Analysis 12) Which mechanism maintains the biological ability of populations to interbreed and can thus prevent evolutionary divergence of populations?

Answer: gene flow Section: 22.5 Skill: Knowledge/Comprehension

13) If the genotype frequency in an admixed population is f(A1) = 0.75, what is the genotypic frequency of heterozygotes in the admixed population?

Answer: 0.375 or 37.5% Section: 22.5 Skill: Application/Analysis 14) A difference in allelic frequencies that is produced when a new population is established by a few individuals is known as what type of genetic drift?

Answer: founder effect Section: 22.6 Skill: Knowledge/Comprehension 15) Of the five aspects of Hardy-Weinberg equilibrium, which one directly affects phenotypic frequency rather than genotypic frequencies?

Answer: nonrandom mating Section: 22.7 Skill: Knowledge/Comprehension 16) What are the two most common types of nonrandom mating?

Answer: inbreeding and assortative mating Section: 22.7 Skill: Knowledge/Comprehension 17) Inbreeding depression, the reduction in fitness of inbred organisms, often results from the reduced level of what (within the gene pool)?

Answer: genetic heterozygosity Section: 22.7 Skill: Knowledge/Comprehension 18) Which two molecular techniques are used in DNA profiling to generate data included in the FBI's CODIS database?

Answer: PCR and gel electrophoresis Section: 22.8 Skill: Knowledge/Comprehension

19) Using the allele frequencies in the table below, what is the probability that a random person would be heterozygous 12/14 for VWA?

Answer: 1.96 × 10-3, or approximately 1 in 2000 Section: 22.8 Skill: Application/Analysis 20) Using the allele frequencies in the table below, what is the probability that a random person who was heterozygous 12/14 for VWA would also be homozygous 18/18 for FGA?

Answer: 4.42 × 10-7, or approximately 1 in 44 million Section: 22.8 Skill: Application/Analysis

22.3 Fill-in-the-Blank Questions 1) ________ is an evolutionary mechanism that favors the reproductive success of certain members of a population over others as a result of differences in anatomical, physiological, or behavioral traits they possess.

Answer: Natural selection Section: 22.3 Skill: Knowledge/Comprehension 2) If the heterozygous genotype is favored (as exhibited by the sickle cell allele in malaria-prone regions), selection results in a ________ in which alleles reach stable equilibrium frequencies that are maintained by the ongoing action of selection against the homozygous genotypes.

Answer: balanced polymorphism Section: 22.3 Skill: Knowledge/Comprehension 3) Forward and reverse mutation can create a ________, given a sufficient number of generations and the absence of other evolutionary processes.

Answer: balanced equilibrium Section: 22.4 Skill: Knowledge/Comprehension 4) The coefficient of inbreeding quantifies the probability that two alleles in a homozygous individual are ________, having descended from the same copy of the allele carried by a common ancestor of the inbred individual.

Answer: identical by descent (IBD) Section: 22.7 Skill: Knowledge/Comprehension 5) The pattern of species evolution known as ________ is one of branching in which an ancestral species gives rise to two or more new species.

Answer: cladogenesis Section: 22.8 Skill: Knowledge/Comprehension

22.4 Essay Questions In the general population, 1 in 300 individuals is a carrier for Tay-Sachs disease, while 1 in 30 individuals of Ashkenazi Jew descent are carriers. Tay-Sachs also affects 1 in 30 individuals with French-Canadian ancestry, although two completely unique mutations are responsible for the Ashkenazi and French-Canadian mutations.

1) What are the chances of two individuals who are carriers having a child with Tay-Sachs if a. both individuals are not of Ashkenazi or French-Canadian descent? b. one individual is of Ashkenazi descent and one is not? c. both individuals are of Ashkenazi descent?

Answer: For each question, you will calculate [Chance of mom being a carrier] × [Chance of dad being a carrier] × [Chance of having an affected child if both parents are carriers] a. (1/300) × (1/300) × 1/4 = 2.77 × 10-6 b. (1/30) × (1/300) × 1/4 = 2.77 × 10-5 c. (1/30) × (1/300) × 1/4 = 2.77 × 10-4 Section: 22.1 Skill: Synthesis/Evaluation 2) If an Ashkenazi carrier and a French-Canadian carrier had a child, what are the chances that they would have a child who is a carrier for both types of the mutation?

Answer: (1/30) × (1/30) = 1.11 × 10-3, or 0.11% Section: 22.1 Skill: Synthesis/Evaluation 3) In an island population, the frequencies of the three MN blood types are given here: M: 550 MN: 725 N: 225 a. Calculate the allele frequency of M and N. b. Assuming random mating, what are the expected frequencies for each blood type? c. Use chi-square analysis to test your hypothesis that mating is random for this population.

Answer: a. p = f(M) = [2(550) + 725]/3000 = 0.608 q = f(N) = [2(225) + 725]/3000 = 0.392 b. f(M) = (0.608)2 (1500) = 555 f(MN) = (2)(0.608)(0.392)(1500) = 715 f(N) = (0.392)2 (1500) = 230 c. Χ2 = [(550 - 555)2/555] + [(725 -715)2/715] + [(225 - 230)2/230] = 0.2936 Section: 22.2 Skill: Synthesis/Evaluation

4) Achrondroplasia is an autosomal dominant genetic disorder, while red-green color blindness is an Xlinked recessive disorder. For a given population, the frequency of achondroplasia is 1 in 25,000. If redgreen color blindness affects 9% of males, what percentage of the population would you expect to be a. colorblind males of normal height? b. colorblind females of normal height? c. males with normal vision and achondroplasia? d. females with normal vision and achondroplasia? e. males with both colorblindness and achondroplasia? f. females with both colorblindness and achondroplasia?

Answer: a. 0.09 × 0.99992 = 0.08999 or 8.999% b. 0.008 × 0.99992 = 0.00799 or 0.799% c. 0.91 × 0.00008 = 0.0000728 or 0.00728% d. 0.9992 × 0.00008 = 0.0000799 or 0.00799% e. 0.09 × 0.00008 = 0.0000072 or 0.00072% f. 0.008 × 0.00008 = 0.00000064 or 0.000064% Calculate the probabilities separately, and use the product rule to solve each question.

Colorblindness q = 0.09, so q2 = 0.008 Colorblind male = 0.09 = 9% Colorblind female = 0.008 = 0.8% Normal vision male = 0.91 = 91% Normal vision female = 0.992 = 99.2% Achondroplasia q = 0.99996 and p = 0.00004 So, the chance of being unaffected is q2 = 0.99992.

Because 99.992% of the population are unaffected, 0.008% of the population have achondroplasia (primarily heterozygotes).

Section: 22.2 Skill: Synthesis/Evaluation 5) What are the three key elements for determining F, a measure that a particular allele is IBD, using pedigree analysis?

Answer: The three key elements to determining F from pedigrees are (1) the number of alleles of a gene carried by common ancestors, (2) the number of transmission events required to produce a genotype that is homozygous IBD, and (3) the probability of transmission for each event linking the allele in a common ancestor to the inbred individual. First-cousin mating is a form of inbreeding that is relatively common in many human societies and is common in mammals in general.

Section: 22.7 Skill: Synthesis/Evaluation

6) Explain why rare recessive alleles are more likely to produce a recessive phenotype as a result of inbreeding.

Answer: Negative genetic outcomes in the form of infants with recessive conditions due to inbreeding occur when a recessive allele is rare in a population. In such cases, there is a meaningful increase in the likelihood that a first-cousin mating will produce a child with a recessive phenotype. When the recessive allele frequency is q = 0.01, for example, the chance of producing a recessive homozygote from a firstcousin mating is only a few times more likely than the chance of producing a recessive homozygote by random mating.

Section: 22.7 Skill: Synthesis/Evaluation