Genetics Practice Test - Chapters 14-17 PDF

Genetics Practice Test - Chapters 14-17 Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. What is the difference between a monohybrid cross and a dihybrid cross? a. A monohybrid cross involves a single parent, whereas a dihybrid cr...

Genetics Practice Test - Chapters 14-17 Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. What is the difference between a monohybrid cross and a dihybrid cross? a. A monohybrid cross involves a single parent, whereas a dihybrid cross involves two parents. b. A monohybrid cross produces a single progeny, whereas a dihybrid cross produces two progeny. c. A dihybrid cross involves organisms that are heterozygous for two characters and a monohybrid only one. d. A monohybrid cross is performed for one generation, whereas a dihybrid cross is performed for two generations. e. A monohybrid cross results in a 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1 ratio. ____ 2. The offspring of Mendel's classic pea cross always looked like one of the two parental varieties because a. one phenotype was completely dominant over another. b. each allele affected phenotypic expression. c. the traits blended together during fertilization. d. no genes interacted to produce the parental phenotype. e. different genes interacted to produce the parental phenotype. ____ 3. How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE? a. 4 b. 8 c. 16 d. 32 e. 64 ____ 4. It was important that Mendel examined not just the generation in his breeding experiments, but the generation as well, because a. he obtained very few progeny, making statistical analysis difficult. b. parental traits that were not observed in the reappeared in the. c. analysis of the progeny would have allowed him to discover the law of segregation, but not the law of independent assortment. d. the dominant phenotypes were visible in the generation, but not in the. e. many of the progeny died. Use Figure 14.1 and the following description to answer the questions below. In a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A true-breeding dark-leaved plant is crossed with a light-leaved one, and the offspring is allowed to self-pollinate. The predicted outcome of the is diagrammed in the Punnett square shown in Figure 14.1, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square. Figure 14.1 ____ 5. Which of the boxes correspond to plants with a heterozygous genotype? a. 1 b. 1 and 2 c. 1, 2, and 3 d. 2 and 3 e. 2, 3, and 4 ____ 6. Which of the following about the law of segregation is false? a. It states that each of two alleles for a given trait segregate into different gametes. b. It can be explained by the segregation of homologous chromosomes during meiosis. c. It can account for the 3:1 ratio seen in the generation of Mendel's crosses. d. It can be used to predict the likelihood of transmission of certain genetic diseases within families. e. It is a method that can be used to determine the number of chromosomes in a plant. ____ 7. The fact that all seven of the pea plant traits studied by Mendel obeyed the principle of independent assortment most probably indicates which of the following? a. None of the traits obeyed the law of segregation. b. The diploid number of chromosomes in the pea plants was 7. c. All of the genes controlling the traits were located on the same chromosome. d. All of the genes controlling the traits behaved as if they were on different chromosomes. e. The formation of gametes in plants occurs by mitosis only. ____ 8. Black fur in mice (B) is dominant to brown fur (b). Short tails (T) are dominant to long tails (t). What fraction of the progeny of the cross BbTt × BBtt will have black fur and long tails? a. 1/16 b. 3/16 c. 3/8 d. 1/2 e. 9/16 ____ 9. Two true-breeding stocks of pea plants are crossed. One parent has red, axial flowers and the other has white, terminal flowers; all individuals have red, axial flowers. The genes for flower color and location assort independently. If 1,000 offspring resulted from the cross, approximately how many of them would you expect to have red, terminal flowers? a. 65 b. 190 c. 250 d. 565 e. 750  Refer to the following to answer the questions below. Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether cactuses have spines. Homozygous recessive nn cactuses have no spines at all. ____ 10. A cross between a true-breeding sharp-spined cactus and a spineless cactus would produce a. all sharp-spined progeny. b. 50% sharp-spined, 50% dull-spined progeny. c. 25% sharp-spined, 50% dull-spined, 25% spineless progeny d. all spineless progeny. e. It is impossible to determine the phenotypes of the progeny. ____ 11. Cystic fibrosis affects the lungs, the pancreas, the digestive system, and other organs, resulting in symptoms ranging from breathing difficulties to recurrent infections. Which of the following terms best describes this? a. Incomplete dominance b. Multiple alleles c. Pleiotropy d. Epistasis ____ 12. Which of the following is an example of polygenic inheritance? a. Pink flowers in snapdragons b. The ABO blood groups in humans c. Huntington's disease in humans d. White and purple flower color in peas e. Skin pigmentation in humans  The following questions refer to the pedigree chart in Figure 14.2 for a family, some of whose members exhibit the dominant trait, wooly hair. Affected individuals are indicated by an open square or circle. Figure 14.2 ____ 13. What is the likelihood that the progeny of IV-3 and IV-4 will have wooly hair? a. 0% b. 25% c. 50% d. 75% e. 100% ____ 14. When Thomas Hunt Morgan crossed his red-eyed generation flies to each other, the 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 X chromosome. b. The gene involved is on the Y chromosome. c. The gene involved is on an autosome. d. Other male-specific factors influence eye color in flies. e. Other female-specific factors influence eye color in flies. ____ 15. SRY is best described in which of the following ways? a. A gene region present on the Y chromosome that triggers male development b. A gene present on the X chromosome that triggers female development c. An autosomal gene that is required for the expression of genes on the Y chromosome d. An autosomal gene that is required for the expression of genes on the X chromosome e. Required for development, and males or females lacking the gene do not survive past early childhood ____ 16. A 0.1% frequency of recombination is observed a. only in sex chromosomes. b. only on genetic maps of viral chromosomes. c. on unlinked chromosomes. d. in any two genes on different chromosomes. e. in genes located very close to one another on the same chromosome. ____ 17. What is the mechanism for the production of genetic recombinants? a. X inactivation b. Methylation of cytosine c. Crossing over and independent assortment d. Nondisjunction e. Deletions and duplications during meiosis Refer to Figure 15.2 to answer the following questions. Figure 15.2 ____ 18. In a series of mapping experiments, the recombination frequencies for four different linked genes of Drosophila were determined as shown in the figure. 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 ____ 19. If a human interphase nucleus contains three Barr bodies, it can be assumed that the person a. has hemophilia. b. is a male. c. has four X chromosomes. d. has Turner syndrome. e. has Down syndrome. ____ 20. If a pair of homologous chromosomes fails to separate during anaphase of meiosis I, what will be the chromosome number of the four resulting gametes with respect to the normal haploid number (n)? a. n + 1; n + 1; n - 1; n - 1 b. n + 1; n - 1; n; n c. n + 1; n - 1; n - 1; n - 1 d. n + 1; n + 1; n; n e. n - 1; n - 1; n; n ____ 21. One possible result of chromosomal breakage is for a fragment to join a nonhomologous chromosome. What is this alteration called? a. Deletion b. Disjunction c. Inversion d. Translocation e. Duplication ____ 22. In order for chromosomes to undergo inversion or translocation, which of the following is required? a. Point mutation b. Immunological insufficiency c. Advanced maternal age d. Chromosome breakage and rejoining e. Meiosis ____ 23. 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. ____ 24. 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, but protein does not. c. DNA contains nitrogen, whereas protein does not. d. DNA contains purines, whereas protein includes pyrimidines. e. RNA includes ribose, while DNA includes deoxyribose sugars. ____ 25. Chargaff's analysis of the relative base composition of DNA was significant because he was able to show that a. the relative proportion of each of the four bases differs within individuals of a species. b. the human genome is more complex than that of other species. c. the amount of A is always equivalent to T, and C to G. d. the amount of ribose is always equivalent to deoxyribose. e. transformation causes protein to be brought into the cell. ____ 26. Why does the DNA double helix have a uniform diameter? a. Purines pair with pyrimidines. b. C nucleotides pair with A nucleotides. c. Deoxyribose sugars bind with ribose sugars. d. Nucleotides bind with nucleosides. e. Nucleotides bind with nucleoside triphosphates. ____ 27. 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 ____ 28. The enzyme telomerase solves the problem of replication at the ends of linear chromosomes by which method? a. adding a single 5' cap structure that resists degradation by nucleases b. causing specific double strand DNA breaks that result in blunt ends on both strands c. causing linear ends of the newly replicated DNA to circularize d. adding numerous short DNA sequences such as TTAGGG, which form a hairpin turn e. adding numerous GC pairs which resist hydrolysis and maintain chromosome integrity ____ 29. The DNA of telomeres has been found to be highly conserved throughout the evolution of eukaryotes. What does this most probably reflect? a. the inactivity of this DNA b. the low frequency of mutations occurring in this DNA c. that new evolution of telomeres continues d. that mutations in telomeres are relatively advantageous e. that the critical function of telomeres must be maintained ____ 30. At a specific area of a chromosome, the sequence of nucleotides below is present where the chain opens to form a replication fork: 3' C C T A G 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' ____ 31. Which of the following separates the DNA strands during replication? a. helicase b. DNA polymerase III c. ligase d. DNA polymerase I e. primase ____ 32. 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. ____ 33. What is the function of topoisomerase? a. relieving strain in the DNA ahead of the replication fork b. elongation of new DNA at a replication fork by addition of nucleotides to the existing chain c. the addition of methyl groups to bases of DNA d. unwinding of the double helix e. stabilizing single-stranded DNA at the replication fork ____ 34. What is the role of DNA ligase in the elongation of the lagging strand during DNA replication? a. synthesize RNA nucleotides to make a primer b. catalyze the lengthening of telomeres c. join Okazaki fragments together d. unwind the parental double helix e. stabilize the unwound parental DNA ____ 35. Which of the following help 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 ____ 36. Which would you expect of a eukaryotic cell lacking telomerase? a. a high probability of becoming cancerous b. production of Okazaki fragments c. inability to repair thymine dimers d. a reduction in chromosome length e. high sensitivity to sunlight ____ 37. Which of the following statements describes histones? a. Each nucleosome consists of two molecules of histone H1. b. Histone H1 is not present in the nucleosome bead; instead it is involved in the formation of higher-level chromatin structures. 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. e. The mass of histone in chromatin is approximately nine times the mass of DNA. ____ 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. ____ 39. The nitrogenous base adenine is found in all members of which group? a. proteins, triglycerides, and testosterone b. proteins, ATP, and DNA c. ATP, RNA, and DNA d. alpha glucose, ATP, and DNA e. proteins, carbohydrates, and ATP  The following questions refer to Figure 17.2, a table of codons. Figure 17.2 ____ 40. A possible sequence of nucleotides in the template strand of DNA that would code for the polypeptide sequence phe-leu-ile-val would be a. 5' TTG-CTA-CAG-TAG 3'. b. 3' AAC-GAC-GUC-AUA 5'. c. 5' AUG-CTG-CAG-TAT 3'. d. 3' AAA-AAT-ATA-ACA 5'. e. 3' AAA-GAA-TAA-CAA 5'. ____ 41. What is the sequence of a peptide based on the following mRNA sequence? 5'... UUUUCUUAUUGUCUU 3' a. leu-cys-tyr-ser-phe b. cyc-phe-tyr-cys-leu c. phe-leu-ile-met-val d. leu-pro-asp-lys-gly e. phe-ser-tyr-cys-leu ____ 42. A transcription unit that is 8,000 nucleotides long may use 1,200 nucleotides to make a protein consisting of approximately 400 amino acids. This is best explained by the fact that a. many noncoding stretches of nucleotides are present in mRNA. b. there is redundancy and ambiguity in the genetic code. c. many nucleotides are needed to code for each amino acid. d. nucleotides break off and are lost during the transcription process. e. there are termination exons near the beginning of mRNA. ____ 43. Once transcribed, eukaryotic mRNA typically undergoes substantial alteration that includes a. union with ribosomes. b. fusion into circular forms known as plasmids. c. linkage to histone molecules. d. excision of introns. e. fusion with other newly transcribed mRNA. ____ 44. Introns are significant to biological evolution because a. their presence allows exons to be shuffled. b. they protect the mRNA from degeneration. c. they are translated into essential amino acids. d. they maintain the genetic code by preventing incorrect DNA base pairings. e. they correct enzymatic alterations of DNA bases. ____ 45. Which of the following is (are) true of snRNPs? a. They are made up of both DNA and RNA. b. They bind to splice sites at each end of the exon. c. They join together to form a large structure called the spliceosome. d. They act only in the cytosol. e. They attach introns to exons in the correct order. ____ 46. Each eukaryotic mRNA, even after post-transcriptional modification, includes 5' and 3' UTRs. Which are these? a. the cap and tail at each end of the mRNA b. the untranslated regions at either end of the coding sequence c. the U attachment sites for the tRNAs d. the U translation sites that signal the beginning of translation e. the U — A pairs that are found in high frequency at the ends  Figure 17.4 ____ 47. Figure 17.4 represents tRNA that recognizes and binds a particular amino acid (in this instance, phenylalanine). Which codon on the mRNA strand codes for this amino acid? a. UGG b. GUG c. GUA d. UUC e. CAU ____ 48. What is the most abundant type of RNA? a. mRNA b. tRNA c. rRNA d. pre-mRNA e. hnRNA ____ 49. Choose the answer that has these events of protein synthesis in the proper sequence. 1. An aminoacyl-tRNA binds to the A site. 2. A peptide bond forms between the new amino acid and a polypeptide chain. 3. tRNA leaves the P site, and the P site remains vacant. 4. A small ribosomal subunit binds with mRNA. 5. tRNA translocates to the P site. a. 1, 3, 2, 4, 5 b. 4, 1, 2, 5, 3 c. 5, 4, 3, 2, 1 d. 4, 1, 3, 2, 5 e. 2, 4, 5, 1, 3 ____ 50. As a ribosome translocates along an mRNA molecule by one codon, which of the following occurs? a. The tRNA that was in the A site moves into the P site. b. The tRNA that was in the P site moves into the A site. c. The tRNA that was in the A site moves to the E site and is released. d. The tRNA that was in the A site departs from the ribosome via a tunnel. e. The polypeptide enters the E site. Use the following information to answer the following questions. A transfer RNA (#1) attached to the amino acid lysine enters the ribosome. The lysine binds to the growing polypeptide on the other tRNA (#2) in the ribosome already. ____ 51. When the ribosome reaches a stop codon on the mRNA, no corresponding tRNA enters the A site. If the translation reaction were to be experimentally stopped at this point, which of the following would you be able to isolate? a. an assembled ribosome with a polypeptide attached to the tRNA in the P site b. separated ribosomal subunits, a polypeptide, and free tRNA c. an assembled ribosome with a separated polypeptide d. separated ribosomal subunits with a polypeptide attached to the tRNA e. a cell with fewer ribosomes ____ 52. Why might a point mutation in DNA make a difference in the level of protein's activity? a. It might result in a chromosomal translocation. b. It might exchange one stop codon for another stop codon. c. It might exchange one serine codon for a different serine codon. d. It might substitute an amino acid in the active site. e. It might substitute the N terminus of the polypeptide for the C terminus. ____ 53. A frameshift mutation could result from a. a base insertion only. b. a base deletion only. c. a base substitution only. d. deletion of three consecutive bases. e. either an insertion or a deletion of a base. Genetics Practice Test - Chapters 14-17 Answer Section MULTIPLE CHOICE 1. ANS: C PTS: 1 2. ANS: A PTS: 1 3. ANS: B PTS: 1 4. ANS: B PTS: 1 5. ANS: D PTS: 1 6. ANS: E PTS: 1 7. ANS: D PTS: 1 8. ANS: D PTS: 1 9. ANS: B PTS: 1 10. ANS: A PTS: 1 11. ANS: C PTS: 1 12. ANS: E PTS: 1 13. ANS: C PTS: 1 14. ANS: A PTS: 1 MSC: Knowledge/Comprehension 15. ANS: A PTS: 1 MSC: Knowledge/Comprehension 16. ANS: E PTS: 1 MSC: Application/Analysis 17. ANS: C PTS: 1 MSC: Knowledge/Comprehension 18. ANS: D PTS: 1 MSC: Application/Analysis 19. ANS: C PTS: 1 MSC: Knowledge/Comprehension 20. ANS: A PTS: 1 MSC: Application/Analysis 21. ANS: D PTS: 1 MSC: Knowledge/Comprehension 22. ANS: D PTS: 1 MSC: Application/Analysis 23. ANS: B PTS: 1 MSC: Knowledge/Comprehension 24. ANS: B PTS: 1 MSC: Knowledge/Comprehension 25. ANS: C PTS: 1 MSC: Knowledge/Comprehension 26. ANS: A PTS: 1 MSC: Knowledge/Comprehension 27. ANS: C PTS: 1 MSC: Knowledge/Comprehension 28. ANS: D PTS: 1 MSC: Application/Analysis 29. ANS: E PTS: 1 MSC: Synthesis/Evaluation 30. ANS: D PTS: 1 MSC: Synthesis/Evaluation 31. ANS: A PTS: 1 MSC: Knowledge/Comprehension 32. ANS: A PTS: 1 MSC: Knowledge/Comprehension 33. ANS: A PTS: 1 MSC: Knowledge/Comprehension 34. ANS: C PTS: 1 MSC: Knowledge/Comprehension 35. ANS: D PTS: 1 MSC: Knowledge/Comprehension 36. ANS: D PTS: 1 MSC: Application/Analysis 37. ANS: B PTS: 1 MSC: Knowledge/Comprehension 38. ANS: A PTS: 1 MSC: Knowledge/Comprehension 39. ANS: C PTS: 1 MSC: Knowledge/Comprehension 40. ANS: E PTS: 1 MSC: Application/Analysis 41. ANS: E PTS: 1 MSC: Application/Analysis 42. ANS: A PTS: 1 MSC: Knowledge/Comprehension 43. ANS: D PTS: 1 MSC: Knowledge/Comprehension 44. ANS: A PTS: 1 MSC: Knowledge/Comprehension 45. ANS: C PTS: 1 MSC: Knowledge/Comprehension 46. ANS: B PTS: 1 MSC: Knowledge/Comprehension 47. ANS: D PTS: 1 MSC: Application/Analysis 48. ANS: C PTS: 1 MSC: Knowledge/Comprehension 49. ANS: B PTS: 1 MSC: Knowledge/Comprehension 50. ANS: A PTS: 1 MSC: Knowledge/Comprehension 51. ANS: A PTS: 1 MSC: Synthesis/Evaluation 52. ANS: D PTS: 1 MSC: Synthesis/Evaluation 53. ANS: E PTS: 1 MSC: Knowledge/Comprehension