Preview (11 of 36 pages)

Preview Extract

This Document Contains Chapters 12 to 13 Chapter 12: The Chromosomal Basis of Inheritance 12.1 Multiple-Choice Questions 1) When Thomas Hunt Morgan crossed his red-eyed F1 generation flies to each other, the F2 generation included both red- and white-eyed flies. Remarkably, all the white-eyed flies were male. What was the explanation for this result? A) The gene involved is on the Y chromosome. B) The gene involved is on the X chromosome. C) The gene involved is on an autosome, but only in males. D) Other male-specific factors influence eye color in flies. E) Other female-specific factors influence eye color in flies. Answer: B 2) Which of the following is the meaning of the chromosome theory of inheritance as expressed in the early 20th century? A) Individuals inherit particular chromosomes attached to genes. B) Mendelian genes are at specific loci on the chromosome and in turn segregate during meiosis. C) Homologous chromosomes give rise to some genes and crossover chromosomes to other genes. D) No more than a single pair of chromosomes can be found in a healthy normal cell. E) Natural selection acts on certain chromosome arrays rather than on genes. Answer: B 3) Males are more often affected by sex-linked traits than females because A) male hormones such as testerone often alter the affects of mutations on the X chromosome. B) female hormones such as estrogen often compensate for the effects of mutations on the X chromosome. C) X chromosomes in males generally have more mutations than X chromosomes in females. D) males are hemizygous for the X chromosome. E) mutations on the Y chromosome often worsen the effects of X-linked mutations. Answer: D 4) SRY is best described in which of the following ways? A) a gene present on the X chromosome that triggers female development B) an autosomal gene that is required for the expression of genes on the Y chromosome C) a gene region present on the Y chromosome that triggers male development D) an autosomal gene that is required for the expression of genes on the X chromosome E) a gene required for development, and males or females lacking the gene do not survive past early childhood Answer: C 5) In cats, black fur color is caused by an X-linked allele; the other allele at this locus causes orange color. The heterozygote is tortoiseshell. What kinds of offspring would you expect from the cross of a black female and an orange male? A) tortoiseshell females; tortoiseshell males B) black females; orange males C) orange females; orange males D) tortoiseshell females; black males E) orange females; black males Answer: D 6) Red-green color blindness is a sex-linked recessive trait in humans. Two people with normal color vision have a color-blind son. What are the genotypes of the parents? A) XnXn and XnY B) XnXn and XNY C) XNXN and XnY D) XNXN and XNY E) XNXn and XNY Answer: E 7) Cinnabar eyes is a sex-linked recessive characteristic in fruit flies. If a female having cinnabar eyes is crossed with a wild-type male, what percentage of the F1 males will have cinnabar eyes? A) 0% B) 25% C) 50% D) 75% E) 100% Answer: E 8) Normally, only female cats have the tortoiseshell phenotype because A) the males die during embryonic development. B) a male inherits only one allele of the X-linked gene controlling hair color. C) the Y chromosome has a gene blocking orange coloration. D) only males can have Barr bodies. E) multiple crossovers on the Y chromosome prevent orange pigment production. Answer: B 9) Sex determination in mammals is due to the SRY region of the Y chromosome. An abnormality of this region could allow which of the following to have a male phenotype? A) Turner syndrome, 45, X B) translocation of SRY to an autosome of a 46, XX individual C) a person with an extra X chromosome D) a person with one normal and one shortened (deleted) X E) Down syndrome, 46, XX Answer: B 10) In humans, clear gender differentiation occurs not at fertilization, but after the second month of gestation. What is the first event of this differentiation? A) formation of testosterone in male embryos B) formation of estrogens in female embryos C) anatomical differentiation of a penis in male embryos D) activation of SRY in male embryos and masculinization of the gonads E) activation of SRY in females and feminization of the gonads Answer: D 11) Duchenne muscular dystrophy is a serious condition caused by a recessive allele of a gene on the human X chromosome. The patients have muscles that weaken over time because they have absent or decreased dystrophin, a muscle protein. They rarely live past their 20s. How likely is it for a woman to have this condition? A) Women can never have this condition. B) One-half of the daughters of an affected man would have this condition. C) One-fourth of the daughters of an affected father and a carrier mother could have this condition. D) Very rarely: it is rare that an affected male would mate with a carrier female. E) Only if a woman is XXX could she have this condition. Answer: D 12) All female mammals have one active X chromosome per cell instead of two. What causes this? A) activation of the XIST gene on the X chromosome that will become the Barr body B) activation of the BARR gene on one X chromosome, which then becomes inactive C) crossing over between the XIST gene on one X chromosome and a related gene on an autosome D) inactivation of the XIST gene on the X chromosome derived from the male parent E) attachment of methyl (CH3) groups to the X chromosome that will remain active Answer: A 13) Which of the following statements is true of linkage? A) The closer two genes are on a chromosome, the lower the probability that a crossover will occur between them. B) The observed frequency of recombination of two genes that are far apart from each other has a maximum value of 100%. C) All of the traits that Mendel studied–seed color, pod shape, flower color, and others–are due to genes linked on the same chromosome. D) Linked genes are found on different chromosomes. E) Crossing over occurs during prophase II of meiosis. Answer: A 14) How would one explain a testcross involving F1 dihybrid flies in which more parental-type offspring than recombinant-type offspring are produced? A) The two genes are closely linked on the same chromosome. B) The two genes are linked but on different chromosomes. C) Recombination did not occur in the cell during meiosis. D) The testcross was improperly performed. E) Both of the characters are controlled by more than one gene. Answer: A 15) What does a frequency of recombination of 50% indicate? A) The two genes are likely to be located on different chromosomes. B) All of the offspring have combinations of traits that match one of the two parents. C) The genes are located on sex chromosomes. D) Abnormal meiosis has occurred. E) Independent assortment is hindered. Answer: A 16) Three genes (A, B, and C) at three loci are being mapped in a particular species. Each gene has two alleles, one of which results in a phenotype that is markedly different from the wild type. The unusual allele of gene A is inherited with the unusual allele of gene B or C about 50% of the time. However, the unusual alleles of genes B and C are inherited together 14.4% of the time. Which of the following describes what is happening? A) The three genes are showing independent assortment. B) The three genes are linked. C) Gene A is linked but genes B and C are not. D) Gene A is assorting independently of genes B and C, which are linked. E) Gene A is located 14.4 map units from genes B and C. Answer: D 17) What is one map unit equivalent to? A) the physical distance between two linked genes B) 1% frequency of recombination between two genes C) 1 nanometer of distance between two genes D) the distance between a pair of homologous chromosomes E) the recombination frequency between two genes assorting independently Answer: B 18) Recombination between linked genes comes about for what reason? A) Mutation on one homolog is different from that on the other homolog. B) Independent assortment sometimes fails because Mendel had not calculated appropriately. C) When genes are linked they always "travel" together at anaphase. D) Crossovers between these genes result in chromosomal exchange. E) Nonrecombinant chromosomes break and then re-join with one another. Answer: D 19) Why does recombination between linked genes continue to occur? A) Recombination is a requirement for independent assortment. B) Recombination must occur or genes will not assort independently. C) New allele combinations are acted upon by natural selection. D) The forces on the cell during meiosis II always result in recombination. E) Without recombination there would be an insufficient number of gametes. Answer: C 20) Map units on a linkage map cannot be relied upon to calculate physical distances on a chromosome for which of the following reasons? A) The frequency of crossing over varies along the length of the chromosome. B) The relationship between recombination frequency and map units is different in every individual. C) Physical distances between genes change during the course of the cell cycle. D) The gene order on the chromosomes is slightly different in every individual. E) Linkage map distances are identical between males and females. Answer: A 21) What is the reason that closely linked genes are typically inherited together? A) The likelihood of a crossover event between these two genes is low. B) The number of genes in a cell is greater than the number of chromosomes. C) Chromosomes are unbreakable. D) Alleles are paired together during meiosis. E) Genes align that way during metaphase I of meiosis. Answer: A 22) Sturtevant provided genetic evidence for the existence of four pairs of chromosomes in Drosophila in which of these ways? A) There are four major functional classes of genes in Drosophila. B) Drosophila genes cluster into four distinct groups of linked genes. C) The overall number of genes in Drosophila is a multiple of four. D) The entire Drosophila genome has approximately 400 map units. E) Drosophila genes have, on average, four different alleles. Answer: B 23) If cell X enters meiosis, and nondisjunction of one chromosome occurs in one of its daughter cells during meiosis II, what will be the result at the completion of meiosis? A) All the gametes descended from cell X will be diploid. B) Half of the gametes descended from cell X will be n + 1, and half will be n - 1. C) One-fourth of the gametes descended from cell X will be n + 1, 1/4 will be n - 1, and 1/2 will be n. D) There will be three extra gametes. E) Two of the four gametes descended from cell X will be haploid, and two will be diploid. Answer: C 24) One possible result of chromosomal breakage is for a fragment to join a nonhomologous chromosome. What is this alteration called? A) deletion B) transversion C) inversion D) translocation E) duplication Answer: D 25) A nonreciprocal crossover causes which of the following products? A) deletion only B) duplication only C) nondisjunction D) deletion and duplication E) duplication and nondisjunction Answer: D 26) Of the following human aneuploidies, which is the one that generally has the most severe impact on the health of the individual? A) 47, +21 B) 47, XXY C) 47, XXX D) 47, XYY E) 45, X Answer: A 27) A phenotypically normal prospective couple seeks genetic counseling because the man knows that he has a translocation of a portion of his chromosome 4 that has been exchanged with a portion of his chromosome 12. Although he is normal because his translocation is balanced, he and his wife want to know the probability that his sperm will be abnormal. What is your prognosis regarding his sperm? A) One-fourth will be normal, 1/4 will have the translocation, and 1/2 will have duplications and deletions. B) All will carry the same translocation as the father. C) None will carry the translocation because abnormal sperm will die. D) His sperm will be sterile and the couple might consider adoption. E) One-half will be normal and the rest will have the father's translocation. Answer: A 28) Abnormal chromosomes are frequently found in malignant tumors. Errors such as translocations may place a gene in close proximity to different control regions. Which of the following might then occur to make the cancer worse? A) an increase in nondisjunction B) expression of inappropriate gene products C) a decrease in mitotic frequency D) death of the cancer cells in the tumor E) sensitivity of the immune system Answer: B 29) An inversion in a human chromosome often results in no demonstrable phenotypic effect in the individual. What else may occur? A) There may be deletions later in life. B) Some abnormal gametes may be formed. C) There is an increased frequency of mutation. D) All inverted chromosomes are deleted. E) The individual is more likely to get cancer. Answer: B 30) What is the source of the extra chromosome 21 in an individual with Down syndrome? A) nondisjunction in the mother only B) nondisjunction in the father only C) duplication of the chromosome D) nondisjunction or translocation in either parent E) It is impossible to detect with current technology. Answer: D 31) Down syndrome has a frequency in the U.S. population of 1/700 live births. In which of the following groups would you expect this frequency to be significantly higher? A) people in Latin or South America B) the Inuit and other peoples in very cold habitats C) people living in equatorial areas of the world D) very small population groups E) No groups have such higher frequency. Answer: E 32) A couple has a child with Down syndrome. The mother is 39 years old at the time of delivery. Which of the following is the most probable cause of the child's condition? A) The woman inherited this tendency from her parents. B) One member of the couple carried a translocation. C) One member of the couple underwent nondisjunction in somatic cell production. D) One member of the couple underwent nondisjunction in gamete production. E) The mother had a chromosomal duplication. Answer: D 33) What is a syndrome? A) a characteristic facial appearance B) a group of traits, all of which must be present if an aneuploidy is to be diagnosed C) a group of traits typically found in conjunction with a particular chromosomal aberration or gene mutation D) a characteristic trait usually given the discoverer's name E) a characteristic that only appears in conjunction with one specific aneuploidy Answer: C 34) Which of the following is known as a Philadelphia chromosome? A) a human chromosome 22 that has had a specific translocation B) a human chromosome 9 that is found only in one type of cancer C) an animal chromosome found primarily in the mid-Atlantic area of the United States D) an imprinted chromosome that always comes from the mother E) a chromosome found not in the nucleus but in mitochondria Answer: A 35) At what point in cell division is a chromosome lost so that, after fertilization with a normal gamete, the result is an embryo with 45, X? I. an error in anaphase I II. an error in anaphase II III. an error of the first postfertilization mitosis IV. an error in pairing A) I or II only B) II or IV only C) III or IV only D) I, II, or III only E) I, II, III, or IV Answer: E 36) Which of the following is true of aneuploidies in general? A) A monosomy is more frequent than a trisomy. B) 45, X is the only known human live-born monosomy. C) Some human aneuploidies have selective advantage in some environments. D) Of all human aneuploidies, only Down syndrome is associated with mental retardation. E) An aneuploidy resulting in the deletion of a chromosome segment is less serious than a duplication. Answer: B 37) A woman is found to have 47 chromosomes, including three X chromosomes. Which of the following describes her expected phenotype? A) masculine characteristics such as facial hair B) enlarged genital structures C) excessive emotional instability D) healthy female of slightly above-average height E) sterile female Answer: D 12.2 Art Questions Figure 12.1 shows a map of four genes on a chromosome. Figure 12.1 1) Between which two genes would you expect the highest frequency of recombination? A) A and W B) W and E C) E and G D) A and E E) A and G Answer: E Figure 12.2 2) In a series of mapping experiments, the recombination frequencies for four different linked genes of Drosophila were determined as shown in Figure 12.2. What is the order of these genes on a chromosome map? A) rb-cn-vg-b B) vg-b-rb-cn C) cn-rb-b-vg D) b-rb-cn-vg E) vg-cn-b-rb Answer: D Use the following information to answer the questions below. A plantlike organism on the planet Pandora can have three recessive genetic traits: bluish leaves, due to an allele (a) of gene A; a feathered stem, due to an allele (b) of gene B; and hollow roots due to an allele (c) of gene C. The three genes are linked and recombine as follows: A geneticist did a testcross with an organism that had been found to be heterozygous for the three recessive traits and she was able to identify progeny of the following phenotypic distribution (+ = wild type): Figure 12.3 3) Which of the following are the phenotypes of the parents in this cross? A) 2 and 5 B) 1 and 6 C) 4 and 8 D) 3 and 7 E) 1 and 2 Answer: C 4) In which progeny phenotypes has there been recombination between genes A and B? A) 1, 2, 5, and 6 B) 1, 3, 6, and 7 C) 2, 4, 5, and 8 D) 2, 3, 5, and 7 E) in all 8 of them Answer: A 5) If recombination frequency is equal to distance in map units, what is the approximate distance between genes A and B? A) 1.5 map units B) 3 map units C) 6 map units D) 15 map units E) 30 map units Answer: B 6) What is the greatest benefit of having used a testcross for this experiment? A) The homozygous recessive parents are obvious to the naked eye. B) The homozygous parents are the only ones whose crossovers make a difference. C) Progeny can be scored by their phenotypes alone. D) All of the progeny will be heterozygous. E) The homozygous recessive parents will be unable to cross over. Answer: C 7) The greatest distance among the three genes is between a and c. What does this mean? A) Gene c is between a and b. B) Genes are in the order: a–b–c. C) Gene a is not recombining with c. D) Gene a is between b and c. E) Distance a–b is equal to distance a–c. Answer: B 12.3 Scenario Questions Refer to the following information to answer the questions below. A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive. 1) How many of their daughters might be expected to be color-blind dwarfs? A) all B) none C) half D) one out of four E) three out of four Answer: B 2) What proportion of their sons would be color-blind and of normal height? A) none B) half C) one out of four D) three out of four E) all Answer: B 3) They have a daughter who is a dwarf with normal color vision. What is the probability that she is heterozygous for both genes? A) 0% B) 25% C) 50% D) 75% E) 100% Answer: E Chapter 13: The Molecular Basis of Inheritance 13.1 Multiple-Choice Questions 1) In his transformation experiments, what did Griffith observe? A) Mutant mice were resistant to bacterial infections. B) Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form. C) Mixing a heat-killed nonpathogenic strain of bacteria with a living pathogenic strain makes the pathogenic strain nonpathogenic. D) Infecting mice with nonpathogenic strains of bacteria makes them resistant to pathogenic strains. E) Mice infected with a pathogenic strain of bacteria can spread the infection to other mice. Answer: B 2) How do we describe transformation in bacteria? A) the creation of a strand of DNA from an RNA molecule B) the creation of a strand of RNA from a DNA molecule C) the infection of cells by a phage DNA molecule D) the type of semiconservative replication shown by DNA E) assimilation of external DNA into a cell Answer: E 3) After mixing a heat-killed, phosphorescent (light-emitting) strain of bacteria with a living, nonphosphorescent strain, you discover that some of the living cells are now phosphorescent. Which observation(s) would provide the best evidence that the ability to phosphoresce is a heritable trait? A) DNA passed from the heat-killed strain to the living strain. B) Protein passed from the heat-killed strain to the living strain. C) The phosphorescence in the living strain is especially bright. D) Descendants of the living cells are also phosphorescent. E) Both DNA and protein passed from the heat-killed strain to the living strain. Answer: D 4) In trying to determine whether DNA or protein is the genetic material, Hershey and Chase made use of which of the following facts? A) DNA contains sulfur, whereas protein does not. B) DNA contains phosphorus, whereas protein does not. C) DNA contains nitrogen, whereas protein does not. D) DNA contains purines, whereas protein includes pyrimidines. E) RNA includes ribose, whereas DNA includes deoxyribose sugars. Answer: B 5) Which of the following investigators was (were) responsible for the following discovery? In DNA from any species, the amount of adenine equals the amount of thymine, and the amount of guanine equals the amount of cytosine. A) Frederick Griffith B) Alfred Hershey and Martha Chase C) Oswald Avery, Maclyn McCarty, and Colin MacLeod D) Erwin Chargaff E) Matthew Meselson and Franklin Stahl Answer: D 6) Cytosine makes up 42% of the nucleotides in a sample of DNA from an organism. Approximately what percentage of the nucleotides in this sample will be thymine? A) 8% B) 16% C) 31% D) 42% E) It cannot be determined from the information provided. Answer: A 7) Which of the following can be determined directly from X-ray diffraction photographs of crystallized DNA? A) the diameter of the helix B) the rate of replication C) the sequence of nucleotides D) the bond angles of the subunits E) the frequency of A vs. T nucleotides Answer: A 8) It became apparent to Watson and Crick after completion of their model that the DNA molecule could carry a vast amount of hereditary information in which of the following? A) sequence of bases B) phosphate-sugar backbones C) complementary pairing of bases D) side groups of nitrogenous bases E) different five-carbon sugars Answer: A 9) In an analysis of the nucleotide composition of DNA, which of the following will be found? A) A = C B) A = G and C = T C) A + C = G + T D) G + C = T + A Answer: C 10) What is meant by the description "antiparallel" regarding the strands that make up DNA? A) The twisting nature of DNA creates nonparallel strands. B) The 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand. C) Base pairings create unequal spacing between the two DNA strands. D) One strand is positively charged and the other is negatively charged. E) One strand contains only purines and the other contains only pyrimidines. Answer: B 11) Replication in prokaryotes differs from replication in eukaryotes for which of the following reasons? A) Prokaryotic chromosomes have histones, whereas eukaryotic chromosomes do not. B) Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have many. C) The rate of elongation during DNA replication is slower in prokaryotes than in eukaryotes. D) Prokaryotes produce Okazaki fragments during DNA replication, but eukaryotes do not. E) Prokaryotes have telomeres, and eukaryotes do not. Answer: B 12) Suppose you are provided with an actively dividing culture of E. coli bacteria to which radioactive thymine has been added. What would happen if a cell replicates once in the presence of this radioactive base? A) One of the daughter cells, but not the other, would have radioactive DNA. B) Neither of the two daughter cells would be radioactive. C) All four bases of the DNA would be radioactive. D) Radioactive thymine would pair with nonradioactive guanine. E) DNA in both daughter cells would be radioactive. Answer: E 13) An Okazaki fragment has which of the following arrangements? A) primase, polymerase, ligase B) 3' RNA nucleotides, DNA nucleotides 5' C) 5' RNA nucleotides, DNA nucleotides 3' D) DNA polymerase I, DNA polymerase III E) 5' DNA to 3' Answer: C 14) In E. coli, there is a mutation in a gene called dnaB that alters the helicase that normally acts at the origin. Which of the following would you expect as a result of this mutation? A) No proofreading will occur. B) No replication fork will be formed. C) The DNA will supercoil. D) Replication will occur via RNA polymerase alone. E) Replication will require a DNA template from another source. Answer: B 15) Which enzyme catalyzes the elongation of a DNA strand in the 5' → 3' direction? A) primase B) DNA ligase C) DNA polymerase III D) topoisomerase E) helicase Answer: C 16) At a specific area of a chromosome, the following sequence of nucleotides is present where the chain opens to form a replication fork: 3' C T A G C T G C A A T C C 5' An RNA primer is formed starting at the underlined T (T) of the template. Which of the following represents the primer sequence? A) 5' G C C T A G G 3' B) 3' G C C T A G G 5' C) 5' A C G T T A G G 3' D) 5' A C G U U A G G 3' E) 5' G C C U A G G 3' Answer: D 17) Polytene chromosomes of Drosophila salivary glands each consist of multiple identical DNA strands that are aligned in parallel arrays. How could these arise? A) replication followed by mitosis B) replication without separation C) meiosis followed by mitosis D) fertilization by multiple sperm E) special association with histone proteins Answer: B 18) To repair a thymine dimer by nucleotide excision repair, in which order do the necessary enzymes act? A) exonuclease, DNA polymerase III, RNA primase B) helicase, DNA polymerase I, DNA ligase C) DNA ligase, nuclease, helicase D) DNA polymerase I, DNA polymerase III, DNA ligase E) endonuclease, DNA polymerase I, DNA ligase Answer: E 19) What is the function of DNA polymerase III? A) to unwind the DNA helix during replication B) to seal together the broken ends of DNA strands C) to add nucleotides to the 3' end of a growing DNA strand D) to degrade damaged DNA molecules E) to rejoin the two DNA strands (one new and one old) after replication Answer: C 20) The difference between ATP and the nucleoside triphosphates used during DNA synthesis is that A) the nucleoside triphosphates have the sugar deoxyribose; ATP has the sugar ribose. B) the nucleoside triphosphates have two phosphate groups; ATP has three phosphate groups. C) ATP contains three high-energy bonds; the nucleoside triphosphates have two. D) ATP is found only in human cells; the nucleoside triphosphates are found in all animal and plant cells. E) triphosphate monomers are active in the nucleoside triphosphates, but not in ATP. Answer: A 21) The leading and the lagging strands differ in that A) the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction. B) the leading strand is synthesized by adding nucleotides to the 3' end of the growing strand, and the lagging strand is synthesized by adding nucleotides to the 5' end. C) the lagging strand is synthesized continuously, whereas the leading strand is synthesized in short fragments that are ultimately stitched together. D) the leading strand is synthesized at twice the rate of the lagging strand. Answer: A 22) A new DNA strand elongates only in the 5' to 3' direction because A) DNA polymerase begins adding nucleotides at the 5' end of the template. B) Okazaki fragments prevent elongation in the 3' to 5' direction. C) the polarity of the DNA molecule prevents addition of nucleotides at the 3' end. D) replication must progress toward the replication fork. E) DNA polymerase can only add nucleotides to the free 3' end. Answer: E 23) What is the function of topoisomerase? A) relieving strain in the DNA ahead of the replication fork B) elongating new DNA at a replication fork by adding nucleotides to the existing chain C) adding methyl groups to bases of DNA D) unwinding of the double helix E) stabilizing single-stranded DNA at the replication fork Answer: A 24) What is the role of DNA ligase in the elongation of the lagging strand during DNA replication? A) It synthesizes RNA nucleotides to make a primer. B) It catalyzes the lengthening of telomeres. C) It joins Okazaki fragments together. D) It unwinds the parental double helix. E) It stabilizes the unwound parental DNA. Answer: C 25) Which of the following help(s) to hold the DNA strands apart while they are being replicated? A) primase B) ligase C) DNA polymerase D) single-strand binding proteins E) exonuclease Answer: D 26) Individuals with the disorder xeroderma pigmentosum are hypersensitive to sunlight. This occurs because their cells are impaired in what way? A) They cannot replicate DNA. B) They cannot undergo mitosis. C) They cannot exchange DNA with other cells. D) They cannot repair thymine dimers. E) They do not recombine homologous chromosomes during meiosis. Answer: D Use the list of choices below for the following questions: I. helicase II. DNA polymerase III III. ligase IV. DNA polymerase I V. primase 27) Which of the enzymes removes the RNA nucleotides from the primer and adds equivalent DNA nucleotides to the 3' end of Okazaki fragments? A) I B) II C) III D) IV E) V Answer: D 28) Which of the enzymes separates the DNA strands during replication? A) I B) II C) III D) IV E) V Answer: A 29) Which of the enzymes covalently connects segments of DNA? A) I B) II C) III D) IV E) V Answer: C 30) Which of the enzymes synthesizes short segments of RNA? A) I B) II C) III D) IV E) V Answer: E 31) Given the damage caused by UV radiation, the kind of gene affected in those with XP is one whose product is involved with A) mending of double-strand breaks in the DNA backbone. B) breakage of cross-strand covalent bonds. C) the ability to excise single-strand damage and replace it. D) the removal of double-strand damaged areas. E) causing affected skin cells to undergo apoptosis. Answer: C 32) Which of the following sets of materials is required by both eukaryotes and prokaryotes for replication? A) double-stranded DNA, four kinds of dNTPs, primers, origins of replication B) topoisomerases, telomerases, polymerases C) G-C rich regions, polymerases, chromosome nicks D) nucleosome loosening, four dNTPs, four rNTPs E) ligase, primers, nucleases Answer: A 33) Studies of nucleosomes have shown that histones (except H1) exist in each nucleosome as two kinds of tetramers: one of 2 H2A molecules and 2 H2B molecules, and the other as 2 H3 and 2 H4 molecules. Which of the following is supported by this data? A) DNA can wind itself around either of the two kinds of tetramers. B) The two types of tetramers associate to form an octamer. C) DNA has to associate with individual histones before they form tetramers. D) Only H2A can form associations with DNA molecules. E) The structure of H3 and H4 molecules is not basic like that of the other histones. Answer: B 34) In a linear eukaryotic chromatin sample, which of the following strands is looped into domains by scaffolding? A) DNA without attached histones B) DNA with H1 only C) the 10-nm chromatin fiber D) the 30-nm chromatin fiber E) the metaphase chromosome Answer: D 35) Which of the following statements describes the eukaryotic chromosome? A) It is composed of DNA alone. B) The nucleosome is its most basic functional subunit. C) The number of genes on each chromosome is different in different cell types of an organism. D) It consists of a single linear molecule of double-stranded DNA plus proteins. E) Active transcription occurs on heterochromatin but not euchromatin. Answer: D 36) If a cell were unable to produce histone proteins, which of the following would be a likely effect? A) There would be an increase in the amount of "satellite" DNA produced during centrifugation. B) The cell's DNA couldn't be packed into its nucleus. C) Spindle fibers would not form during prophase. D) Amplification of other genes would compensate for the lack of histones. E) Pseudogenes would be transcribed to compensate for the decreased protein in the cell. Answer: B 37) Which of the following statements is true of histones? A) Each nucleosome consists of two molecules of histone H1. B) Histone H1 is not present in the nucleosome bead; instead, it draws the nucleosomes together. C) The carboxyl end of each histone extends outward from the nucleosome and is called a "histone tail." D) Histones are found in mammals, but not in other animals or in plants or fungi. E) The mass of histone in chromatin is approximately nine times the mass of DNA. Answer: B 38) Why do histones bind tightly to DNA? A) Histones are positively charged, and DNA is negatively charged. B) Histones are negatively charged, and DNA is positively charged. C) Both histones and DNA are strongly hydrophobic. D) Histones are covalently linked to the DNA. E) Histones are highly hydrophobic, and DNA is hydrophilic. Answer: A 39) Which of the following represents the order of increasingly higher levels of organization of chromatin? A) nucleosome, 30-nm chromatin fiber, looped domain B) looped domain, 30-nm chromatin fiber, nucleosome C) looped domain, nucleosome, 30-nm chromatin fiber D) nucleosome, looped domain, 30-nm chromatin fiber E) 30-nm chromatin fiber, nucleosome, looped domain Answer: A 40) Which of the following statements describes chromatin? A) Heterochromatin is composed of DNA, whereas euchromatin is made of DNA and RNA. B) Both heterochromatin and euchromatin are found in the cytoplasm. C) Heterochromatin is highly condensed, whereas euchromatin is less compact. D) Euchromatin is not transcribed, whereas heterochromatin is transcribed. E) Only euchromatin is visible under the light microscope. Answer: C 41) Which of the following modifications is least likely to alter the rate at which a DNA fragment moves through a gel during electrophoresis? A) altering the nucleotide sequence of the DNA fragment without adding or removing nucleotides B) acetylating the cytosine bases within the DNA fragment C) increasing the length of the DNA fragment D) decreasing the length of the DNA fragment E) neutralizing the negative charges within the DNA fragment Answer: A 42) Assume that you are trying to insert a gene into a plasmid. Someone gives you a preparation of genomic DNA that has been cut with restriction enzyme X. The gene you wish to insert has sites on both ends for cutting by restriction enzyme Y. You have a plasmid with a single site for Y, but not for X. Your strategy should be to A) insert the fragments cut with restriction enzyme X directly into the plasmid without cutting the plasmid. B) cut the plasmid with restriction enzyme X and insert the fragments cut with restriction enzyme Y into the plasmid. C) cut the DNA again with restriction enzyme Y and insert these fragments into the plasmid cut with the same enzyme. D) cut the plasmid twice with restriction enzyme Y and ligate the two fragments onto the ends of the DNA fragments cut with restriction enzyme X. E) cut the plasmid with restriction enzyme X and then insert the gene into the plasmid. Answer: C 43) How does a bacterial cell protect its own DNA from restriction enzymes? A) by adding methyl groups to adenines and cytosines B) by using DNA ligase to seal the bacterial DNA into a closed circle C) by adding histones to protect the double-stranded DNA D) by forming "sticky ends" of bacterial DNA to prevent the enzyme from attaching E) by reinforcing the bacterial DNA structure with covalent phosphodiester bonds Answer: A 44) What is the most logical sequence of steps for splicing foreign DNA into a plasmid and inserting the plasmid into a bacterium? I. Transform bacteria with a recombinant DNA molecule. II. Cut the plasmid DNA using restriction enzymes. III. Extract plasmid DNA from bacterial cells. IV. Hydrogen-bond the plasmid DNA to nonplasmid DNA fragments. V. Use ligase to seal plasmid DNA to nonplasmid DNA. A) I, II, IV, III, V B) II, III, V, IV, I C) III, II, IV, V, I D) III, IV, V, I, II E) IV, V, I, II, III Answer: C 45) Why is it so important to be able to amplify DNA fragments when studying genes? A) DNA fragments are too small to use individually. B) A gene may represent only a millionth of the cell's DNA. C) Restriction enzymes cut DNA into fragments that are too small. D) A clone requires multiple copies of each gene per clone. E) It is important to have multiple copies of DNA in the case of laboratory error. Answer: B 46) The reason for using Taq polymerase for PCR is that A) it is heat stable and can withstand the heating step of PCR. B) only minute amounts are needed for each cycle of PCR. C) it binds more readily than other polymerases to the primers. D) it has regions that are complementary to the primers. E) it is heat stable, and it binds more readily than other polymerases to the primers. Answer: A 13.2 Art Questions Figure 13.1 1) In the late 1950s, Meselson and Stahl grew bacteria in a medium containing "heavy" nitrogen (15N) and then transferred them to a medium containing 14N. Which of the results in Figure 13.1 would be expected after one round of DNA replication in the presence of 14N? A) A B) B C) C D) D E) E Answer: D 2) A space probe returns with a culture of a microorganism found on a distant planet. Analysis shows that it is a carbon-based life-form that has DNA. You grow the cells in 15N medium for several generations and then transfer them to 14N medium. Which pattern in Figure 13.1 would you expect if the DNA was replicated in a conservative manner? A) A B) B C) C D) D E) E Answer: B 3) Once the pattern found after one round of replication was observed, Meselson and Stahl could be confident of which of the following conclusions? A) Replication is semi-conservative. B) Replication is not dispersive. C) Replication is not semi-conservative. D) Replication is not conservative. E) Replication is neither dispersive nor conservative. Answer: D Figure 13.2 4) In an experiment, DNA is allowed to replicate in an environment with all necessary enzymes, dATP, dCTP, dGTP, and radioactively labeled dTTP (3H thymidine). After several minutes, the DNA is switched to nonradioactive medium and is then viewed by electron microscopy and autoradiography. Figure 13.2 represents the results. It shows a replication bubble, and the dots represent radioactive material. Which of the following is the most likely interpretation of the results? A) There are two replication forks going in opposite directions. B) Thymidine is being added only where the DNA strands are farthest apart. C) Thymidine is being added only at the very beginning of replication. D) Replication proceeds in one direction only. Answer: A Use Figure 13.3 to answer the following question. Figure 13.3 5) Which enzyme was used to produce the molecule in Figure 13.3? A) ligase B) transcriptase C) a restriction enzyme D) RNA polymerase E) DNA polymerase Answer: C Use Figure 13.4 to answer the following question. Figure 13.4 6) The segment of DNA shown in Figure 13.4 has restriction sites I and II, which create restriction fragments A, B, and C. Which of the gels produced by electrophoresis shown below best represents the separation and identity of these fragments? A) B) C) D) E) Answer: B 13.3 Scenario Questions 1) For a science fair project, two students decided to repeat the Hershey and Chase experiment, with modifications. They decided to label the nitrogen of the DNA, rather than the phosphate. They reasoned that each nucleotide has only one phosphate and two to five nitrogens. Thus, labeling the nitrogens would provide a stronger signal than labeling the phosphates. Why won't this experiment work? A) There is no radioactive isotope of nitrogen. B) Radioactive nitrogen has a half-life of 100,000 years, and the material would be too dangerous for too long. C) Avery et al. have already concluded that this experiment showed inconclusive results. D) Although there are more nitrogens in a nucleotide, labeled phosphates actually have 16 extra neutrons; therefore, they are more radioactive. E) Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins. Answer: E 2) You briefly expose bacteria undergoing DNA replication to radioactively labeled nucleotides. When you centrifuge the DNA isolated from the bacteria, the DNA separates into two classes. One class of labeled DNA includes very large molecules (thousands or even millions of nucleotides long), and the other includes short stretches of DNA (several hundred to a few thousand nucleotides in length). These two classes of DNA probably represent A) leading strands and Okazaki fragments. B) lagging strands and Okazaki fragments. C) Okazaki fragments and RNA primers. D) leading strands and RNA primers. E) RNA primers and mitochondrial DNA. Answer: A Use the following information to answer the next few questions. A eukaryotic gene has "sticky ends" produced by the restriction endonuclease EcoRI. The gene is added to a mixture containing EcoRI and a bacterial plasmid that carries two genes conferring resistance to ampicillin and tetracycline. The plasmid has one recognition site for EcoRI located in the tetracycline resistance gene. This mixture is incubated for several hours, exposed to DNA ligase, and then added to bacteria growing in nutrient broth. The bacteria are allowed to grow overnight and are streaked on a plate using a technique that produces isolated colonies that are clones of the original. Samples of these colonies are then grown in four different media: nutrient broth plus ampicillin, nutrient broth plus tetracycline, nutrient broth plus ampicillin and tetracycline, and nutrient broth without antibiotics. 3) Bacteria that contain the plasmid, but not the eukaryotic gene, would grow A) in the nutrient broth plus ampicillin, but not in the broth containing tetracycline. B) only in the broth containing both antibiotics. C) in the broth containing tetracycline, but not in the broth containing ampicillin. D) in all four types of broth. E) in the nutrient broth without antibiotics only. Answer: D 4) Bacteria containing a plasmid into which the eukaryotic gene has integrated would grow A) in the nutrient broth only. B) in the nutrient broth and the tetracycline broth only. C) in the nutrient broth, the ampicillin broth, and the tetracycline broth. D) in all four types of broth. E) in the ampicillin broth and the nutrient broth. Answer: E 5) Bacteria that do not take up any plasmids would grow on which media? A) the nutrient broth only B) the nutrient broth and the tetracycline broth C) the nutrient broth and the ampicillin broth D) the tetracycline broth and the ampicillin broth E) all three broths Answer: A Use the following information to answer the next few questions. A group of six students has taken samples of their own cheek cells, purified the DNA, and used a restriction enzyme known to cut at zero, one, or two sites in a particular gene of interest. 6) Why might they be conducting such an experiment? A) to find the location of this gene in the human genome B) to prepare to isolate the chromosome on which the gene of interest is found C) to find which of the students has which alleles D) to collect population data that can be used to assess natural selection E) to collect population data that can be used to study genetic drift Answer: C 7) Analysis of the data obtained shows that two students each have two fragments, two students each have three fragments, and two students each have one only. What does this demonstrate? A) Each pair of students has a different gene for this function. B) The two students who have two fragments have one restriction site in this region. C) The two students who have two fragments have two restriction sites within this gene. D) The students with three fragments are said to have "fragile sites." E) Each of these students is heterozygous for this gene. Answer: B 13.4 End-of-Chapter Questions 1) In his work with pneumonia-causing bacteria and mice, Griffith found that A) the protein coat from pathogenic cells was able to transform nonpathogenic cells. B) heat-killed pathogenic cells caused pneumonia. C) some substance from pathogenic cells was transferred to nonpathogenic cells, making them pathogenic. D) the polysaccharide coat of bacteria caused pneumonia. E) bacteriophages injected DNA into bacteria Answer: C 2) What is the basis for the difference in how the leading and lagging strands of DNA molecules are synthesized? A) The origins of replication occur only at the 5' end. B) Helicases and single-strand binding proteins work at the 5' end. C) DNA polymerase can join new nucleotides only to the 3' end of a growing strand. D) DNA ligase works only in the 3' → 5' direction. E) Polymerase can work on only one strand at a time. Answer: C 3) In analyzing the number of different bases in a DNA sample, which result would be consistent with the base-pairing rules? A) A = G B) A + G = C + T C) A + T = G + T D) A = C E) G = T Answer: B 4) The elongation of the leading strand during DNA synthesis A) progresses away from the replication fork. B) occurs in the 3' → 5' direction. C) produces Okazaki fragments. D) depends on the action of DNA polymerase. E) does not require a template strand. Answer: D 5) In a nucleosome, the DNA is wrapped around A) polymerase molecules. B) ribosomes. C) histones. D) a thymine dimer. E) satellite DNA. Answer: C 6) Which of the following sequences in double-stranded DNA is most likely to be recognized as a cutting site for a restriction enzyme A) B) C) D) E) Answer: C 7) E. coli cells grown on 15N medium are transferred to 14N medium and allowed to grow for two more generations (two rounds of DNA replication). DNA extracted from these cells is centrifuged. What density distribution of DNA would you expect in this experiment? A) one high-density and one low-density band B) one intermediate-density band C) one high-density and one intermediate-density band D) one low-density and one intermediate-density band E) one low-density band Answer: D 8) A biochemist isolates, purifies, and combines in a test tube a variety of molecules needed for DNA replication. When she adds some DNA to the mixture, replication occurs, but each DNA molecule consists of a normal strand paired with numerous segments of DNA a few hundred nucleotides long. What has she probably left out of the mixture? A) DNA polymerase B) DNA ligase C) nucleotides D) Okazaki fragments E) primase Answer: B 9) The spontaneous loss of amino groups from adenine in DNA results in hypoxanthine, an uncommon base, opposite thymine. What combination of proteins could repair such damage? A) nuclease, DNA polymerase, DNA ligase B) topoisomerase, primase, DNA polymerase C) topoisomerase, helicase, single-strand binding protein D) DNA ligase, replication fork proteins, adenylyl cyclase E) nuclease, topoisomerase, primase Answer: A Test Bank for Campbell Biology in Focus Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, Robert B. Jackson, Jane B. Reece 9780321813664, 9780321962751, 9780134710679

Document Details

Related Documents

person
Charlotte Scott View profile
Close

Send listing report

highlight_off

You already reported this listing

The report is private and won't be shared with the owner

rotate_right
Close
rotate_right
Close

Send Message

image
Close

My favorites

image
Close

Application Form

image
Notifications visibility rotate_right Clear all Close close
image
image
arrow_left
arrow_right