Molecular Biology: Principles And Practice, Second Edition Test Bank

Molecular Biology- Principles and Practice 2e Cox et. al Chapter 2 1. Mendel’s success in formulating his fundamental principles of inheritance can be attributed to which of the following? A. B. C. D. E. Mendel focused on the overall appearance of the plant rather than on individual traits. Mendel focused on individual traits of the plant rather than on the overall appearance. Mendel chose to study complex traits that result from interactions between multiple genes. Mendel used an organism that grew slowly, therefore having long generation times. Mendel used fruit flies to study inheritance. Answer: B Section: 2.1 Level: Easy Blooms: Comprehension 2. If a pea plant strain is true-breeding for a particular trait, then: A. B. C. D. E. crosses between two true-breeding plants with the same trait will have offspring identical to the parents. it will be homozygous for all genes. it must be a recessive trait. it is not useful for studying inheritance. it only self-fertilizes. Answer: A Section: 2.1 Level: Easy Blooms: Comprehension 3. True-breeding, purple-flowered pea plants were crossed with true-breeding, white-flowered pea plants. The F1 progeny all had purple flowers. If the F1 offspring are self-fertilized, what progeny phenotypes would be expected? A. B. C. D. E. 1 purple : 2 pink : 1 white 3 white : 1 purple 3 purple : 1 white all purple all white Answer: C Section: 2.1 Level: Easy Blooms: Application 4. A cross of pea plants, both heterozygous for seed shape and color traits, resulted in progeny in the ratio 9/16 round yellow (R_Y_), 3/16 round green (R_yy), 3/16 wrinkled yellow (rrY_), and 1/16 wrinkled green (rryy). This result demonstrates that the: A. B. C. D. E. alleles responsible for seed color are segregating but those responsible for shape are not. alleles of seed shape and color are segregating independently. phenotypes corresponding to seed shape are segregating 1:2:1. alleles governing seed color display codominance. alleles for seed shape and color are linked. Answer: B Section: 2.1 Level: Medium Blooms: Comprehension 5. True-breeding pea plants with inflated, yellow pods were crossed with true-breeding plants with constricted, green pods. The F1 progeny all had inflated, green pods. When the F1 is allowed to self-fertilize, what fraction of the F2 progeny will have constricted, yellow pods? A. B. C. D. E. 1/4 3/4 1/16 3/16 9/16 Answer: C Section: 2.1 Level: Medium Blooms: Application 6. True-breeding tall snapdragons with red flowers were crossed with short snapdragons with white flowers. The resulting F1 progeny were all of medium height and had pink flowers. When the F1 progeny are crossed with each other, what proportion of the F2 progeny would be of medium height and have pink flowers? A. B. C. D. E. 1/4 1/2 1/8 1/16 3/4 Answer: A Section: 2.1 Level: Medium Blooms: Application 7. Which of the following traits will show Mendelian behavior? A. B. C. D. E. dominance semidominance codominance epistasis linkage Answer: A Section: 2.1 Level: Easy Blooms: Knowledge AA AB 8. The child of a mother with I I blood and father with I I blood will have: AA A. 25% chance of being I I . A A. B. 50% chance of being I I AA C. 75% chance of being I I . AB D. 25% chance of being I I . BB E. 50% chance of being I I . Answer: B Section: 2.1 Level: Medium Blooms: Application 9. A fruit fly homozygous for wild-type eyes and straight wings was crossed with a fly homozygous for rosy eyes and wrinkled wings. The progeny all showed wild-type eyes and straight wings. When crossing the F 1, the resulting offspring showed 3 wild-type eyes/straight wings : 1 rosy eyes/wrinkled wings. This data suggest: A. rosy eyes is dominant. B. straight wings is recessive. C. these two genes are linked. D. these two genes are sex-linked. E. these two genes are codominant. Answer: C Section: 2.1 Level: Hard Blooms: Analysis 10. Using the great lubber grasshopper as a model, Walter Sutton observed: A. separation of maternal and paternal chromosomes during mitosis. B. replication of chromosomes during meiosis. C. crossing over of homologous chromosomes. D. pairing of homologous chromosomes during mitosis. E. pairing of homologous chromosomes during meiosis. Answer: E Section: 2.1 Level: Medium Blooms: Comprehension 11. Photographs of cells from onion root tips are often used to show chromosomes primarily because these cells are: A. diploid. B. rapidly undergoing mitosis. C. rapidly undergoing meiosis. D. in stationary phase. E. lacking organelles. Answer: B Section: 2.2 Level: Medium Blooms: Application 12. Tetrads are: A. B. C. D. E. composed of a pair of nonhomologous chromosomes. composed of two pairs of sister chromatids. formed in prophase II of meiosis. lined up along the cell axis during metaphase II. formed during DNA synthesis. Answer: B Section: 2.2 Level: Easy Blooms: Knowledge 13. The pair of sister-chromatids, formed by replication of a chromosome, are held together at a specific region of the chromosome called the: A. B. C. D. E. replication fork. telomere. centromere. centriole. origin. Answer: C Section: 2.2 Level: Easy Blooms: Knowledge 14. Differentiated cells, such as fat cells, have acquired their specialized function and remain in: A. B. C. D. E. S phase. G1 phase. G2 phase. M phase. G0 phase. Answer: E Section: 2.2 Level: Easy Blooms: Knowledge 15. The mitotic stage in which the chromosomes condense, nuclei disappear, and the mitotic spindle begins to form is known as: A. B. C. D. E. prophase. metaphase. anaphase. telophase. interphase. Answer: A Section: 2.2 Level: Easy Blooms: Knowledge 16. What technical advance allowed Walther Flemming and other scientists of his generation to better understand the stages of mitosis and meiosis? A. B. C. D. E. the ability to obtain single cells the ability to slice cells thinly microscopes with two eyepieces microscopes with high-quality oil immersion lens electron microscopes Answer: D Section: 2.2 Level: Medium Blooms: Comprehension 17. The mitotic stage in which the chromosomes decondense and the nuclear membrane reforms is: A. B. C. D. E. prophase. metaphase. anaphase. telophase. interphase. Answer: D Section: 2.2 Level: Easy Blooms: Knowledge 18. The centromeres of sister chromatids uncouple and the chromatids separate in what meiotic phase? A. B. C. D. E. prophase I metaphase II anaphase I telophase II anaphase II Answer: E Section: 2.2 Level: Easy Blooms: Knowledge 19. The primary result of meiosis is: A. B. C. D. E. gamete formation. two daughter cells from one parental cell. four daughter cells from one parental cell. gamete formation, two from one parental cell. gamete formation, four from one parental cell. Answer: E Section: 2.2 Level: Easy Blooms: Knowledge 20. In meiosis, the complexity of DNA (amount of information encoded) per cell is essentially halved as compared to the original cell. This reduction in information occurs immediately after: A. B. C. D. metaphase I. metaphase II. telophase I. telophase II. Answer: C Section: 2.2 Level: Medium Blooms: Analysis 21. In the ZW system of sex determination (as in birds), a male is _____________ while a female is ______________. A. ZZ : ZW B. ZW : ZZ C. Z0 : ZW D. Z : ZW Answer: A Section: 2.2 Level: Medium Blooms: Knowledge 22. The wild-type allele of a gene is: A. dominant to other alleles. B. recessive to other alleles. C. found at the lowest frequency in a population. D. found at the greatest frequency in a population. Answer: D Section: 2.3 Level: Easy Blooms: Knowledge 23. Genes are said to be linked when: A. B. C. D. E. they lie on the same chromosome. they lie on the same chromosome and are greater than 50 map units apart. they lie on the same chromosome and are less than 50 map units apart. they are alleles of the same gene. None of these choices are correct. Answer: C Section: 2.3 Level: Medium Blooms: Comprehension 24. What observation by Thomas Hunt Morgan led to the understanding that genes were on chromosomes? A. The white-eyed trait was recessive and disappeared in the F1 generation. B. In a cross of a red-eyed female and a white-eyed male, only females in the F2 generation were white-eyed. C. The inheritance of the white-eyed trait in fruit flies followed the same pattern of inheritance as that of the X chromosome. D. The inheritance of the white-eyed trait in fruit flies followed the same pattern of inheritance as that of the Y chromosome. E. Inheritance of the white-eyed trait was the same for female and male flies. Answer: C Section: 2.3 Level: Hard Blooms: Comprehension 25. Calvin Bridges, an associate of Thomas Hunt Morgan’s, found further evidence that genes were located on chromosomes. Which of the following is not a component of his study? w w W A. He crossed white-eyed females (X X ) with red-eyed males (X Y). B. The F1 progeny were mostly the red-eyed males and white-eyed females expected. C. A few rare white-eyed females and red-eyed males were observed, which he called primary exceptionals. D. He proposed that the rare white-eyed females were the result of abnormal chromosome number. E. Abnormal chromosome number was a result of nondisjunction in the female fly. Answer: B Section: 2.3 Level: Medium Blooms: Comprehension 26. The chromosomal abnormality Calvin Bridges observed as a characteristic of “primary exceptionals” is a result of: A. mutation. B. nondisjunction. C. chiasmata. D. recombination. E. linkage. Answer: B Section: 2.3 Level: Medium Blooms: Comprehension 27. The test cross ab/ab × AB/ab is performed. The following number of progeny of each genotype are obtained: 38% AaBb, 12% Aabb, 12% aaBb, 38% aabb. What is the distance (in map units) between the two genes in question? A. B. C. D. E. 38 centimorgans (cM) 24 centimorgans (cM) 19 centimorgans (cM) 12 centimorgans (cM) 6 centimorgans (cM) Answer: D Section: 2.3 Level: Hard Blooms: Analysis 28. Five two-point crosses involving the four linked genes A, B, C, and D were performed in yeast with the following results: AB × ab: 41% AB 39% ab 9% Ab 11% aB AC × ac: 45% AC 48% ac 4% Ac 4% aC BC × bc: 44% BC 44% bc 6% Bc 6% bC AD × ad: 48% AD 46% ad 3% Ad 3% aD BD × bd: 37% BD 37% bd 12% bD 13% Bd Which is the best map for these four genes? A. B. C. D. E. B 6mu A 4mu C 3mu B 12mu C 8mu A 6mu D 3mu A 4mu C 6mu A 4mu C 6mu B 12mu The data suggest that the genes are not linked. Answer: C Section: 2.3 Level: Hard Blooms: Analysis D D B D 29. Oswald Avery found that DNA was necessary and sufficient for bacterial transformation. Which of the following experimental results support this statement (select all that apply)? A. B. C. D. Removal of DNA from virulent Streptococcus pneumoniae cell extracts eliminated transformation. Removal of RNA from virulent Streptococcus pneumoniae cell extracts eliminated transformation. Protein alone will cause bacterial transformation to occur. DNA alone will cause bacterial transformation to occur. Answer: A, D Section: 2.4 Level: Medium Blooms: Comprehension 30. In the following pathway, mutation of enzyme II would result in the accumulation of ______________ and the mutant would require _____________ in the media in order to grow. A. B. C. D. E. Citrulline : arginine Ornithine : citrulline Precursor : ornithine Ornithine : precursor Arginine : citrulline Answer: B Section: 2.4 Level: Easy Blooms: Comprehension 31. Which of the following has the information flow of the central dogma in the correct order? A. B. C. D. E. DNA, translation, RNA, replication, protein DNA, transcription, RNA, replication, protein RNA, translation, DNA, transcription, protein DNA, transcription, RNA, translation, protein RNA, transcription, DNA, translation, protein Answer: D Section: 2.4 Level: Easy Blooms: Knowledge 32. When considering alleles of a gene, DNA is to genotype as ______________ is/are to phenotype. A. RNA B. C. D. E. mutation proteins lipids nucleotides Answer: C Section: 2.4 Level: Easy Blooms: Knowledge 33. In the early 1900s, determining the chemical basis of heredity was difficult because chromosomes are composed of: A. B. C. D. E. DNA. RNA. protein. DNA and protein. DNA, RNA, and protein. Answer: D Section: 2.4 Level: Medium Blooms: Comprehension 34. If protein had been the genetic material, then what would Oswald Avery have observed in his experiments? 1. 2. 3. 4. Bacterial extracts treated with proteinase would transform nonvirulent bacteria into the virulent strain. Bacterial extracts treated with proteinase would not transform nonvirulent bacteria into the virulent strain. Bacterial extracts treated with DNAse would transform nonvirulent bacteria into the virulent strain. Bacterial extracts treated with DNAse would not transform nonvirulent bacteria into the virulent strain. A. B. C. D. E. Statement 1 Statement 2 Statements 1 and 3 Statements 2 and 3 Statements 2 and 4 Answer. D Section: 2.4 Level: Hard Blooms: Synthesis 35. Archibald Garrod’s evaluations of alkaptonuria (black urine in patients) led him to conclude: A. B. C. D. a mutation can result in a nonfunctional enzyme. enzymes are required for biochemical pathways. this defect is sex-linked and recessive. disjunction can result in a nonfunctional enzyme. Answer:. A Section: 2.4 Level: Medium Blooms: Comprehension 36. George Beadle and Edward Tatum used _____________as a model organisms and created mutants using ______________. A. Drosophila melanogaster : irradiation B. Escherichia coli : irradiation C. Neurospora crassa : irradiation D. Drosophila melanogaster : carcinogenic chemicals E. Neurospora crassa : carcinogenic chemicals Answer. C Section: 2.4 Level: Medium Blooms: Knowledge 37. George Beadle and Edward Tatum used auxotrophs to determine their “one gene–one enzyme” hypothesis. An auxotroph is best described as: A. a defective enzyme. B. a result of nondisjunction. C. a mutant unable to grow on minimal media. D. a wild-type version of a model organism. E. an organism with a dominant lethal mutation. Answer. C Section: 2.4 Level: Medium Blooms: Knowledge 38. The “one gene–one enzyme” hypothesis of George Beadle and Edward Tatum was an oversimplification because: A. mutations may not completely inactivate an enzyme. B. some enzymes contain more than one polypeptide. C. some enzymes can be phosphorylated. D. auxotrophs can be corrected. E. some RNAs are independently catalytic. Answer. B Section: 2.4 Level: Hard Blooms: Comprehension 39. George Beadle and Edward Tatum created a collection of mutants to study the following pathway. A mutant shows no growth on minimal media; however it shows growth if citrulline is provided. These results indicate the mutation would most likely be in the gene for: A. Enzyme I. B. Enzyme II. C. Enzyme III D. either Enzyme I or Enzyme II. E. either Enzyme II or Enzyme III. Answer. D Section: 2.4 Level: Hard Blooms: Analysis 40. rRNA is (select all that apply): A. associated with ribosomes. B. the least abundant RNA in the cell. C. the most abundant RNA in the cell. D. relatively consistent in sequence between organisms. E. considerably variable in sequence between organisms. Answer: A, C, D Section: 2.4 Level: Hard Blooms: Synthesis 41. Crick’s proposed “adaptor molecule” is now known to be: A. mRNA. B. rRNA. C. tRNA. D. imRNA. Answer: C Section: 2.4 Level: Medium Blooms: Synthesis 42. A mutation in a tRNA gene could potentially decrease the ability of that tRNA to (select all that apply): A. complex with ribosomes. B. bind to an amino acid. C. bind multiple amino acids simultaneously. D. control DNA replication. E. code for protein. Answer: A, B Section: 2.4 Level: Hard Blooms: Analysis 43. Epigenetic inheritance may depend on all the following except: A. the DNA sequence from a parent. B. inherited chemical modification of parental nucleotides. C. inherited nucleosome structure of chromosome. D. inherited chemical modification of parental proteins. Answer: A Section: 2.4 Level: Hard Blooms: Analysis 44. Hemophilia in humans results from an X-linked recessive gene. Which of these statements regarding the inheritance of this trait is true? A. Heterozygous females will exhibit a milder form of hemophilia. B. All daughters of an affected mother will have hemophilia. C. Carrier females will always pass the hemophilia allele to their sons. D. Males who inherit the recessive allele from their fathers will exhibit hemophilia. E. All sons of an affected mother will have hemophilia. Answer: E Section: 2.4 Level: Medium Blooms: Analysis 45. A woman with apparently normal health (mother) and a colorblind man (father) had a son with hemophilia and a normal daughter. If the genes for both traits are on the X chromosome, which statement about this family is true? A. B. C. D. E. The maternal grandfather is colorblind. The daughter cannot be a carrier for colorblindness. The maternal grandfather is not hemophilic. There is a 50% chance that the maternal grandfather is hemophilic. The father is a carrier of hemophilia. Answer: D Section: 2.4 Level: Hard Blooms: Analysis 46. A patient in his forties is diagnosed with Huntington disease. The patient’s mother is healthy and a genetic evaluation shows 25 and 26 CAG repeats within her HTT alleles. The patient’s father died at age 24 when the patient was 2 and no genetic evaluation can be done. It is most likely that: A. B. C. D. the father was heterozygous for the defective HTT allele. the father was homozygous for the defective HTT allele. the mother is a carrier for the defective HTT allele. the HTT gene acquired additional CAG repeats during the 40 years of the patient’s life. Answer: A Section: 2.4 Level: Hard Blooms: Synthesis 47. A newborn is identified as having cystic fibrosis. While both parents are healthy, a genetic screening of both parents will most likely reveal: A. B. C. D. the father is a carrier of the mutant CF allele. the mother is a carrier of the mutant CF allele. both mother and father are carriers of a mutant CF allele. neither mother nor father are carriers; spontaneous mutation caused the baby’s disease. Answer: C Section: 2.4 Level: Hard Blooms: Evaluation 48. Cystic fibrosis is most often caused by the CFTR∆F508 mutation. This mutation: A. reduces the function of a chloride ion channel. B. increases ATP binding to chloride. C. is autosomal dominant. D. causes symptoms in carriers. Answer: A Section: 2.4 Level: Easy Blooms: Knowledge 49. A newborn is identified as having sickle-cell anemia. While both parents are healthy, a screening of both parents will most likely reveal: A. both mother and father are carriers of a mutant α chain gene. B. only one parent has a mutant α chain gene since mutation is dominant. C. both mother and father are carriers of a mutant β chain gene. D. only one parent has a mutant β chain gene since mutation is dominant. E. both parents carry a mutant hemoglobin gene and are highly susceptible to malaria. Answer: C Section: 2.4 Level: Medium Blooms: Application 50. In order to prove genetic recombination occurred through material (DNA) exchange between homologous maize (corn) chromosomes, Harriet Creighton and Barbara McClintock: A. stained individual chromosomes to see banding, and then observed segregation of bands. B. observed specific morphologic features of one chromosome being “traded” with its homolog. C. created mutations of chromosomes via irradiation and observed transfer of damaged DNA between homologs. 32 D. labeled one chromosome with P and observed the transfer of radioactivity to the homolog. Answer: B Section: 2.4 Level: Hard Blooms: Comprehension